NCERT Solutions Class 12 Mathematics Chapter 7

Q.1

$$\begin{gathered} \mathrm{Find}\mathrm{the}\mathrm{integral}\mathrm{of}\mathrm{the}\mathrm{function}\mathrm{given}\mathrm{below}. \\ \frac{\mathrm{cosx}-\mathrm{sinx}}{1+\sin 2\mathrm{x}} \end{gathered}$$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\\ \frac{\mathrm{cosx}-\mathrm{sinx}}{1+\sin 2\mathrm{x}}=\frac{\mathrm{cosx}-\mathrm{sinx}}{{\sin }^2\mathrm{x}+{\cos }^2\mathrm{x}+2\mathrm{sinxcosx}}\\ =\frac{\mathrm{cosx}-\mathrm{sinx}}{{(\mathrm{sinx}+\mathrm{cosx})}^2}\\ \therefore \int \frac{\mathrm{cosx}-\mathrm{sinx}}{1+\sin 2\mathrm{x}}\mathrm{dx}\\ =\int \frac{\mathrm{cosx}-\mathrm{sinx}}{{(\mathrm{sinx}+\mathrm{cosx})}^2}\mathrm{dx}\\ \mathrm{Let}\text{t}=\mathrm{sinx}+\mathrm{cosx}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=\mathrm{cosx}-\mathrm{sinx}\\ \Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{\mathrm{cosx}-\mathrm{sinx}}\\ \int \frac{\mathrm{cosx}-\mathrm{sinx}}{1+\sin 2\mathrm{x}}\mathrm{dx}=\int \frac{\mathrm{cosx}-\mathrm{sinx}}{{\left(\mathrm{t}\right)}^2}\frac{\mathrm{dt}}{\mathrm{cosx}-\mathrm{sinx}}\\ =\int {\mathrm{t}}^{-2}\mathrm{dt}\\ =-\frac{1}{\mathrm{t}}+\mathrm{C}\\ =-\frac{1}{\mathrm{sinx}+\mathrm{cosx}}+\mathrm{C}\end{array}$

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Q.2 Find the integral of the function tan³ 2x sec 2x.

Ans.

$\begin{array}{l}\int {\tan }^{3}2\mathrm{xsec}2\mathrm{x}\mathrm{dx}=\int {\tan }^{2}2\mathrm{xtan}2\mathrm{xsec}2\mathrm{x}\mathrm{dx}\\ =\int {\tan }^{2}2\mathrm{xtan}2\mathrm{xsec}2\mathrm{x}\mathrm{dx}\\ =\int \left({\sec }^{2}2\mathrm{x}-1\right)\tan 2\mathrm{xsec}2\mathrm{x}\mathrm{dx}\\ =\int {\sec }^{2}2\mathrm{x}.\tan 2\mathrm{xsec}2\mathrm{x}\mathrm{dx}-\int \tan 2\mathrm{xsec}2\mathrm{x}\mathrm{dx}\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}t}=\sec 2\mathrm{x}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=2\sec 2\mathrm{xtan}2\mathrm{x}\\ \Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{2\sec 2\mathrm{xtan}2\mathrm{x}}\\ \mathrm{So},\\ \int {\tan }^{3}2\mathrm{xsec}2\mathrm{x}\mathrm{dx}=\int {\mathrm{t}}^2.\tan 2\mathrm{xsec}2\mathrm{x}.\frac{\mathrm{dt}}{2\sec 2\mathrm{xtan}2\mathrm{x}}-\frac{1}{2}\sec 2\mathrm{x}+\mathrm{C}\\ =\frac{1}{2}\int {\mathrm{t}}^2\mathrm{dt}-\frac{1}{2}\sec 2\mathrm{x}+\mathrm{C}\\ =\frac{1}{2}\times \frac{{\mathrm{t}}^3}{3}-\frac{1}{2}\sec 2\mathrm{x}+\mathrm{C}\\ =\frac{1}{6}{\sec }^{3}2\mathrm{x}-\frac{1}{2}\sec 2\mathrm{x}+\mathrm{C}\end{array}$

Q.3 Find the integral of the function tan⁴x.

Ans.

$\begin{array}{l}\int {\tan }^4\mathrm{x}\mathrm{dx}=\int {\tan }^2{\mathrm{xtan}}^2\mathrm{x}\mathrm{dx}\\ =\int ({\sec }^2\mathrm{x}-1){\tan }^2\mathrm{x}\mathrm{dx}\\ =\int {\sec }^2\mathrm{x}{\tan }^2\mathrm{x}\mathrm{dx}-\int {\tan }^2\mathrm{x}\mathrm{dx}\\ =\int {\sec }^2\mathrm{x}{\tan }^2\mathrm{x}\mathrm{dx}-\int ({\sec }^2\mathrm{x}-1)\mathrm{dx}\\ =\int {\sec }^2\mathrm{x}{\tan }^2\mathrm{x}\mathrm{dx}-\int {\sec }^2\mathrm{x}\mathrm{dx}+\int 1\mathrm{dx}\\ \mathrm{Let}\text{t}=\mathrm{tanx}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}={\sec }^2\mathrm{x}\Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{{\sec }^2\mathrm{x}}\\ \int {\tan }^4\mathrm{x}\mathrm{dx}=\int {\sec }^2\mathrm{x}.{\mathrm{t}}^2\frac{\mathrm{dt}}{{\sec }^2\mathrm{x}}-\mathrm{tanx}+\mathrm{x}+\mathrm{C}\\ =\int {\mathrm{t}}^2\mathrm{dt}-\mathrm{tanx}+\mathrm{x}+\mathrm{C}\\ =\frac{{\mathrm{t}}^3}{3}-\mathrm{tanx}+\mathrm{x}+\mathrm{C}\\ =\frac{{\tan }^3\mathrm{x}}{3}-\mathrm{tanx}+\mathrm{x}+\mathrm{C}\end{array}$

Q.4

$$\begin{gathered} \mathrm{Find}\mathrm{the}\mathrm{integral}\mathrm{of}\mathrm{the}\mathrm{function}\mathrm{given}\mathrm{below}. \\ \frac{{\sin }^3\mathrm{x}+{\cos }^3\mathrm{x}}{{\sin }^2{\mathrm{xcos}}^2\mathrm{x}} \end{gathered}$$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\\ \frac{{\text{sin}}^{\text{3}}{\text{x+cos}}^{\text{3}}\text{x}}{{\text{sin}}^{\text{2}}{\text{x cos}}^{\text{2}}\text{x}}=\frac{{\text{sin}}^{\text{3}}\text{x}}{{\text{sin}}^{\text{2}}{\text{x cos}}^{\text{2}}\text{x}}+\frac{{\text{cos}}^{\text{3}}\text{x}}{{\text{sin}}^{\text{2}}{\text{x cos}}^{\text{2}}\text{x}}\\ =\frac{\text{sinx}}{{\text{cos}}^{\text{2}}\text{x}}+\frac{\text{cosx}}{{\text{sin}}^{\text{2}}\text{x}}\\ =\mathrm{tanxsecx}+\mathrm{cotxcosecx}\\ \therefore \int \frac{{\text{sin}}^{\text{3}}{\text{x+cos}}^{\text{3}}\text{x}}{{\text{sin}}^{\text{2}}{\text{x cos}}^{\text{2}}\text{x}}\mathrm{dx}=\int \mathrm{tanxsecx}\mathrm{dx}+\int \mathrm{cotxcosecx}\mathrm{dx}\\ =\mathrm{secx}-\mathrm{cosecx}+\mathrm{C}\end{array}$

Q.5

$$\begin{gathered} \mathrm{Find}\mathrm{the}\mathrm{integral}\mathrm{of}\mathrm{the}\mathrm{function}\mathrm{given}\mathrm{below}. \\ \frac{\cos 2\mathrm{x}+2{\sin }^2\mathrm{x}}{{\cos }^2\mathrm{x}} \end{gathered}$$ $\begin{array}{l}\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\\ \frac{\cos 2\mathrm{x}+2{\sin }^2\mathrm{x}}{{\cos }^2\mathrm{x}}=\frac{\cos 2\mathrm{x}+(1-\cos 2\mathrm{x})}{{\cos }^2\mathrm{x}}\\ =\frac{1}{{\cos }^2\mathrm{x}}\\ ={\sec }^2\mathrm{x}\\ \therefore \int \frac{\cos 2\mathrm{x}+2{\sin }^2\mathrm{x}}{{\cos }^2\mathrm{x}}\mathrm{dx}=\int {\sec }^2\mathrm{x}\mathrm{dx}\\ =\mathrm{tanx}+\mathrm{C}\end{array}$

Q.6

$\text{Find the integral of the function\hspace{0.17em}\hspace{0.17em}}\frac{1}{{\mathrm{sinxcos}}^3\mathrm{x}}.$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\frac{1}{\mathrm{sinx}{\cos }^3\mathrm{x}}=\frac{{\sin }^2\mathrm{x}+{\cos }^2\mathrm{x}}{{\mathrm{sinxcos}}^3\mathrm{x}}\\ =\frac{{\sin }^2\mathrm{x}}{{\mathrm{sinxcos}}^3\mathrm{x}}+\frac{{\cos }^2\mathrm{x}}{{\mathrm{sinxcos}}^3\mathrm{x}}\\ =\mathrm{tanx}{\sec }^2\mathrm{x}+\frac{1}{\mathrm{sinx}\mathrm{cosx}}\\ =\mathrm{tanx}{\sec }^2\mathrm{x}+\frac{{\sec }^2\mathrm{x}}{\mathrm{tanx}}\\ \therefore \int \frac{1}{{\mathrm{sinxcos}}^3\mathrm{x}}\mathrm{dx}=\int \mathrm{tanx}{\sec }^2\mathrm{x}\mathrm{dx}+\int \frac{{\sec }^2\mathrm{x}}{\mathrm{tanx}}\mathrm{dx}\\ \mathrm{Let}\text{t}=\text{tanx}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}={\sec }^2\mathrm{x}\Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{{\sec }^2\mathrm{x}}\\ \mathrm{So},\int \frac{1}{\mathrm{sinx}{\cos }^3\mathrm{x}}\mathrm{dx}=\int \mathrm{t}{\sec }^2\mathrm{x}.\frac{\mathrm{dt}}{{\sec }^2\mathrm{x}}+\int \frac{{\sec }^2\mathrm{x}}{\mathrm{t}}.\frac{\mathrm{dt}}{{\sec }^2\mathrm{x}}\\ =\int \mathrm{t}\mathrm{dt}+\int \frac{1}{\mathrm{t}}\mathrm{dt}\\ =\frac{{\mathrm{t}}^2}{2}+\log \left|\mathrm{t}\right|+\mathrm{C}\\ \therefore \int \frac{1}{{\mathrm{sinxcos}}^3\mathrm{x}}\mathrm{dx}=\frac{{\tan }^2\mathrm{x}}{2}+\log \left|\mathrm{tanx}\right|+\mathrm{C}\end{array}$

Q.7

$$\begin{gathered} \mathrm{Find}\mathrm{the}\mathrm{integral}\mathrm{of}\mathrm{the}\mathrm{function}\mathrm{given}\mathrm{below}. \\ \frac{\cos 2\mathrm{x}}{{(\mathrm{cosx}+\mathrm{sinx})}^2} \end{gathered}$$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\\ \frac{\cos 2\mathrm{x}}{{(\mathrm{cosx}+\mathrm{sinx})}^2}=\frac{\cos 2\mathrm{x}}{({\cos }^2\mathrm{x}+{\sin }^2\mathrm{x}+2\mathrm{sinxcosx})}\\ =\frac{\cos 2\mathrm{x}}{(1+\sin 2\mathrm{x})}\\ \therefore \int \frac{\cos 2\mathrm{x}}{{(\mathrm{cosx}+\mathrm{sinx})}^2}\mathrm{dx}=\int \frac{\cos 2\mathrm{x}}{(1+\sin 2\mathrm{x})}\mathrm{dx}\\ \mathrm{Let}\mathrm{t}=1+\sin 2\mathrm{x}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=2\cos 2\mathrm{x}\Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{2\cos 2\mathrm{x}}\\ \Rightarrow \int \frac{\cos 2\mathrm{x}}{{(\mathrm{cosx}+\mathrm{sinx})}^2}\mathrm{dx}=\int \frac{\cos 2\mathrm{x}}{\mathrm{t}}.\frac{\mathrm{dt}}{2\cos 2\mathrm{x}}\\ =\frac{1}{2}\int \frac{1}{\mathrm{t}}\mathrm{dt}\\ =\frac{1}{2}\log \left|\mathrm{t}\right|+\mathrm{C}\\ =\frac{1}{2}\log \left|1+\sin 2\mathrm{x}\right|+\mathrm{C}\\ =\frac{1}{2}\log \left|{(\mathrm{cosx}+\mathrm{sinx})}^2\right|+\mathrm{C}\\ =\frac{1}{2}\times 2\log (\mathrm{cosx}+\mathrm{sinx})+\mathrm{C}\\ \text{Hence},\int \frac{\cos 2\mathrm{x}}{{\left(\mathrm{cosx}+\mathrm{sinx}\right)}^2}\mathrm{dx}=\log (\mathrm{cosx}+\mathrm{sinx})+\mathrm{C}\end{array}$

Q.8

Find the integral of the function given below.
sin^-1(cosx)

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\\ \int {\sin }^{-1}\left(\mathrm{cosx}\right)\mathrm{dx}\\ \mathrm{Let}\text{t}=\text{cosx}\Rightarrow \mathrm{sinx}=\sqrt{1-{\mathrm{t}}^2}\\ \mathrm{and}\frac{\mathrm{dt}}{\mathrm{dx}}=-\mathrm{sinx}\\ \Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{-\mathrm{sinx}}=\frac{-\mathrm{dt}}{\sqrt{1-{\mathrm{t}}^2}}\\ \therefore \int {\sin }^{-1}\left(\mathrm{cosx}\right)\mathrm{dx}=\int {\sin }^{-1}\mathrm{t}\times \frac{-\mathrm{dt}}{\sqrt{1-{\mathrm{t}}^2}}\\ =-\int \frac{{\sin }^{-1}\mathrm{t}}{\sqrt{1-{\mathrm{t}}^2}}\mathrm{dt}\\ =-\int \frac{\mathrm{u}}{\sqrt{1-{\mathrm{t}}^2}}\sqrt{1-{\mathrm{t}}^2}\mathrm{du}\left[\begin{array}{l}\mathrm{Let}\text{u}={\sin }^{-1}\mathrm{t}\\ \Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=\frac{1}{\sqrt{1-{\mathrm{t}}^2}}\\ \mathrm{dx}=\sqrt{1-{\mathrm{t}}^2}\mathrm{du}\end{array}\right]\\ =-\int \mathrm{u}\mathrm{du}\\ =-\frac{{\mathrm{u}}^2}{2}+\mathrm{C}\\ =-\frac{{\left({\sin }^{-1}\mathrm{t}\right)}^2}{2}+\mathrm{C}\\ \int {\sin }^{-1}\left(\mathrm{cosx}\right)\mathrm{dx}=-\frac{{\left\{{\sin }^{-1}\left(\text{cosx}\right)\right\}}^2}{2}+\mathrm{C}\\ =-\frac{{\left\{\frac{\mathrm{\pi }}{2}-{\cos }^{-1}\left(\text{cosx}\right)\right\}}^2}{2}+\mathrm{C}\left[{\sin }^{-1}\mathrm{x}+{\cos }^{-1}\mathrm{x}=\frac{\mathrm{\pi }}{2}\right]\\ =-\frac{{\left\{\frac{\mathrm{\pi }}{2}-\mathrm{x}\right\}}^2}{2}+\mathrm{C}\\ =-\frac{1}{2}\left(\frac{{\mathrm{\pi }}^2}{4}-\mathrm{\pi x}+{\mathrm{x}}^2\right)+\mathrm{C}\\ =-\frac{{\mathrm{\pi }}^2}{8}+\frac{\mathrm{\pi x}}{2}-\frac{{\mathrm{x}}^2}{2}+\mathrm{C}\\ =-\frac{{\mathrm{x}}^2}{2}+\frac{\mathrm{\pi x}}{2}+\left(\mathrm{C}-\frac{{\mathrm{\pi }}^2}{8}\right)\\ =\frac{\mathrm{\pi x}}{2}-\frac{{\mathrm{x}}^2}{2}+\mathrm{C}‘,\text{where C’=}\mathrm{C}+\frac{{\mathrm{\pi }}^2}{8}\end{array}$

Q.9

$$\begin{gathered} \mathrm{Find}\mathrm{the}\mathrm{integral}\mathrm{of}\mathrm{the}\mathrm{function}\mathrm{given}\mathrm{below}. \\ \frac{1}{\cos (\mathrm{x}-\mathrm{a})\cos (\mathrm{x}-\mathrm{b})} \end{gathered}$$

Ans.

$\begin{array}{l}\mathrm{We}\text{have}\frac{1}{\cos \left(\mathrm{x}-\mathrm{a}\right)\cos \left(\mathrm{x}-\mathrm{b}\right)}\\ =\frac{1}{\sin \left(\mathrm{a}-\mathrm{b}\right)}\left[\frac{\sin \left(\mathrm{a}-\mathrm{b}\right)}{\cos \left(\mathrm{x}-\mathrm{a}\right)\cos \left(\mathrm{x}-\mathrm{b}\right)}\right]\\ =\frac{1}{\sin \left(\mathrm{a}-\mathrm{b}\right)}\left[\frac{\sin \left(\mathrm{x}-\mathrm{b}-\mathrm{x}+\mathrm{a}\right)}{\cos \left(\mathrm{x}-\mathrm{a}\right)\cos \left(\mathrm{x}-\mathrm{b}\right)}\right]\\ =\frac{1}{\sin \left(\mathrm{a}-\mathrm{b}\right)}\left[\frac{\sin \left\{\left(\mathrm{x}-\mathrm{b}\right)-\left(\mathrm{x}-\mathrm{a}\right)\right\}}{\cos \left(\mathrm{x}-\mathrm{a}\right)\cos \left(\mathrm{x}-\mathrm{b}\right)}\right]\\ =\frac{1}{\sin \left(\mathrm{a}-\mathrm{b}\right)}\left\{\frac{\sin \left(\mathrm{x}-\mathrm{b}\right)\cos \left(\mathrm{x}-\mathrm{a}\right)-\cos \left(\mathrm{x}-\mathrm{b}\right)\sin \left(\mathrm{x}-\mathrm{a}\right)}{\cos \left(\mathrm{x}-\mathrm{a}\right)\cos \left(\mathrm{x}-\mathrm{b}\right)}\right\}\\ =\frac{1}{\sin \left(\mathrm{a}-\mathrm{b}\right)}\left\{\frac{\sin \left(\mathrm{x}-\mathrm{b}\right)\cos \left(\mathrm{x}-\mathrm{a}\right)}{\cos \left(\mathrm{x}-\mathrm{a}\right)\cos \left(\mathrm{x}-\mathrm{b}\right)}-\frac{\cos \left(\mathrm{x}-\mathrm{b}\right)\sin \left(\mathrm{x}-\mathrm{a}\right)}{\cos \left(\mathrm{x}-\mathrm{a}\right)\cos \left(\mathrm{x}-\mathrm{b}\right)}\right\}\\ =\frac{1}{\sin \left(\mathrm{a}-\mathrm{b}\right)}\left\{\frac{\sin \left(\mathrm{x}-\mathrm{b}\right)}{\cos \left(\mathrm{x}-\mathrm{b}\right)}-\frac{\sin \left(\mathrm{x}-\mathrm{a}\right)}{\cos \left(\mathrm{x}-\mathrm{a}\right)}\right\}\\ =\frac{1}{\sin \left(\mathrm{a}-\mathrm{b}\right)}\left\{\tan \left(\mathrm{x}-\mathrm{b}\right)-\tan \left(\mathrm{x}-\mathrm{a}\right)\right\}\\ \therefore \int \frac{1}{\cos \left(\mathrm{x}-\mathrm{a}\right)\cos \left(\mathrm{x}-\mathrm{b}\right)}\mathrm{dx}\\ =\frac{1}{\sin \left(\mathrm{a}-\mathrm{b}\right)}\left\{\int \tan \left(\mathrm{x}-\mathrm{b}\right)\mathrm{dx}-\int \tan \left(\mathrm{x}-\mathrm{a}\right)\mathrm{dx}\right\}\\ =\frac{1}{\sin \left(\mathrm{a}-\mathrm{b}\right)}\left\{-\log \left|\cos \left(\mathrm{x}-\mathrm{b}\right)\right|+\log \left|\cos \left(\mathrm{x}-\mathrm{a}\right)\right|\right\}+\mathrm{C}\\ =\frac{1}{\sin \left(\mathrm{a}-\mathrm{b}\right)}\log \left|\frac{\cos \left(\mathrm{x}-\mathrm{a}\right)}{\cos \left(\mathrm{x}-\mathrm{b}\right)}\right|+\mathrm{C}\end{array}$

Q.10

$\begin{array}{l}\mathrm{Choose}\mathrm{the}\mathrm{correct}\mathrm{answer}\\ \int \frac{{\sin }^2\text{x}-{\cos }^2\text{x}}{{\sin }^2{\mathrm{xcos}}^2\text{x}}\mathrm{dx}\text{\hspace{0.17em}is equal to}\\ \left(\text{A}\right)\mathrm{tanx}+\mathrm{cotx}+\text{C}\\ \left(\text{B}\right)\mathrm{tanx}+\mathrm{cosec}\text{x}+\text{C}\\ \left(\text{C}\right)-\mathrm{tanx}+\mathrm{cotx}+\text{C}\\ \left(\text{D}\right)\mathrm{tanx}+\mathrm{secx}+\text{C}\end{array}$

Ans.

$\begin{array}{l}\int \frac{{\sin }^2\mathrm{x}-{\cos }^2\mathrm{x}}{{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx}=\int \frac{{\sin }^2\mathrm{x}}{{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx}-\int \frac{{\cos }^2\mathrm{x}}{{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{\hspace{0.17em}}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\int {\sec }^2\mathrm{x}\mathrm{\hspace{0.17em}}\mathrm{dx}-\int {\mathrm{cosec}}^2\mathrm{x}\mathrm{\hspace{0.17em}}\mathrm{dx}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{tanx}+\mathrm{cotx}+\mathrm{C}\\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\mathrm{A}\mathrm{.}\end{array}$

Q.11

$\begin{array}{l}\mathrm{Choose}\mathrm{the}\mathrm{correct}\mathrm{answer}\\ \int \frac{{\mathrm{e}}^{\mathrm{x}}\left(1+\mathrm{x}\right)}{{\cos }^2\left({\mathrm{e}}^{\mathrm{x}}\mathrm{x}\right)}\mathrm{dx}\mathrm{equals}\\ \left(\mathrm{A}\right)\text{\hspace{0.33em}}-\cot \left({\mathrm{ex}}^{\mathrm{x}}\right)+\mathrm{C}\\ \left(\mathrm{B}\right)\tan \left({\mathrm{xe}}^{\mathrm{x}}\right)+\mathrm{C}\\ \left(\mathrm{C}\right)\tan \left({\mathrm{e}}^{\mathrm{x}}\right)+\mathrm{C}\\ \left(\mathrm{D}\right)\cot \left({\mathrm{e}}^{\mathrm{x}}\right)+\mathrm{C}\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have}\int \frac{{\mathrm{e}}^{\mathrm{x}}(1+\mathrm{x})}{{\cos }^2\left({\mathrm{e}}^{\mathrm{x}}\mathrm{x}\right)}\mathrm{dx}\\ \mathrm{Let}\text{t}={\mathrm{e}}^{\mathrm{x}}\mathrm{x}\\ \Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}={\mathrm{e}}^{\mathrm{x}}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{x}+\mathrm{x}\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{e}}^{\mathrm{x}}\\ ={\mathrm{e}}^{\mathrm{x}}+{\mathrm{xe}}^{\mathrm{x}}\\ ={\mathrm{e}}^{\mathrm{x}}(1+\mathrm{x})\\ \Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{{\mathrm{e}}^{\mathrm{x}}(1+\mathrm{x})}\\ \therefore \int \frac{{\mathrm{e}}^{\mathrm{x}}(1+\mathrm{x})}{{\cos }^2\left({\mathrm{e}}^{\mathrm{x}}\mathrm{x}\right)}\mathrm{dx}=\int \frac{{\mathrm{e}}^{\mathrm{x}}(1+\mathrm{x})}{{\cos }^2\mathrm{t}}\frac{\mathrm{dt}}{{\mathrm{e}}^{\mathrm{x}}(1+\mathrm{x})}\\ =\int \frac{1}{{\cos }^2\mathrm{t}}\mathrm{dt}\\ =\int {\sec }^2\mathrm{t}\mathrm{dx}\\ =\mathrm{tant}+\mathrm{C}\\ =\tan \left({\mathrm{xe}}^{\mathrm{x}}\right)+\mathrm{C}\\ \mathrm{Hence},\text{the correct option is B.}\end{array}$

Q.12

$$\begin{gathered} \mathrm{Integrate}\mathrm{the}\mathrm{function}\mathrm{given}\mathrm{below}. \\ \frac{3{\mathrm{x}}^2}{{\mathrm{x}}^6+1} \end{gathered}$$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{t}={\mathrm{x}}^3\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=3{\mathrm{x}}^2\\ \Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{3{\mathrm{x}}^2}\\ \therefore \int \frac{3{\mathrm{x}}^2}{{\mathrm{x}}^6+1}\mathrm{dx}=\int \frac{3{\mathrm{x}}^2}{{\left({\mathrm{x}}^3\right)}^2+1}\mathrm{dx}\\ =\int \frac{3{\mathrm{x}}^2}{{\mathrm{t}}^2+1}.\frac{\mathrm{dt}}{3{\mathrm{x}}^2}\\ =\int \frac{1}{{\mathrm{t}}^2+1}\mathrm{dt}\\ ={\tan }^{-1}\mathrm{t}+\mathrm{C}\\ ={\tan }^{-1}\left({\mathrm{x}}^3\right)+\mathrm{C}\end{array}$

Q.13

$$\begin{gathered} \mathrm{Integrate}\mathrm{the}\mathrm{function}\mathrm{given}\mathrm{below}. \\ \frac{1}{\sqrt{1+4{\mathrm{x}}^2}} \end{gathered}$$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em}have,}\frac{1}{\sqrt{1+4{\mathrm{x}}^2}}=\frac{1}{\sqrt{1+{\left(2\mathrm{x}\right)}^2}}\\ \mathrm{Let}\mathrm{t}=2\mathrm{x}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=2\\ \Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{2}\\ \therefore \int \frac{1}{\sqrt{1+4{\mathrm{x}}^2}}\mathrm{dx}=\int \frac{1}{\sqrt{1+{\left(2\mathrm{x}\right)}^2}}\mathrm{dx}\\ =\int \frac{1}{\sqrt{1+{\mathrm{t}}^2}}.\frac{\mathrm{dt}}{2}\\ =\frac{1}{2}\int \frac{1}{\sqrt{1+{\mathrm{t}}^2}}\mathrm{dt}\\ =\frac{1}{2}\log |\mathrm{t}+\sqrt{{\mathrm{t}}^2+1}|+\mathrm{C}\\ =\frac{1}{2}\log |2\mathrm{x}+\sqrt{{\left(2\mathrm{x}\right)}^2+1}|+\mathrm{C}\\ =\frac{1}{2}\log |2\mathrm{x}+\sqrt{4{\mathrm{x}}^2+1}|+\mathrm{C}\end{array}$

Q.14

$$\begin{gathered} \mathrm{Integrate}\mathrm{the}\mathrm{function}\mathrm{given}\mathrm{below}. \\ \frac{1}{\sqrt{{(2-\mathrm{x})}^2+1}} \end{gathered}$$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em}have,}\frac{1}{\sqrt{{(2-\mathrm{x})}^2+1}}\\ \mathrm{Let}\mathrm{t}=2-\mathrm{x}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=-1\\ \Rightarrow \mathrm{dx}=-\mathrm{dt}\\ \therefore \int \frac{1}{\sqrt{{(2-\mathrm{x})}^2+1}}\mathrm{dx}=\int \frac{1}{\sqrt{{\mathrm{t}}^2+1}}(-\mathrm{dt})\\ =-\int \frac{1}{\sqrt{1+{\mathrm{t}}^2}}.\mathrm{dt}\\ =-\log |\mathrm{t}+\sqrt{{\mathrm{t}}^2+1}|+\mathrm{C}\left[\begin{array}{l}\because \int \frac{1}{\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}}\mathrm{dx}\\ =\log |\mathrm{x}+\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}|\end{array}\right]\\ =-\log |(2-\mathrm{x})+\sqrt{{(2-\mathrm{x})}^2+1}|+\mathrm{C}\\ =\log \left|\frac{1}{(2-\mathrm{x})+\sqrt{{(2-\mathrm{x})}^2+1}}\right|+\mathrm{C}[\because -\mathrm{logx}=\log \left(\frac{1}{\mathrm{x}}\right)]\\ =\log \left|\frac{1}{(2-\mathrm{x})+\sqrt{4-4\mathrm{x}+{\mathrm{x}}^2+1}}\right|+\mathrm{C}\\ =\log \left|\frac{1}{(2-\mathrm{x})+\sqrt{{\mathrm{x}}^2-4\mathrm{x}+5}}\right|+\mathrm{C}\end{array}$

Q.15

$$\begin{gathered} \mathrm{Integrate}\mathrm{the}\mathrm{function}\mathrm{given}\mathrm{below}. \\ \frac{1}{\sqrt{9-25{\mathrm{x}}^2}} \end{gathered}$$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.33em}have,}\frac{1}{\sqrt{9-25{\mathrm{x}}^2}}=\frac{1}{\sqrt{9-{\left(5\mathrm{x}\right)}^2}}\\ \mathrm{Let}\mathrm{t}=5\mathrm{x}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=5\\ \Rightarrow \mathrm{dx}=\frac{1}{5}\mathrm{dt}\\ \therefore \int \frac{1}{\sqrt{9-25{\mathrm{x}}^2}}\mathrm{dx}=\frac{1}{5}\int \frac{1}{\sqrt{9-{\left(5\mathrm{x}\right)}^2}}\mathrm{dt}\\ =\frac{1}{5}\int \frac{1}{\sqrt{3^2-{\mathrm{t}}^2}}.\mathrm{dt}\\ =\frac{1}{5}{\sin }^{-1}\left(\frac{\mathrm{t}}{3}\right)+\mathrm{C}\\ =\frac{1}{5}{\sin }^{-1}\left(\frac{5\mathrm{x}}{3}\right)+\mathrm{C}\end{array}$

Q.16

$\mathrm{Integrate}\mathrm{the}\mathrm{function}\frac{3\mathrm{x}}{1+2{\mathrm{x}}^4}$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em}have,}\frac{3\mathrm{x}}{1+2{\mathrm{x}}^4}=\frac{3\mathrm{x}}{1+{\left(\sqrt{2}{\mathrm{x}}^2\right)}^2}\\ \mathrm{Let}\mathrm{t}=\sqrt{2}{\mathrm{x}}^2\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=2\sqrt{2}\mathrm{x}\\ \Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{2\sqrt{2}\mathrm{x}}\\ \therefore \int \frac{3\mathrm{x}}{1+2{\mathrm{x}}^4}\mathrm{dx}=\int \frac{3\mathrm{x}}{1+{\left(\sqrt{2}{\mathrm{x}}^2\right)}^2}\mathrm{dx}\\ =\int \frac{3\mathrm{x}}{1+{\mathrm{t}}^2}.\frac{\mathrm{dt}}{2\sqrt{2}\mathrm{x}}\\ =\frac{3}{2\sqrt{2}}\int \frac{1}{1+{\mathrm{t}}^2}\mathrm{dt}\\ =\frac{3}{2\sqrt{2}}{\tan }^{-1}\mathrm{t}+\mathrm{C}=\frac{3}{2\sqrt{2}}{\tan }^{-1}\left(\sqrt{2}{\mathrm{x}}^2\right)+\mathrm{C}\end{array}$

Q.17

$\mathrm{Integrate}\mathrm{the}\mathrm{function}\frac{{\mathrm{x}}^2}{1-{\mathrm{x}}^6}$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em}have,}\frac{{\mathrm{x}}^2}{1-{\mathrm{x}}^6}=\frac{{\mathrm{x}}^2}{1-{\left({\mathrm{x}}^3\right)}^2}\\ \mathrm{Let}\mathrm{t}={\mathrm{x}}^3\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=3{\mathrm{x}}^2\\ \Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{3{\mathrm{x}}^2}\\ \therefore \int \frac{{\mathrm{x}}^2}{1-{\mathrm{x}}^6}\mathrm{dx}=\int \frac{{\mathrm{x}}^2}{1-{\left({\mathrm{x}}^3\right)}^2}\mathrm{dx}\\ =\int \frac{{\mathrm{x}}^2}{1-{\mathrm{t}}^2}.\frac{\mathrm{dt}}{3{\mathrm{x}}^2}\\ =\frac{1}{3}\int \frac{1}{1-{\mathrm{t}}^2}\mathrm{dt}\\ =\frac{1}{3}\left(\frac{1}{2}\log \left|\frac{1+\mathrm{t}}{1-\mathrm{t}}\right|\right)+\mathrm{C}\\ =\frac{1}{6}\log \left|\frac{1+{\mathrm{x}}^3}{1-{\mathrm{x}}^3}\right|+\mathrm{C}\end{array}$

Q.18

$\mathrm{Integrate}\mathrm{the}\mathrm{function}\frac{\mathrm{x}-1}{\sqrt{{\mathrm{x}}^2-1}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em}have,}\frac{\mathrm{x}-1}{\sqrt{{\mathrm{x}}^2-1}}=\frac{\mathrm{x}-1}{\sqrt{{\mathrm{x}}^2-1}}\\ \mathrm{Let}\mathrm{t}={\mathrm{x}}^2-1\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=2\mathrm{x}\\ \Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{2\mathrm{x}}\\ \therefore \int \frac{\mathrm{x}-1}{\sqrt{{\mathrm{x}}^2-1}}\mathrm{dx}=\int \frac{\mathrm{x}}{\sqrt{{\mathrm{x}}^2-1}}\mathrm{dx}-\int \frac{1}{\sqrt{{\mathrm{x}}^2-1}}\mathrm{dx}\\ =\int \frac{\mathrm{x}}{\sqrt{\mathrm{t}}}.\frac{\mathrm{dt}}{2\mathrm{x}}-\int \frac{1}{\sqrt{{\mathrm{x}}^2-1}}\mathrm{dx}\\ =\frac{1}{2}\int {\mathrm{t}}^{-\frac{1}{2}}.\mathrm{dt}-\log |\mathrm{x}+\sqrt{{\mathrm{x}}^2-1}|+\mathrm{C}\\ =\frac{1}{2}\left(2\sqrt{\mathrm{t}}\right)-\log |\mathrm{x}+\sqrt{{\mathrm{x}}^2-1}|+\mathrm{C}\\ =\sqrt{\mathrm{t}}-\log |\mathrm{x}+\sqrt{{\mathrm{x}}^2-1}|+\mathrm{C}\\ =\sqrt{{\mathrm{x}}^2-1}-\log |\mathrm{x}+\sqrt{{\mathrm{x}}^2-1}|+\mathrm{C}\end{array}$

Q.19

$\mathrm{Integrate}\mathrm{the}\mathrm{function}\frac{{\mathrm{x}}^2}{\sqrt{{\mathrm{x}}^6+{\mathrm{a}}^6}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em}have,}\frac{{\mathrm{x}}^2}{\sqrt{{\mathrm{x}}^6+{\mathrm{a}}^6}}=\frac{{\mathrm{x}}^2}{\sqrt{{\mathrm{a}}^6+{\left({\mathrm{x}}^3\right)}^2}}\\ \mathrm{Let}\mathrm{t}={\mathrm{x}}^3\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=3{\mathrm{x}}^2\\ \Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{3{\mathrm{x}}^2}\\ \therefore \int \frac{{\mathrm{x}}^2}{\sqrt{{\mathrm{x}}^6+{\mathrm{a}}^6}}\mathrm{dx}=\int \frac{{\mathrm{x}}^2}{\sqrt{{\left({\mathrm{a}}^3\right)}^2+{\left({\mathrm{x}}^3\right)}^2}}\mathrm{dx}\\ =\int \frac{{\mathrm{x}}^2}{\sqrt{{\left({\mathrm{a}}^3\right)}^2+{\mathrm{t}}^2}}.\frac{\mathrm{dt}}{3{\mathrm{x}}^2}\\ =\frac{1}{3}\int \frac{1}{\sqrt{{\left({\mathrm{a}}^3\right)}^2+{\mathrm{t}}^2}}\mathrm{dt}\\ =\frac{1}{3}\log |\mathrm{t}+\sqrt{{\mathrm{t}}^2+{\left({\mathrm{a}}^3\right)}^2}|+\mathrm{C}\\ =\frac{1}{3}\log |{\mathrm{x}}^3+\sqrt{{\left({\mathrm{x}}^3\right)}^2+{\left({\mathrm{a}}^3\right)}^2}|+\mathrm{C}\\ =\frac{1}{3}\log |{\mathrm{x}}^3+\sqrt{{\mathrm{x}}^6+{\mathrm{a}}^6}|+\mathrm{C}\end{array}$

Q.20

$\mathrm{Integrate}\mathrm{the}\mathrm{function}\frac{{\sec }^2\mathrm{x}}{\sqrt{{\tan }^2\mathrm{x}+4}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em}have,}\frac{{\sec }^2\mathrm{x}}{\sqrt{{\tan }^2\mathrm{x}+4}}=\frac{{\sec }^2\mathrm{x}}{\sqrt{{\tan }^2\mathrm{x}+2^2}}\\ \mathrm{Let}\mathrm{t}=\mathrm{tanx}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}={\sec }^2\mathrm{x}\\ \Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{{\sec }^2\mathrm{x}}\\ \therefore \int \frac{{\sec }^2\mathrm{x}}{\sqrt{{\tan }^2\mathrm{x}+4}}\mathrm{dx}=\int \frac{{\sec }^2\mathrm{x}}{\sqrt{{\tan }^2\mathrm{x}+2^2}}\mathrm{dx}\\ =\int \frac{{\sec }^2\mathrm{x}}{\sqrt{{\mathrm{t}}^2+2^2}}.\frac{\mathrm{dt}}{{\sec }^2\mathrm{x}}\\ =\int \frac{1}{\sqrt{{\mathrm{t}}^2+2^2}}\mathrm{dt}\\ =\log |\mathrm{t}+\sqrt{{\mathrm{t}}^2+{\left(2\right)}^2}|+\mathrm{C}\\ =\log |\mathrm{tanx}+\sqrt{{\tan }^2\mathrm{x}+4}|+\mathrm{C}\end{array}$

Q.21

$\text{Integrate the function\hspace{0.17em}}\frac{1}{\sqrt{{\mathrm{x}}^2+2\mathrm{x}+2}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\\ \frac{1}{\sqrt{{\mathrm{x}}^2+2\mathrm{x}+2}}=\frac{1}{\sqrt{{(\mathrm{x}+1)}^2+1}}\\ \mathrm{Let}\mathrm{t}=\mathrm{x}+1\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=1\\ \Rightarrow \mathrm{dx}=\mathrm{dt}\\ \therefore \int \frac{1}{\sqrt{{\mathrm{x}}^2+2\mathrm{x}+2}}\mathrm{dx}=\int \frac{1}{\sqrt{{(\mathrm{x}+1)}^2+1}}\mathrm{dx}\\ =\int \frac{1}{\sqrt{{\mathrm{t}}^2+1^2}}.\mathrm{dt}\\ =\log |\mathrm{t}+\sqrt{{\mathrm{t}}^2+{\left(1\right)}^2}|+\mathrm{C}\\ =\log |(\mathrm{x}+1)+\sqrt{{(\mathrm{x}+1)}^2+1}|+\mathrm{C}\\ =\log |(\mathrm{x}+1)+\sqrt{{\mathrm{x}}^2+2\mathrm{x}+2}|+\mathrm{C}\end{array}$

Q.22

$\text{Integrate the function\hspace{0.17em}}\frac{1}{9{\mathrm{x}}^2+6\mathrm{x}+5}$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em}have,}\frac{1}{9{\mathrm{x}}^2+6\mathrm{x}+5}=\frac{1}{{(3\mathrm{x}+1)}^2+2^2}\\ \mathrm{Let}\mathrm{t}=3\mathrm{x}+1\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=3\\ \Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{3}\\ \therefore \int \frac{1}{9{\mathrm{x}}^2+6\mathrm{x}+5}\mathrm{dx}=\int \frac{1}{{(3\mathrm{x}+1)}^2+2^2}\mathrm{dx}\\ =\int \frac{1}{{\mathrm{t}}^2+2^2}.\frac{\mathrm{dt}}{3}\\ =\frac{1}{3}\int \frac{1}{{\mathrm{t}}^2+2^2}\mathrm{dt}\\ =\frac{1}{3}\left(\frac{1}{2}{\tan }^{-1}\frac{\mathrm{t}}{2}\right)+\mathrm{C}\\ =\frac{1}{6}{\tan }^{-1}\frac{(3\mathrm{x}+1)}{2}+\mathrm{C}\end{array}$

Q.23

$\text{Integrate the function\hspace{0.17em}}\frac{1}{\sqrt{7-6\mathrm{x}-{\mathrm{x}}^2}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em}have,}\frac{1}{\sqrt{7-6\mathrm{x}-{\mathrm{x}}^2}}=\frac{1}{\sqrt{16-{(\mathrm{x}+3)}^2}}\\ \mathrm{Let}\mathrm{t}=\mathrm{x}+3\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=1\\ \Rightarrow \mathrm{dx}=\mathrm{dt}\\ \therefore \int \frac{1}{\sqrt{7-6\mathrm{x}-{\mathrm{x}}^2}}\mathrm{dx}=\int \frac{1}{\sqrt{16-{(\mathrm{x}+3)}^2}}\mathrm{dx}\\ =\int \frac{1}{\sqrt{4^2-{\mathrm{t}}^2}}\mathrm{dt}\\ ={\sin }^{-1}\left(\frac{\mathrm{t}}{4}\right)+\mathrm{C}\\ ={\sin }^{-1}\left(\frac{\mathrm{x}+3}{4}\right)+\mathrm{C}\end{array}$

Q.24

$\text{Integrate the function\hspace{0.17em}}\frac{1}{\sqrt{(\mathrm{x}-1)(\mathrm{x}-2)}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em}have,}\frac{1}{\sqrt{(\mathrm{x}-1)(\mathrm{x}-2)}}=\frac{1}{\sqrt{{\mathrm{x}}^2-3\mathrm{x}+2}}\\ =\frac{1}{\sqrt{{(\mathrm{x}-\frac{3}{2})}^2-\frac{1}{4}}}\\ \mathrm{Let}\mathrm{t}=\mathrm{x}-\frac{3}{2}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=1\\ \Rightarrow \mathrm{dx}=\mathrm{dt}\\ \therefore \int \frac{1}{\sqrt{(\mathrm{x}-1)(\mathrm{x}-2)}}\mathrm{dx}=\int \frac{1}{\sqrt{{(\mathrm{x}-\frac{3}{2})}^2-\frac{1}{4}}}\mathrm{dx}\\ =\int \frac{1}{\sqrt{{\mathrm{t}}^2-{\left(\frac{1}{2}\right)}^2}}\mathrm{dt}\\ =\log |\mathrm{t}+\sqrt{{\mathrm{t}}^2-{\left(\frac{1}{2}\right)}^2}|+\mathrm{C}\\ =\log |(\mathrm{x}-\frac{3}{2})+\sqrt{{(\mathrm{x}-\frac{3}{2})}^2-{\left(\frac{1}{2}\right)}^2}|+\mathrm{C}\\ =\log |(\mathrm{x}-\frac{3}{2})+\sqrt{{\mathrm{x}}^2-3\mathrm{x}+2}|+\mathrm{C}\end{array}$

Q.25

$\text{Integrate the function\hspace{0.17em}}\frac{1}{\sqrt{8+3\mathrm{x}-{\mathrm{x}}^2}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em}have,}\frac{1}{\sqrt{8+3\mathrm{x}-{\mathrm{x}}^2}}=\frac{1}{\sqrt{8-({\mathrm{x}}^2-3\mathrm{x}+\frac{9}{4}-\frac{9}{4})}}\\ =\frac{1}{\sqrt{8+\frac{9}{4}-{(\mathrm{x}-\frac{3}{2})}^2}}\\ =\frac{1}{\sqrt{\frac{41}{4}-{(\mathrm{x}-\frac{3}{2})}^2}}\\ \mathrm{Let}\mathrm{t}=\mathrm{x}-\frac{3}{2}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=1\\ \Rightarrow \mathrm{dx}=\mathrm{dt}\\ \therefore \int \frac{1}{\sqrt{8+3\mathrm{x}-{\mathrm{x}}^2}}\mathrm{dx}=\int \frac{1}{\sqrt{\frac{41}{4}-{(\mathrm{x}-\frac{3}{2})}^2}}\mathrm{dx}\\ =\int \frac{1}{\sqrt{{\left(\frac{\sqrt{41}}{2}\right)}^2-{\mathrm{t}}^2}}\mathrm{dt}\\ ={\sin }^{-1}\left(\frac{\mathrm{t}}{\frac{\sqrt{41}}{2}}\right)+\mathrm{C}\\ ={\sin }^{-1}\left(\frac{2\mathrm{t}}{\sqrt{41}}\right)+\mathrm{C}\\ ={\sin }^{-1}\left\{\frac{2(\mathrm{x}-\frac{3}{2})}{\sqrt{41}}\right\}+\mathrm{C}\\ ={\sin }^{-1}\left\{\frac{(2\mathrm{x}-3)}{\sqrt{41}}\right\}+\mathrm{C}\end{array}$

Q.26

$\text{Integrate the function}\frac{1}{\sqrt{(\mathrm{x}–\mathrm{a})(\mathrm{x}–\mathrm{b})}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em}have,}\frac{1}{\sqrt{(\mathrm{x}-\mathrm{a})(\mathrm{x}-\mathrm{b})}}=\frac{1}{\sqrt{{\mathrm{x}}^2-(\mathrm{a}+\mathrm{b})\mathrm{x}+\mathrm{ab}}}\\ =\frac{1}{\sqrt{{(\mathrm{x}-\frac{\mathrm{a}+\mathrm{b}}{2})}^2-{\left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)}^2}}\\ \mathrm{Let}\mathrm{t}=\mathrm{x}-\frac{\mathrm{a}+\mathrm{b}}{2}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=1\\ \Rightarrow \mathrm{dx}=\mathrm{dt}\\ \therefore \int \frac{1}{\sqrt{(\mathrm{x}-\mathrm{a})(\mathrm{x}-\mathrm{b})}}\mathrm{dx}=\int \frac{1}{\sqrt{{(\mathrm{x}-\frac{\mathrm{a}+\mathrm{b}}{2})}^2-{\left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)}^2}}\mathrm{dx}\\ =\int \frac{1}{\sqrt{{\mathrm{t}}^2-{\left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)}^2}}\mathrm{dt}\\ =\int \frac{1}{\sqrt{{\mathrm{t}}^2-{\left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)}^2}}\mathrm{dt}\\ =\log |\mathrm{t}+\sqrt{{\mathrm{t}}^2-{\left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)}^2}|+\mathrm{C}\\ =\log |(\mathrm{x}-\frac{\mathrm{a}+\mathrm{b}}{2})+\sqrt{{(\mathrm{x}-\frac{\mathrm{a}+\mathrm{b}}{2})}^2-{\left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)}^2}|+\mathrm{C}\\ =\log |(\mathrm{x}-\frac{\mathrm{a}+\mathrm{b}}{2})+\sqrt{{\mathrm{x}}^2-(\mathrm{a}+\mathrm{b})\mathrm{x}+\mathrm{ab}}|+\mathrm{C}\end{array}$

Q.27

$\text{Integrate the function\hspace{0.17em}}\frac{4\mathrm{x}+1}{\sqrt{2{\mathrm{x}}^2+\mathrm{x}-3}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{4\mathrm{x}+1}{\sqrt{2{\mathrm{x}}^2+\mathrm{x}-3}}\\ \mathrm{Here},4\mathrm{x}+1=\mathrm{A}\frac{\mathrm{d}}{\mathrm{dx}}(2{\mathrm{x}}^2+\mathrm{x}-3)+\mathrm{B}\\ =\mathrm{A}(4\mathrm{x}+1)+\mathrm{B}\\ \mathrm{Equating}\text{the coefficients of x and the contant terms from}\\ \text{both sides, we get}\\ 4=4\mathrm{A}\text{and A}+\mathrm{B}=1\Rightarrow \mathrm{A}=1\text{and}\mathrm{B}=1-1=0\\ \mathrm{Now},\text{\hspace{0.17em}\hspace{0.17em}let t}=2{\mathrm{x}}^2+\mathrm{x}-3\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=4\mathrm{x}+1\\ \Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{4\mathrm{x}+1}\\ \therefore \int \frac{4\mathrm{x}+1}{\sqrt{2{\mathrm{x}}^2+\mathrm{x}-3}}\mathrm{dx}=\int \frac{4\mathrm{x}+1}{\sqrt{2{\mathrm{x}}^2+\mathrm{x}-3}}.\frac{\mathrm{dt}}{4\mathrm{x}+1}\\ =\int \frac{1}{\sqrt{\mathrm{t}}}.\mathrm{dt}\\ =2\sqrt{\mathrm{t}}+\mathrm{C}\\ =2\sqrt{2{\mathrm{x}}^2+\mathrm{x}-3}+\mathrm{C}\end{array}$

Q.28

$\text{Integrate the function\hspace{0.17em}}\frac{\mathrm{x}+2}{\sqrt{{\mathrm{x}}^2-1}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}+2}{\sqrt{{\mathrm{x}}^2-1}}\\ \mathrm{Here},\mathrm{x}+2=\mathrm{A}\frac{\mathrm{d}}{\mathrm{dx}}({\mathrm{x}}^2-1)+\mathrm{B}\\ =\mathrm{A}\left(2\mathrm{x}\right)+\mathrm{B}\\ \mathrm{Equating}\text{the coefficients of x and the contant terms from}\\ \text{both sides, we get}\\ 1=2\mathrm{A}\text{and}\mathrm{B}=2\Rightarrow \mathrm{A}=\frac{1}{2}\text{and}\mathrm{B}=2\\ \mathrm{Therefore},\mathrm{x}+2=\frac{1}{2}\left(2\mathrm{x}\right)+2\\ \therefore \int \frac{\mathrm{x}+2}{\sqrt{{\mathrm{x}}^2-1}}\mathrm{dx}=\int \frac{\frac{1}{2}\left(2\mathrm{x}\right)+2}{\sqrt{{\mathrm{x}}^2-1}}.\mathrm{dx}\\ =\int \frac{\frac{1}{2}\left(2\mathrm{x}\right)}{\sqrt{{\mathrm{x}}^2-1}}\mathrm{dx}+\int \frac{2}{\sqrt{{\mathrm{x}}^2-1}}\mathrm{dx}\\ \mathrm{Now},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}let t}={\mathrm{x}}^2-1\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=2\mathrm{x}\\ \Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{2\mathrm{x}}\\ \mathrm{then}\\ \int \frac{\mathrm{x}+2}{\sqrt{{\mathrm{x}}^2-1}}\mathrm{dx}=\frac{1}{2}\int \frac{\left(2\mathrm{x}\right)}{\sqrt{\mathrm{t}}}.\frac{\mathrm{dt}}{2\mathrm{x}}+\int \frac{2}{\sqrt{{\mathrm{x}}^2-1}}\mathrm{dx}\\ =\int \frac{1}{\sqrt{\mathrm{t}}}.\mathrm{dt}+2\log |\mathrm{x}+\sqrt{{\mathrm{x}}^2-1}|\\ =\sqrt{\mathrm{t}}+2\log |\mathrm{x}+\sqrt{{\mathrm{x}}^2-1}|+\mathrm{C}\\ =\sqrt{{\mathrm{x}}^2-1}+2\log |\mathrm{x}+\sqrt{{\mathrm{x}}^2-1}|+\mathrm{C}\end{array}$

Q.29

$\text{Integrate the function\hspace{0.17em}}\frac{5\mathrm{x}-2}{1+2\mathrm{x}+3{\mathrm{x}}^2}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{5\mathrm{x}-2}{1+2\mathrm{x}+3{\mathrm{x}}^2}\\ \mathrm{Here},5\mathrm{x}-2=\mathrm{A}\frac{\mathrm{d}}{\mathrm{dx}}(1+2\mathrm{x}+3{\mathrm{x}}^2)+\mathrm{B}\\ =\mathrm{A}(6\mathrm{x}+2)+\mathrm{B}\\ \mathrm{Equating}\text{the coefficients of x and the contant terms from}\\ \text{both sides, we get}\\ 5=6\mathrm{A}\text{and 2A}+\mathrm{B}=-2\\ \Rightarrow \mathrm{A}=\frac{5}{6}\text{and}\mathrm{B}=-2-2\left(\frac{5}{6}\right)=-2-\frac{5}{3}=-\frac{11}{3}\\ \therefore 5\mathrm{x}-2=\frac{5}{6}(6\mathrm{x}+2)-\frac{11}{3}\\ \therefore \int \frac{5\mathrm{x}-2}{1+2\mathrm{x}+3{\mathrm{x}}^2}\mathrm{dx}=\int \frac{\frac{5}{6}(6\mathrm{x}+2)-\frac{11}{3}}{\sqrt{2{\mathrm{x}}^2+\mathrm{x}-3}}\mathrm{dx}\\ =\frac{5}{6}\int \frac{(6\mathrm{x}+2)}{\sqrt{1+2\mathrm{x}+3{\mathrm{x}}^2}}\mathrm{dx}-\frac{11}{3}\int \frac{1}{\sqrt{1+2\mathrm{x}+3{\mathrm{x}}^2}}\mathrm{dx}\\ \mathrm{Now},\text{\hspace{0.17em}\hspace{0.17em}let t}=1+2\mathrm{x}+3{\mathrm{x}}^2\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=6\mathrm{x}+2\\ \Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{6\mathrm{x}+2},\text{\hspace{0.17em}}\mathrm{then}\\ \int \frac{5\mathrm{x}-2}{1+2\mathrm{x}+3{\mathrm{x}}^2}\mathrm{dx}=\frac{5}{6}\int \frac{(6\mathrm{x}+2)}{1+2\mathrm{x}+3{\mathrm{x}}^2}.\frac{\mathrm{dt}}{6\mathrm{x}+2}-\frac{11}{3}\int \frac{1}{1+2\mathrm{x}+3{\mathrm{x}}^2}.\mathrm{dx}\\ =\frac{5}{6}\int \frac{1}{\mathrm{t}}.\mathrm{dt}-\frac{11}{3}\int \frac{1}{3\{{(\mathrm{x}+\frac{1}{3})}^2+{\left(\frac{\sqrt{2}}{3}\right)}^2\}}.\mathrm{dx}\\ =\frac{5}{6}\log \left|\mathrm{t}\right|-\frac{11}{9}\int \frac{1}{{(\mathrm{x}+\frac{1}{3})}^2+{\left(\frac{\sqrt{2}}{3}\right)}^2}.\mathrm{dx}\\ =\frac{5}{6}\log |1+2\mathrm{x}+3{\mathrm{x}}^2|-\frac{11}{9}\times \frac{1}{\left(\frac{\sqrt{2}}{3}\right)}{\tan }^{-1}\left(\frac{\mathrm{x}+\frac{1}{3}}{\frac{\sqrt{2}}{3}}\right)+\mathrm{C}\\ =\frac{5}{6}\log |1+2\mathrm{x}+3{\mathrm{x}}^2|-\frac{11}{3\sqrt{2}}{\tan }^{-1}\left(\frac{3\mathrm{x}+1}{\sqrt{2}}\right)+\mathrm{C}\end{array}$

Q.30

$\text{Integrate the function\hspace{0.17em}}\frac{6\mathrm{x}+7}{\sqrt{(\mathrm{x}-5)(\mathrm{x}-4)}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{6\mathrm{x}+7}{\sqrt{\left(\mathrm{x}-5\right)\left(\mathrm{x}-4\right)}}=\frac{6\mathrm{x}+7}{\sqrt{{\mathrm{x}}^2-9\mathrm{x}+20}}\\ \mathrm{Here},6\mathrm{x}+7=\mathrm{A}\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^2-9\mathrm{x}+20\right)+\mathrm{B}\\ 6\mathrm{x}+7=\mathrm{A}\left(2\mathrm{x}-9\right)+\mathrm{B}\\ \mathrm{Equating}\text{the coefficients of x and the contant terms from}\\ \text{both sides, we get}\\ 6=2\mathrm{A}\text{and}-\text{9A}+\mathrm{B}=7\\ \Rightarrow \mathrm{A}=\frac{6}{2}=\text{3 and}\mathrm{B}=7+9\left(3\right)=34\\ \therefore 6\mathrm{x}+7=3\left(2\mathrm{x}-9\right)+34\\ \therefore \int \frac{6\mathrm{x}+7}{\sqrt{\left(\mathrm{x}-5\right)\left(\mathrm{x}-4\right)}}\mathrm{dx}=\int \frac{6\mathrm{x}+7}{\sqrt{{\mathrm{x}}^2-9\mathrm{x}+20}}.\mathrm{dx}\\ =\int \frac{3\left(2\mathrm{x}-9\right)+34}{\sqrt{{\mathrm{x}}^2-9\mathrm{x}+20}}.\mathrm{dx}\\ =3\int \frac{\left(2\mathrm{x}-9\right)}{\sqrt{{\mathrm{x}}^2-9\mathrm{x}+20}}\mathrm{dx}+\int \frac{34}{\sqrt{{\mathrm{x}}^2-9\mathrm{x}+20}}\mathrm{dx}\\ \mathrm{Now},\text{\hspace{0.17em}\hspace{0.17em}let t}={\mathrm{x}}^2-9\mathrm{x}+20\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=2\mathrm{x}-9\\ \Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{2\mathrm{x}-9},\text{\hspace{0.17em}}\mathrm{then}\\ \int \frac{6\mathrm{x}+7}{\sqrt{\left(\mathrm{x}-5\right)\left(\mathrm{x}-4\right)}}\mathrm{dx}=3\int \frac{\left(2\mathrm{x}-9\right)}{\sqrt{\mathrm{t}}}.\frac{\mathrm{dt}}{2\mathrm{x}-9}+34\int \frac{1}{\sqrt{{\mathrm{x}}^2-9\mathrm{x}+20}}.\mathrm{dx}\\ =3\int \frac{1}{\sqrt{\mathrm{t}}}.\mathrm{dt}+34\int \frac{1}{\sqrt{\left\{{\left(\mathrm{x}-\frac{9}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2\right\}}}.\mathrm{dx}\\ =3\times 2\sqrt{\mathrm{t}}+34\log \left|\left(\mathrm{x}-\frac{9}{2}\right)+\sqrt{\left\{{\left(\mathrm{x}-\frac{9}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2\right\}}\right|+\mathrm{C}\\ =6\sqrt{{\mathrm{x}}^2-9\mathrm{x}+20}+34\log \left|\left(\mathrm{x}-\frac{9}{2}\right)+\sqrt{{\mathrm{x}}^2-9\mathrm{x}+20}\right|+\mathrm{C}\end{array}$

Q.31

$\text{Integrate the function\hspace{0.17em}}\frac{\mathrm{x}+2}{\sqrt{4\mathrm{x}-{\mathrm{x}}^2}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}+2}{\sqrt{4\mathrm{x}-{\mathrm{x}}^2}}\\ \mathrm{Here},\mathrm{x}+2=\mathrm{A}\frac{\mathrm{d}}{\mathrm{dx}}(4\mathrm{x}-{\mathrm{x}}^2)+\mathrm{B}\\ \mathrm{x}+2=\mathrm{A}(4-2\mathrm{x})+\mathrm{B}\\ \mathrm{Equating}\text{the coefficients of x and the contant terms from}\\ \text{both sides, we get}\\ 1=-2\mathrm{A}\text{and 4A}+\mathrm{B}=2\\ \Rightarrow \mathrm{A}=-\frac{1}{2}\text{and}\mathrm{B}=2-4(-\frac{1}{2})=4\\ \therefore \mathrm{x}+2=-\frac{1}{2}(4-2\mathrm{x})+4\\ \therefore \int \frac{\mathrm{x}+2}{\sqrt{4\mathrm{x}-{\mathrm{x}}^2}}\mathrm{dx}=\int \frac{-\frac{1}{2}(4-2\mathrm{x})+4}{\sqrt{4\mathrm{x}-{\mathrm{x}}^2}}\mathrm{dx}\\ =-\frac{1}{2}\int \frac{(4-2\mathrm{x})}{\sqrt{4\mathrm{x}-{\mathrm{x}}^2}}\mathrm{dx}+\int \frac{4}{\sqrt{4\mathrm{x}-{\mathrm{x}}^2}}\mathrm{dx}\\ =-\frac{1}{2}\int \frac{(4-2\mathrm{x})}{\sqrt{4\mathrm{x}-{\mathrm{x}}^2}}\mathrm{dx}+\int \frac{4}{\sqrt{4-(4-4\mathrm{x}+{\mathrm{x}}^2)}}.\mathrm{dx}\\ =-\frac{1}{2}\int \frac{(4-2\mathrm{x})}{\sqrt{4\mathrm{x}-{\mathrm{x}}^2}}\mathrm{dx}+\int \frac{4}{\sqrt{2^2-{(2-\mathrm{x})}^2}}\mathrm{dx}\\ \mathrm{Now},\text{\hspace{0.17em}\hspace{0.17em}let t}=4\mathrm{x}-{\mathrm{x}}^2\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=4-2\mathrm{x}\\ \Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{4-2\mathrm{x}},\text{\hspace{0.17em}}\mathrm{then}\\ \int \frac{\mathrm{x}+2}{\sqrt{4\mathrm{x}-{\mathrm{x}}^2}}\mathrm{dx}=-\frac{1}{2}\int \frac{(4-2\mathrm{x})}{\sqrt{\mathrm{t}}}\frac{\mathrm{dt}}{4-2\mathrm{x}}+\int \frac{4}{\sqrt{2^2-{(2-\mathrm{x})}^2}}\mathrm{dx}\\ =-\frac{1}{2}\times 2\sqrt{\mathrm{t}}-4{\sin }^{-1}\left(\frac{2-\mathrm{x}}{2}\right)+\mathrm{C}\\ =-\sqrt{4\mathrm{x}-{\mathrm{x}}^2}-4{\sin }^{-1}\left(\frac{2-\mathrm{x}}{2}\right)+\mathrm{C}\end{array}$

Q.32

$\text{Integrate the function\hspace{0.17em}}\frac{\mathrm{x}+2}{\sqrt{{\mathrm{x}}^2+2\mathrm{x}+3}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em}have,}\frac{\mathrm{x}+2}{\sqrt{{\mathrm{x}}^2+2\mathrm{x}+3}}\\ \int \frac{\mathrm{x}+2}{\sqrt{{\mathrm{x}}^2+2\mathrm{x}+3}}\mathrm{dx}=\frac{1}{2}\int \frac{2(\mathrm{x}+2)}{\sqrt{{\mathrm{x}}^2+2\mathrm{x}+3}}\mathrm{dx}\\ =\frac{1}{2}\int \frac{2\mathrm{x}+4}{\sqrt{{\mathrm{x}}^2+2\mathrm{x}+3}}\mathrm{dx}\\ =\frac{1}{2}\int \frac{2\mathrm{x}+2}{\sqrt{{\mathrm{x}}^2+2\mathrm{x}+3}}\mathrm{dx}+\frac{1}{2}\int \frac{2}{\sqrt{{\mathrm{x}}^2+2\mathrm{x}+3}}\mathrm{dx}\\ \mathrm{Let}\text{t}={\mathrm{x}}^2+2\mathrm{x}+3\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=2\mathrm{x}+2\\ \Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{2\mathrm{x}+2},\mathrm{then}\\ \int \frac{\mathrm{x}+2}{\sqrt{{\mathrm{x}}^2+2\mathrm{x}+3}}\mathrm{dx}=\frac{1}{2}\int \frac{2\mathrm{x}+2}{\sqrt{\mathrm{t}}}\frac{\mathrm{dt}}{2\mathrm{x}+2}+\frac{1}{2}\int \frac{2}{\sqrt{{\mathrm{x}}^2+2\mathrm{x}+3}}\mathrm{dx}\\ =\frac{1}{2}\int \frac{1}{\sqrt{\mathrm{t}}}\mathrm{dt}+\int \frac{1}{\sqrt{{(\mathrm{x}+1)}^2+{\left(\sqrt{2}\right)}^2}}\mathrm{dx}\\ =\frac{1}{2}.2\sqrt{\mathrm{t}}+\log |(\mathrm{x}+1)+\sqrt{{(\mathrm{x}+1)}^2+{\left(\sqrt{2}\right)}^2}|+\mathrm{C}\\ =\sqrt{{\mathrm{x}}^2+2\mathrm{x}+3}+\log |(\mathrm{x}+1)+\sqrt{{\mathrm{x}}^2+2\mathrm{x}+3}|+\mathrm{C}\end{array}$

Q.33

$\text{Integrate the function\hspace{0.17em}}\frac{\mathrm{x}+3}{{\mathrm{x}}^2-2\mathrm{x}-5}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}+3}{{\mathrm{x}}^2-2\mathrm{x}-5}\\ \mathrm{Here},\mathrm{x}+3=\mathrm{A}\frac{\mathrm{d}}{\mathrm{dx}}({\mathrm{x}}^2-2\mathrm{x}-5)+\mathrm{B}\\ \mathrm{x}+3=\mathrm{A}(2\mathrm{x}-2)+\mathrm{B}\\ \mathrm{Equating}\text{the coefficients of x and the contant terms from}\\ \text{both sides, we get}\\ 1=2\mathrm{A}\text{and}-\text{2A}+\mathrm{B}=3\\ \Rightarrow \mathrm{A}=\frac{1}{2}\text{and}\mathrm{B}=3+2\left(\frac{1}{2}\right)=4\\ \therefore \mathrm{x}+3=\frac{1}{2}(2\mathrm{x}-2)+4\\ \Rightarrow \int \frac{\mathrm{x}+3}{{\mathrm{x}}^2-2\mathrm{x}-5}\mathrm{dx}=\int \frac{\frac{1}{2}(2\mathrm{x}-2)+4}{{\mathrm{x}}^2-2\mathrm{x}-5}\mathrm{dx}\\ =\frac{1}{2}\int \frac{(2\mathrm{x}-2)}{{\mathrm{x}}^2-2\mathrm{x}-5}\mathrm{dx}+4\int \frac{1}{{\mathrm{x}}^2-2\mathrm{x}-5}\mathrm{dx}\\ \mathrm{Let}\text{t}={\mathrm{x}}^2-2\mathrm{x}-5\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=2\mathrm{x}-2\\ \Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{2\mathrm{x}-2},\mathrm{then}\\ \Rightarrow \int \frac{\mathrm{x}+3}{{\mathrm{x}}^2-2\mathrm{x}-5}\mathrm{dx}=\frac{1}{2}\int \frac{(2\mathrm{x}-2)}{\mathrm{t}}\frac{\mathrm{dt}}{2\mathrm{x}-2}+4\int \frac{1}{{(\mathrm{x}-1)}^2-6}\mathrm{dx}\\ =\frac{1}{2}\int \frac{1}{\mathrm{t}}\mathrm{dt}+4\int \frac{1}{{(\mathrm{x}-1)}^2-{\left(\sqrt{6}\right)}^2}\mathrm{dx}\\ =\frac{1}{2}\log \left|\mathrm{t}\right|+4\times \frac{1}{2\sqrt{6}}\log \left(\frac{\mathrm{x}-1-\sqrt{6}}{\mathrm{x}-1+\sqrt{6}}\right)+\mathrm{C}\\ =\frac{1}{2}\log |{\mathrm{x}}^2-2\mathrm{x}-5|+\frac{2}{\sqrt{6}}\log \left(\frac{\mathrm{x}-1-\sqrt{6}}{\mathrm{x}-1+\sqrt{6}}\right)+\mathrm{C}\end{array}$

Q.34

$\text{Integrate the function\hspace{0.17em}}\frac{5\mathrm{x}+3}{\sqrt{{\mathrm{x}}^2+4\mathrm{x}+10}}$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em}have,}\frac{5\mathrm{x}+3}{\sqrt{{\mathrm{x}}^2+4\mathrm{x}+10}}\\ \mathrm{Here},5\mathrm{x}+3=\mathrm{A}\frac{\mathrm{d}}{\mathrm{dx}}({\mathrm{x}}^2+4\mathrm{x}+10)+\mathrm{B}\\ 5\mathrm{x}+3=\mathrm{A}(2\mathrm{x}+4)+\mathrm{B}\\ \mathrm{Equating}\text{the coefficients of x and the contant terms from}\\ \text{both sides, we get}\\ 5=2\mathrm{A}\text{and 4A}+\mathrm{B}=3\\ \Rightarrow \mathrm{A}=\frac{5}{2}\text{and}\mathrm{B}=3-4\left(\frac{5}{2}\right)=-7\\ \therefore 5\mathrm{x}+3=\frac{5}{2}(2\mathrm{x}+4)-7\\ \therefore \int \frac{5\mathrm{x}+3}{\sqrt{{\mathrm{x}}^2+4\mathrm{x}+10}}\mathrm{dx}=\int \frac{\frac{5}{2}(2\mathrm{x}+4)-7}{\sqrt{{\mathrm{x}}^2+4\mathrm{x}+10}}\mathrm{dx}\\ =\frac{5}{2}\int \frac{(2\mathrm{x}+4)}{\sqrt{{\mathrm{x}}^2+4\mathrm{x}+10}}\mathrm{dx}-\int \frac{7}{\sqrt{{\mathrm{x}}^2+4\mathrm{x}+10}}\mathrm{dx}\\ =\frac{5}{2}\int \frac{(2\mathrm{x}+4)}{\sqrt{{\mathrm{x}}^2+4\mathrm{x}+10}}\mathrm{dx}-\int \frac{7}{\sqrt{{(\mathrm{x}+2)}^2+6}}\mathrm{dx}\\ =\frac{5}{2}\int \frac{(2\mathrm{x}+4)}{\sqrt{{\mathrm{x}}^2+4\mathrm{x}+10}}\mathrm{dx}-7\int \frac{1}{\sqrt{{(\mathrm{x}+2)}^2+{\left(\sqrt{6}\right)}^2}}\mathrm{dx}\\ \mathrm{Now},\text{\hspace{0.17em}\hspace{0.17em}let t}={\mathrm{x}}^2+4\mathrm{x}+10\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=2\mathrm{x}+4\\ \Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{2\mathrm{x}+4},\text{\hspace{0.17em}}\mathrm{then}\\ \int \frac{5\mathrm{x}+3}{\sqrt{{\mathrm{x}}^2+4\mathrm{x}+10}}\mathrm{dx}=\frac{5}{2}\int \frac{(2\mathrm{x}+4)}{\sqrt{\mathrm{t}}}\frac{\mathrm{dt}}{2\mathrm{x}+4}-7\int \frac{1}{\sqrt{{(\mathrm{x}+2)}^2+{\left(\sqrt{6}\right)}^2}}\mathrm{dx}\\ =\frac{5}{2}\times 2\sqrt{\mathrm{t}}-7\log |(\mathrm{x}+2)+\sqrt{{(\mathrm{x}+2)}^2+{\left(\sqrt{6}\right)}^2}|+\mathrm{C}\\ =5\sqrt{{\mathrm{x}}^2+4\mathrm{x}+10}-7\log |(\mathrm{x}+2)+\sqrt{{\mathrm{x}}^2+4\mathrm{x}+10}|+\mathrm{C}\end{array}$

Q.35

$\begin{array}{l}\mathrm{Choose}\mathrm{the}\mathrm{correct}\mathrm{answer}\\ \int \frac{\mathrm{dx}}{{\mathrm{x}}^2+2\mathrm{x}+2}\\ \left(\mathrm{A}\right)\text{\hspace{0.33em}}\mathrm{x}{\tan }^{-1}\left(\mathrm{x}+1\right)+\mathrm{C}\\ \left(\mathrm{B}\right){\tan }^{-1}\left(\mathrm{x}+1\right)+\mathrm{C}\\ \left(\mathrm{C}\right)\left(\mathrm{x}+1\right){\tan }^{-1}\mathrm{x}+\mathrm{C}\\ \left(\mathrm{D}\right){\tan }^{-1}\mathrm{x}+\mathrm{C}\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\\ \int \frac{\mathrm{dx}}{{\mathrm{x}}^2+2\mathrm{x}+2}=\int \frac{\mathrm{dx}}{{(\mathrm{x}+1)}^2+1^2}\\ ={\tan }^{-1}\left(\frac{\mathrm{x}+1}{1}\right)+\mathrm{C}\\ ={\tan }^{-1}(\mathrm{x}+1)+\mathrm{C}\\ \mathrm{Hence},\text{ the correct option is B.}\end{array}$

Q.36

$\begin{array}{l}\mathrm{Choose}\mathrm{the}\mathrm{correct}\mathrm{answer}\\ \int \frac{\mathrm{dx}}{\sqrt{9\mathrm{x}-4{\mathrm{x}}^2}}\mathrm{equals}\\ \left(\mathrm{A}\right)\text{\hspace{0.33em}}\frac{1}{9}{\sin }^{-1}\left(\frac{9\mathrm{x}-8}{8}\right)+\mathrm{C}\\ \left(\mathrm{B}\right)\frac{1}{2}{\sin }^{-1}\left(\frac{8\mathrm{x}-9}{9}\right)+\mathrm{C}\\ \left(\mathrm{C}\right)\frac{1}{3}{\sin }^{-1}\left(\frac{9\mathrm{x}-8}{8}\right)+\mathrm{C}\\ \left(\mathrm{D}\right)\frac{1}{2}{\sin }^{-1}\left(\frac{9\mathrm{x}-8}{9}\right)+\mathrm{C}\end{array}$

Ans.

$\begin{array}{l}\int \frac{\mathrm{dx}}{\sqrt{9\mathrm{x}-4{\mathrm{x}}^2}}=\int \frac{\mathrm{dx}}{\sqrt{-4({\mathrm{x}}^2-\frac{9}{4}\mathrm{x})}}\\ =\int \frac{\mathrm{dx}}{\sqrt{-4\{{\mathrm{x}}^2-\frac{9}{4}\mathrm{x}+{\left(\frac{9}{8}\right)}^2-{\left(\frac{9}{8}\right)}^2\}}}\\ =\frac{1}{2}\int \frac{\mathrm{dx}}{\sqrt{{\left(\frac{9}{8}\right)}^2-{(\mathrm{x}-\frac{9}{8})}^2}}\\ =\frac{1}{2}{\sin }^{-1}\left(\frac{\mathrm{x}-\frac{9}{8}}{\frac{9}{8}}\right)+\mathrm{C}\\ =\frac{1}{2}{\sin }^{-1}\left(\frac{8\mathrm{x}-9}{9}\right)+\mathrm{C}\\ \mathrm{Hence},\text{\hspace{0.17em}the correct option is B.}\end{array}$

Q.37

$\text{Integrate the rational function\hspace{0.17em}}\frac{\mathrm{x}}{(\mathrm{x}+1)(\mathrm{x}+2)}$

Ans.

$\begin{array}{l}\mathrm{Let}\frac{\mathrm{x}}{(\mathrm{x}+1)(\mathrm{x}+2)}=\frac{\mathrm{A}}{\mathrm{x}+1}+\frac{\mathrm{B}}{\mathrm{x}+2}\\ \mathrm{x}=\mathrm{A}(\mathrm{x}+2)+\mathrm{B}(\mathrm{x}+1)\\ \mathrm{x}=(\mathrm{A}+\mathrm{B})\mathrm{x}+2\mathrm{A}+\mathrm{B}\\ \Rightarrow \mathrm{A}+\mathrm{B}=1\text{and}2\mathrm{A}+\mathrm{B}=0\\ \Rightarrow \mathrm{A}=-1\text{and B}=2\\ \mathrm{So},\frac{\mathrm{x}}{(\mathrm{x}+1)(\mathrm{x}+2)}=\frac{-1}{\mathrm{x}+1}+\frac{2}{\mathrm{x}+2}\\ \therefore \int \frac{\mathrm{x}}{(\mathrm{x}+1)(\mathrm{x}+2)}\mathrm{dx}=\int \frac{-1}{\mathrm{x}+1}\mathrm{dx}+\int \frac{2}{\mathrm{x}+2}\mathrm{dx}\\ =\log {(\mathrm{x}+1)}^{-1}+\log {(\mathrm{x}+2)}^2+\mathrm{C}\\ =\log \frac{{(\mathrm{x}+2)}^2}{(\mathrm{x}+1)}+\mathrm{C}\end{array}$

Q.38

$\text{Integrate the rational function\hspace{0.17em}}\frac{1}{{\mathrm{x}}^2-9}$

Ans.

$\begin{array}{l}\mathrm{Let}\frac{1}{{\mathrm{x}}^2-9}=\frac{1}{(\mathrm{x}+3)(\mathrm{x}-3)}\\ =\frac{\mathrm{A}}{\mathrm{x}+3}+\frac{\mathrm{B}}{\mathrm{x}-3}\\ 1=\mathrm{A}(\mathrm{x}-3)+\mathrm{B}(\mathrm{x}+3)\\ 1=(\mathrm{A}+\mathrm{B})\mathrm{x}+3(-\mathrm{A}+\mathrm{B})\\ \Rightarrow \mathrm{A}+\mathrm{B}=0\text{and}3(-\mathrm{A}+\mathrm{B})=1\\ \Rightarrow \mathrm{A}=-\frac{1}{6}\text{and B}=\frac{1}{6}\\ \mathrm{So},\frac{1}{(\mathrm{x}+3)(\mathrm{x}-3)}=\frac{-1}{6(\mathrm{x}+3)}+\frac{1}{6(\mathrm{x}-3)}\\ \therefore \int \frac{1}{(\mathrm{x}+3)(\mathrm{x}-3)}\mathrm{dx}=\int \frac{-1}{6(\mathrm{x}+3)}\mathrm{dx}+\int \frac{1}{6(\mathrm{x}-3)}\mathrm{dx}\\ =-\frac{1}{6}\log |\mathrm{x}+3|+\frac{1}{6}\log |\mathrm{x}-3|+\mathrm{C}\\ =\frac{1}{6}\log \left|\frac{\mathrm{x}-3}{\mathrm{x}+3}\right|+\mathrm{C}\end{array}$

Q.39

$\text{Integrate the rational function\hspace{0.17em}}\frac{3\mathrm{x}-1}{(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)}$

Ans.

$\begin{array}{l}\mathrm{Let}\\ \frac{3\mathrm{x}-1}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)}\\ =\frac{\mathrm{A}}{\left(\mathrm{x}-1\right)}+\frac{\mathrm{B}}{\left(\mathrm{x}-2\right)}+\frac{\mathrm{C}}{\left(\mathrm{x}-3\right)}\\ 3\mathrm{x}-1=\mathrm{A}\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)+\mathrm{B}\left(\mathrm{x}-1\right)\left(\mathrm{x}-3\right)+\mathrm{C}\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)...\left(\mathrm{i}\right)\\ \mathrm{Substituting}\text{x}=1,2\mathrm{and}3\text{respectively in equation}\left(\mathrm{i}\right),\text{we get}\\ \mathrm{A}=1,\text{B}=-5\text{and C}=\text{4}\\ \therefore \frac{3\mathrm{x}-1}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)}=\frac{1}{\mathrm{x}-1}+\frac{-5}{\mathrm{x}-2}+\frac{4}{\left(\mathrm{x}-3\right)}\\ \therefore \int \frac{3\mathrm{x}-1}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)}\mathrm{dx}=\int \frac{1}{\mathrm{x}-1}\mathrm{dx}-\int \frac{5}{\mathrm{x}-2}\mathrm{dx}+\int \frac{4}{\left(\mathrm{x}-3\right)}\mathrm{dx}\\ =\log \left|\mathrm{x}-1\right|-5\log \left|\mathrm{x}-2\right|+4\log \left|\mathrm{x}-3\right|+\mathrm{C}\end{array}$

Q.40

$\text{Integrate the rational function\hspace{0.17em}}\frac{\mathrm{x}}{(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)}$

Ans.

$\begin{array}{l}\mathrm{Let}\frac{\mathrm{x}}{(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)}\\ =\frac{\mathrm{A}}{(\mathrm{x}-1)}+\frac{\mathrm{B}}{(\mathrm{x}-2)}+\frac{\mathrm{C}}{(\mathrm{x}-3)}\\ \mathrm{x}=\mathrm{A}(\mathrm{x}-2)(\mathrm{x}-3)+\mathrm{B}(\mathrm{x}-1)(\mathrm{x}-3)+\mathrm{C}(\mathrm{x}-1)(\mathrm{x}-2)...\left(\mathrm{i}\right)\\ \mathrm{Substituting}\text{x}=1,2\mathrm{and}3\text{respectively in equation}\left(\mathrm{i}\right),\text{we get}\\ \mathrm{A}=\frac{1}{2},\text{B}=-2\text{and C}=\frac{3}{2}\\ \Rightarrow \frac{\mathrm{x}}{(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)}=\frac{1}{2(\mathrm{x}-1)}+\frac{-2}{\mathrm{x}-2}+\frac{3}{2(\mathrm{x}-3)}\\ \therefore \int \frac{\mathrm{x}}{(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)}\mathrm{dx}=\frac{1}{2}\int \frac{1}{\mathrm{x}-1}\mathrm{dx}-2\int \frac{1}{\mathrm{x}-2}\mathrm{dx}+\frac{3}{2}\int \frac{1}{(\mathrm{x}-3)}\mathrm{dx}\\ =\frac{1}{2}\log |\mathrm{x}-1|-2\log |\mathrm{x}-2|+\frac{3}{2}\log |\mathrm{x}-3|+\mathrm{C}\end{array}$

Q.41

$\text{Integrate the rational function\hspace{0.17em}}\frac{2\mathrm{x}}{{\mathrm{x}}^2+3\mathrm{x}+2}$

Ans.

$\begin{array}{l}\mathrm{Let}\frac{2\mathrm{x}}{{\mathrm{x}}^2+3\mathrm{x}+2}=\frac{2\mathrm{x}}{(\mathrm{x}+1)(\mathrm{x}+2)}\\ =\frac{\mathrm{A}}{\mathrm{x}+1}+\frac{\mathrm{B}}{\mathrm{x}+2}\\ 2\mathrm{x}=\mathrm{A}(\mathrm{x}+2)+\mathrm{B}(\mathrm{x}+1)...\left(\mathrm{i}\right)\\ \mathrm{Substituting}\text{x}=-1\mathrm{and}-2\text{respectively in equation}\left(\mathrm{i}\right),\text{we get}\\ \mathrm{A}=-2,\text{B}=4\\ \therefore \frac{2\mathrm{x}}{(\mathrm{x}+1)(\mathrm{x}+2)}=\frac{-2}{\mathrm{x}+1}+\frac{4}{\mathrm{x}+2}\\ \therefore \int \frac{2\mathrm{x}}{(\mathrm{x}+1)(\mathrm{x}+2)}\mathrm{dx}=\int \frac{-2}{\mathrm{x}+1}\mathrm{dx}+\int \frac{4}{\mathrm{x}+2}\mathrm{dx}\\ =-2\log |\mathrm{x}+1|+4\log |\mathrm{x}+2|+\mathrm{C}\end{array}$

Q.42

$\text{Integrate the rational function\hspace{0.17em}}\frac{1-{\mathrm{x}}^2}{\mathrm{x}(1-2\mathrm{x})}$

Ans.

$\begin{array}{l}\frac{1-{\mathrm{x}}^2}{\mathrm{x}\left(1-2\mathrm{x}\right)}=\frac{1}{2}+\frac{1}{2}\left\{\frac{2-\mathrm{x}}{\mathrm{x}\left(1-2\mathrm{x}\right)}\right\}...\left(\mathrm{i}\right)\\ \mathrm{Let}\frac{2-\mathrm{x}}{\mathrm{x}\left(1-2\mathrm{x}\right)}=\frac{\mathrm{A}}{\mathrm{x}}+\frac{\mathrm{B}}{1-2\mathrm{x}}\\ \Rightarrow 2-\mathrm{x}=\mathrm{A}\left(1-2\mathrm{x}\right)+\mathrm{Bx}\\ \mathrm{Substituting}\text{x}=0\mathrm{and}\frac{1}{2}\text{respectively in equation}\left(\mathrm{i}\right),\text{we get}\\ \mathrm{A}=2,\text{B}=3\\ \therefore \frac{2-\mathrm{x}}{\mathrm{x}\left(1-2\mathrm{x}\right)}=\frac{2}{\mathrm{x}}+\frac{3}{1-2\mathrm{x}}\\ \mathrm{From}\text{ equation}\left(\mathrm{i}\right),\mathrm{we}\text{ have}\\ \int \frac{1-{\mathrm{x}}^2}{\mathrm{x}\left(1-2\mathrm{x}\right)}\mathrm{dx}=\int \frac{1}{2}\mathrm{dx}+\frac{1}{2}\int \frac{2}{\mathrm{x}}\mathrm{dx}+\frac{1}{2}\int \frac{3}{1-2\mathrm{x}}\mathrm{dx}\\ =\frac{1}{2}\mathrm{x}+\frac{1}{2}\left(2\log \left|\mathrm{x}\right|-\frac{3}{4}\log \left|1-2\mathrm{x}\right|\right)+\mathrm{C}\\ =\frac{1}{2}\mathrm{x}+\log \left|\mathrm{x}\right|-\frac{3}{4}\log \left|1-2\mathrm{x}\right|+\mathrm{C}\end{array}$

Q.43

$\text{Integrate the rational function\hspace{0.17em}}\frac{\mathrm{x}}{({\mathrm{x}}^2+1)(\mathrm{x}-1)}$

Ans.

$\begin{array}{l}\mathrm{Let}\frac{\mathrm{x}}{({\mathrm{x}}^2+1)(\mathrm{x}-1)}=\frac{\mathrm{Ax}+\mathrm{B}}{{\mathrm{x}}^2+1}+\frac{\mathrm{C}}{\mathrm{x}-1}...\left(\mathrm{i}\right)\\ \mathrm{x}=(\mathrm{Ax}+\mathrm{B})(\mathrm{x}-1)+\mathrm{C}({\mathrm{x}}^2+1)\\ ={\mathrm{Ax}}^2-\mathrm{Ax}+\mathrm{Bx}-\mathrm{B}+{\mathrm{Cx}}^2+\mathrm{C}\\ =(\mathrm{A}+\mathrm{C}){\mathrm{x}}^2+(-\mathrm{A}+\mathrm{B})\mathrm{x}+(-\mathrm{B}+\mathrm{C})\\ \mathrm{Equating}{\text{coefficients of x}}^{\text{2}}\text{, x and constant term from both sides,}\\ \text{we get}\\ \mathrm{A}+\mathrm{C}=0,-\text{A}+\text{B}=1\text{and\hspace{0.17em}\hspace{0.17em}}-\mathrm{B}+\mathrm{C}=0\\ \mathrm{On}\text{ solving these equations, we get}\\ \text{A}=-\frac{1}{2}\text{,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}B}=\frac{1}{2}\text{,\hspace{0.17em}\hspace{0.17em}C}=\frac{1}{2}\\ \mathrm{So},\text{from equation}\left(\mathrm{i}\right)\text{, we get}\\ \therefore \frac{\mathrm{x}}{({\mathrm{x}}^2+1)(\mathrm{x}-1)}=\frac{\frac{1}{2}(-\mathrm{x}+1)}{{\mathrm{x}}^2+1}+\frac{\frac{1}{2}}{\mathrm{x}-1}\\ \therefore \int \frac{\mathrm{x}}{({\mathrm{x}}^2+1)(\mathrm{x}-1)}\mathrm{dx}=\frac{1}{2}\int \frac{(-\mathrm{x})}{{\mathrm{x}}^2+1}\mathrm{dx}+\frac{1}{2}\int \frac{1}{{\mathrm{x}}^2+1}\mathrm{dx}+\frac{1}{2}\int \frac{1}{\mathrm{x}-1}\mathrm{dx}\\ =-\frac{1}{4}\log |{\mathrm{x}}^2+1|+\frac{1}{2}{\tan }^{-1}\mathrm{x}+\frac{1}{2}\log |\mathrm{x}-1|+\mathrm{C}\\ =\frac{1}{2}\log |\mathrm{x}-1|-\frac{1}{4}\log |{\mathrm{x}}^2+1|+\frac{1}{2}{\tan }^{-1}\mathrm{x}+\mathrm{C}\end{array}$

Q.44

$\text{Integrate the rational function\hspace{0.17em}}\frac{\mathrm{x}}{{(\mathrm{x}-1)}^2(\mathrm{x}+2)}$

Ans.

$\begin{array}{l}\mathrm{Let}\frac{\mathrm{x}}{{(\mathrm{x}-1)}^2(\mathrm{x}+2)}=\frac{\mathrm{A}}{(\mathrm{x}-1)}+\frac{\mathrm{B}}{{(\mathrm{x}-1)}^2}+\frac{\mathrm{C}}{(\mathrm{x}+2)}\\ \mathrm{x}=\mathrm{A}(\mathrm{x}-1)(\mathrm{x}+2)+\mathrm{B}(\mathrm{x}+2)+\mathrm{C}{(\mathrm{x}-1)}^2...\left(\mathrm{i}\right)\\ \mathrm{Substituting}\text{x}=1\text{in equation}\left(\mathrm{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} B}=\frac{1}{3}\\ \mathrm{Equating}{\text{coefficients of x}}^{\text{2}}\text{and constant term, we get}\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}A}+\text{C}=\text{0}\\ -\text{2A}+2\mathrm{B}+\mathrm{C}=0\\ \mathrm{On}\text{solving, we get}\\ \text{A}=\frac{2}{9}\text{and C}=\frac{-2}{9}\\ \Rightarrow \frac{\mathrm{x}}{{(\mathrm{x}-1)}^2(\mathrm{x}+2)}=\frac{2}{9(\mathrm{x}-1)}+\frac{1}{3{(\mathrm{x}-1)}^2}-\frac{2}{9(\mathrm{x}+2)}\\ \therefore \int \frac{\mathrm{x}}{{(\mathrm{x}-1)}^2(\mathrm{x}+2)}\mathrm{dx}=\frac{2}{9}\int \frac{1}{\mathrm{x}-1}\mathrm{dx}+\frac{1}{3}\int \frac{1}{{(\mathrm{x}-1)}^2}\mathrm{dx}-\frac{2}{9}\int \frac{1}{(\mathrm{x}+2)}\mathrm{dx}\\ =\frac{2}{9}\log |\mathrm{x}-1|+\frac{1}{3}\frac{-1}{(\mathrm{x}-1)}-\frac{2}{9}\log |\mathrm{x}+2|+\mathrm{C}\\ =\frac{2}{9}\log \left|\frac{\mathrm{x}-1}{\mathrm{x}+2}\right|-\frac{1}{3}\frac{1}{(\mathrm{x}-1)}+\mathrm{C}\end{array}$

Q.45

$\text{Integrate the rational function\hspace{0.17em}}\frac{3\mathrm{x}+5}{{\mathrm{x}}^3-{\mathrm{x}}^2-\mathrm{x}+1}$

Ans.

$\begin{array}{l}\mathrm{Given},\frac{3\mathrm{x}+5}{{\mathrm{x}}^3-{\mathrm{x}}^2-\mathrm{x}+1}=\frac{3\mathrm{x}+5}{{\mathrm{x}}^2(\mathrm{x}-1)-1(\mathrm{x}-1)}\\ =\frac{3\mathrm{x}+5}{(\mathrm{x}-1)({\mathrm{x}}^2-1)}\\ =\frac{3\mathrm{x}+5}{(\mathrm{x}-1)(\mathrm{x}-1)(\mathrm{x}+1)}\\ =\frac{3\mathrm{x}+5}{{(\mathrm{x}-1)}^2(\mathrm{x}+1)}\\ \mathrm{Let},\frac{3\mathrm{x}+5}{{(\mathrm{x}-1)}^2(\mathrm{x}+1)}=\frac{\mathrm{A}}{(\mathrm{x}-1)}+\frac{\mathrm{B}}{{(\mathrm{x}-1)}^2}+\frac{\mathrm{C}}{(\mathrm{x}+1)}\\ 3\mathrm{x}+5=\mathrm{A}(\mathrm{x}-1)(\mathrm{x}+1)+\mathrm{B}(\mathrm{x}+1)+\mathrm{C}{(\mathrm{x}-1)}^2\\ =\mathrm{A}({\mathrm{x}}^2-1)+\mathrm{B}(\mathrm{x}+1)+\mathrm{C}({\mathrm{x}}^2-2\mathrm{x}+1)...\left(\mathrm{i}\right)\\ \mathrm{Substituting}\text{x}=\text{1 in equation}\left(\mathrm{i}\right)\text{, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}B}=\frac{8}{2}=4\\ \mathrm{Equating}{\text{coefficients of x}}^{\text{2}}\text{, x and constant term,we get}\\ \text{A}+\text{C}=\text{0 and B}-2\mathrm{C}=3\\ \mathrm{On}\text{solving these equations, we get}\\ \text{A}=-\frac{1}{2}\text{and C}=\frac{1}{2}\\ \therefore \frac{3\mathrm{x}+5}{{(\mathrm{x}-1)}^2(\mathrm{x}+1)}=\frac{-1}{2(\mathrm{x}-1)}+\frac{4}{{(\mathrm{x}-1)}^2}+\frac{1}{2(\mathrm{x}+1)}\\ \Rightarrow \int \frac{3\mathrm{x}+5}{{(\mathrm{x}-1)}^2(\mathrm{x}+1)}\mathrm{dx}=-\frac{1}{2}\int \frac{1}{(\mathrm{x}-1)}\mathrm{dx}+4\int \frac{1}{{(\mathrm{x}-1)}^2}\mathrm{dx}+\frac{1}{2}\int \frac{1}{(\mathrm{x}+1)}\mathrm{dx}\\ =-\frac{1}{2}\log |\mathrm{x}-1|+4\left(\frac{-1}{\mathrm{x}-1}\right)+\frac{1}{2}\log |\mathrm{x}+1|+\mathrm{C}\\ =\frac{1}{2}\log \left|\frac{\mathrm{x}+1}{\mathrm{x}-1}\right|-\frac{4}{\mathrm{x}-1}+\mathrm{C}\end{array}$

Q.46

$\text{Integrate the rational function\hspace{0.17em}}\frac{2\mathrm{x}-3}{({\mathrm{x}}^2-1)(2\mathrm{x}+3)}$

Ans.

$\begin{array}{l}\frac{2\mathrm{x}-3}{({\mathrm{x}}^2-1)(2\mathrm{x}+3)}=\frac{2\mathrm{x}-3}{(\mathrm{x}-1)(\mathrm{x}+1)(2\mathrm{x}+3)}\\ =\frac{\mathrm{A}}{(\mathrm{x}+1)}+\frac{\mathrm{B}}{(\mathrm{x}-1)}+\frac{\mathrm{C}}{(2\mathrm{x}+3)}\\ 2\mathrm{x}-3=\mathrm{A}(\mathrm{x}-1)(2\mathrm{x}+3)+\mathrm{B}(\mathrm{x}-1)(2\mathrm{x}+3)+\mathrm{C}(\mathrm{x}-1)(\mathrm{x}+1)...\left(\mathrm{i}\right)\\ \mathrm{Substituting}\text{x}=-1,1\text{and}-\frac{3}{2}\text{respectively in equation}\left(\mathrm{i}\right),\text{we get}\\ \text{A}=\frac{5}{2}\text{, \hspace{0.17em}B}=-\frac{1}{10}\text{and C}=-\frac{24}{5}\\ \Rightarrow \frac{2\mathrm{x}-3}{(\mathrm{x}-1)(\mathrm{x}+1)(2\mathrm{x}+3)}=\frac{5}{2(\mathrm{x}+1)}-\frac{1}{10(\mathrm{x}-1)}-\frac{24}{5(2\mathrm{x}+3)}\\ \therefore \int \frac{2\mathrm{x}-3}{({\mathrm{x}}^2-1)(2\mathrm{x}+3)}\mathrm{dx}=\frac{5}{2}\int \frac{1}{(\mathrm{x}+1)}\mathrm{dx}-\frac{1}{10}\int \frac{1}{(\mathrm{x}-1)}\mathrm{dx}\\ -\frac{24}{5}\int \frac{1}{(2\mathrm{x}+3)}\mathrm{dx}\\ =\frac{5}{2}\log \left|(\mathrm{x}+1)\right|-\frac{1}{10}\log \left|(\mathrm{x}-1)\right|\\ -\frac{24}{5\times 2}\log \left|(2\mathrm{x}+3)\right|+\mathrm{D}\\ =\frac{5}{2}\log \left|(\mathrm{x}+1)\right|-\frac{1}{10}\log \left|(\mathrm{x}-1)\right|\\ -\frac{12}{5}\log \left|(2\mathrm{x}+3)\right|+\mathrm{D}\end{array}$

Q.47

$\text{Integrate the rational function\hspace{0.17em}}\frac{5\mathrm{x}}{(\mathrm{x}+1)({\mathrm{x}}^2-4)}$

Ans.

$\begin{array}{l}\frac{5\mathrm{x}}{\left(\mathrm{x}+1\right)\left({\mathrm{x}}^2-4\right)}=\frac{5\mathrm{x}}{\left(\mathrm{x}+1\right)\left(\mathrm{x}-2\right)\left(\mathrm{x}+2\right)}\\ =\frac{\mathrm{A}}{\left(\mathrm{x}+1\right)}+\frac{\mathrm{B}}{\left(\mathrm{x}+2\right)}+\frac{\mathrm{C}}{\left(\mathrm{x}-2\right)}\\ 5\mathrm{x}=\mathrm{A}\left(\mathrm{x}-2\right)\left(\mathrm{x}+2\right)+\mathrm{B}\left(\mathrm{x}+1\right)\left(\mathrm{x}-2\right)+\mathrm{C}\left(\mathrm{x}+1\right)\left(\mathrm{x}+2\right)...\left(\mathrm{i}\right)\\ \mathrm{Substituting}\text{x}=-1,-2\text{and\hspace{0.17em}\hspace{0.17em}}2\text{respectively in equation}\left(\mathrm{i}\right),\text{we get}\\ \text{A}=\frac{5}{3}\text{,\hspace{0.17em}\hspace{0.17em}B}=-\frac{5}{2}\text{and C}=\frac{5}{6}\\ \Rightarrow \frac{5\mathrm{x}}{\left(\mathrm{x}+1\right)\left(\mathrm{x}-2\right)\left(\mathrm{x}+2\right)}=\frac{5}{3\left(\mathrm{x}+1\right)}-\frac{5}{2\left(\mathrm{x}+2\right)}+\frac{5}{6\left(\mathrm{x}-2\right)}\\ \therefore \int \frac{2\mathrm{x}-3}{\left({\mathrm{x}}^2-1\right)\left(2\mathrm{x}+3\right)}\mathrm{dx}=\frac{5}{3}\int \frac{1}{\left(\mathrm{x}+1\right)}\mathrm{dx}-\frac{5}{2}\int \frac{1}{\left(\mathrm{x}+2\right)}\mathrm{dx}\\ +\frac{5}{6}\int \frac{1}{\left(\mathrm{x}-2\right)}\mathrm{dx}\\ =\frac{5}{3}\log \left|\left(\mathrm{x}+1\right)\right|-\frac{5}{2}\log \left|\left(\mathrm{x}+2\right)\right|\\ +\frac{5}{6}\log \left|\left(\mathrm{x}-2\right)\right|+\mathrm{D}\end{array}$

Q.48

$\text{Integrate the rational function\hspace{0.17em}}\frac{{\mathrm{x}}^3+\mathrm{x}+1}{{\mathrm{x}}^2-1}$

Ans.

$\begin{array}{l}\because \frac{{\mathrm{x}}^3+\mathrm{x}+1}{{\mathrm{x}}^2-1}=\mathrm{x}+\frac{2\mathrm{x}+1}{{\mathrm{x}}^2-1}\\ \mathrm{Let},\frac{2\mathrm{x}+1}{{\mathrm{x}}^2-1}=\frac{\mathrm{A}}{\mathrm{x}+1}+\frac{\mathrm{B}}{\mathrm{x}-1}\\ 2\mathrm{x}+1=\mathrm{A}(\mathrm{x}-1)+\mathrm{B}(\mathrm{x}+1)...\left(\mathrm{i}\right)\\ \mathrm{Substituting}\text{x}=-1\text{and 1 respectively in equation\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{i}\right),\text{we get}\\ \text{2}(-1)+1=\mathrm{A}(-1-1)+\mathrm{B}(-1+1)\\ -1=\mathrm{A}(-2)\Rightarrow \mathrm{A}=\frac{1}{2}\\ \mathrm{and}\text{B}=\frac{3}{2}\\ \Rightarrow \frac{{\mathrm{x}}^3+\mathrm{x}+1}{{\mathrm{x}}^2-1}=\mathrm{x}+\frac{1}{2(\mathrm{x}+1)}+\frac{3}{2(\mathrm{x}-1)}\\ \int \frac{{\mathrm{x}}^3+\mathrm{x}+1}{{\mathrm{x}}^2-1}\mathrm{dx}=\int \mathrm{x}\mathrm{dx}+\frac{1}{2}\int \frac{1}{(\mathrm{x}+1)}\mathrm{dx}+\frac{3}{2}\int \frac{1}{\mathrm{x}-1}\mathrm{dx}\\ =\frac{{\mathrm{x}}^2}{2}+\frac{1}{2}\log |\mathrm{x}+1|+\frac{3}{2}\log |\mathrm{x}-1|+\mathrm{C}\end{array}$

Q.49

$\text{Integrate the rational function\hspace{0.17em}}\frac{2}{(1-\mathrm{x})(1+{\mathrm{x}}^2)}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\frac{2}{(1-\mathrm{x})(1+{\mathrm{x}}^2)}\\ \mathrm{Let}\frac{2}{(1-\mathrm{x})(1+{\mathrm{x}}^2)}=\frac{\mathrm{A}}{1-\mathrm{x}}+\frac{\mathrm{Bx}+\mathrm{C}}{1+{\mathrm{x}}^2}\\ 2=\mathrm{A}(1+{\mathrm{x}}^2)+(\mathrm{Bx}+\mathrm{C})(1-\mathrm{x})\\ 2={\mathrm{x}}^2(\mathrm{A}-\mathrm{B})+\mathrm{x}(\mathrm{B}-\mathrm{C})+(\mathrm{A}+\mathrm{C})\\ \mathrm{Equating}{\text{the coefficients of x}}^{\text{2}}\text{, x and constant term, we get}\\ \mathrm{A}-\mathrm{B}=0\\ \mathrm{B}-\mathrm{C}=0\\ \mathrm{A}+\mathrm{C}=2\\ \text{On solving these equations, we get}\\ \mathrm{A}=1,\mathrm{B}=1\mathrm{and}\mathrm{C}=1\\ \therefore \frac{2}{(1-\mathrm{x})(1+{\mathrm{x}}^2)}=\frac{1}{1-\mathrm{x}}+\frac{\mathrm{x}+1}{1+{\mathrm{x}}^2}\\ \Rightarrow \int \frac{2}{(1-\mathrm{x})(1+{\mathrm{x}}^2)}\mathrm{dx}=\int \frac{1}{1-\mathrm{x}}\mathrm{dx}+\int \frac{\mathrm{x}+1}{1+{\mathrm{x}}^2}\mathrm{dx}\\ =\int \frac{1}{1-\mathrm{x}}\mathrm{dx}+\int \frac{\mathrm{x}}{1+{\mathrm{x}}^2}\mathrm{dx}+\int \frac{1}{1+{\mathrm{x}}^2}\mathrm{dx}\\ =-\log |1-\mathrm{x}|+\frac{1}{2}\log |1+{\mathrm{x}}^2|+{\tan }^{-1}\mathrm{x}+\mathrm{C}\end{array}$

Q.50

$\text{Integrate the rational function}\frac{3\mathrm{x}-1}{{(\mathrm{x}+2)}^2}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\frac{3\mathrm{x}-1}{{(\mathrm{x}+2)}^2}\\ \mathrm{Let}\frac{3\mathrm{x}-1}{{(\mathrm{x}+2)}^2}=\frac{\mathrm{A}}{(\mathrm{x}+2)}+\frac{\mathrm{B}}{{(\mathrm{x}+2)}^2}\\ 3\mathrm{x}-1=\mathrm{A}(\mathrm{x}+2)+\mathrm{B}\\ \mathrm{Equating}{\text{the coefficients of x}}^{\text{2}}\text{, x and constant term, we get}\\ \mathrm{A}=3\\ 2\mathrm{A}+\mathrm{B}=-1\\ \mathrm{Solving},\text{the above equations, we get}\\ \text{A}=3\mathrm{and}\mathrm{B}=-7\\ \therefore \frac{3\mathrm{x}-1}{{(\mathrm{x}+2)}^2}=\frac{3}{(\mathrm{x}+2)}-\frac{7}{{(\mathrm{x}+2)}^2}\\ \Rightarrow \int \frac{3\mathrm{x}-1}{{(\mathrm{x}+2)}^2}\mathrm{dx}=\int \frac{3}{(\mathrm{x}+2)}\mathrm{dx}-\int \frac{7}{{(\mathrm{x}+2)}^2}\mathrm{dx}\\ =3\log |\mathrm{x}+2|-7\left(\frac{-1}{\mathrm{x}+2}\right)+\mathrm{C}\\ =3\log |\mathrm{x}+2|+\frac{7}{\mathrm{x}+2}+\mathrm{C}\end{array}$

Q.51

$\text{Integrate the rational function\hspace{0.17em}}\frac{1}{{\mathrm{x}}^4-1}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\frac{1}{{\mathrm{x}}^4-1}=\frac{1}{(\mathrm{x}+1)(\mathrm{x}-1)({\mathrm{x}}^2+1)}\\ \mathrm{Let}\frac{1}{{\mathrm{x}}^4-1}=\frac{\mathrm{A}}{(\mathrm{x}+1)}+\frac{\mathrm{B}}{(\mathrm{x}-1)}+\frac{\mathrm{Cx}+\mathrm{D}}{({\mathrm{x}}^2+1)}\\ 1=\mathrm{A}(\mathrm{x}-1)({\mathrm{x}}^2+1)+\mathrm{B}(\mathrm{x}+1)({\mathrm{x}}^2+1)+({\mathrm{x}}^2-1)(\mathrm{Cx}+\mathrm{D})\\ 1=\mathrm{A}({\mathrm{x}}^3+\mathrm{x}-{\mathrm{x}}^2-1)+\mathrm{B}({\mathrm{x}}^3+\mathrm{x}+{\mathrm{x}}^2+1)+\mathrm{C}({\mathrm{x}}^3-\mathrm{x})+\mathrm{D}({\mathrm{x}}^2-1)\\ 1={\mathrm{x}}^3(\mathrm{A}+\mathrm{B}+\mathrm{C})+{\mathrm{x}}^2(-\mathrm{A}+\mathrm{B}+\mathrm{D})+\mathrm{x}(\mathrm{A}+\mathrm{B}-\mathrm{C})+(-\mathrm{A}+\mathrm{B}-\mathrm{D})\\ \mathrm{Equating}{\text{the coefficients of x}}^{\text{2}}\text{, x and constant term, we get}\\ \mathrm{A}+\mathrm{B}+\mathrm{C}=0\\ -\mathrm{A}+\mathrm{B}+\mathrm{D}=0\\ \mathrm{A}+\mathrm{B}-\mathrm{C}=0\\ -\mathrm{A}+\mathrm{B}-\mathrm{D}=1\\ \mathrm{Solving},\text{the above equations, we get}\\ \text{A}=-\frac{1}{4},\mathrm{B}=\frac{1}{4},\mathrm{C}=0\text{and D}=-\frac{1}{2}\\ \therefore \frac{1}{{\mathrm{x}}^4-1}=\frac{-1}{4(\mathrm{x}+1)}+\frac{1}{4(\mathrm{x}-1)}+\frac{-1}{2({\mathrm{x}}^2+1)}\\ \Rightarrow \int \frac{1}{{\mathrm{x}}^4-1}\mathrm{dx}=-\frac{1}{4}\int \frac{1}{\mathrm{x}+1}\mathrm{dx}+\frac{1}{4}\int \frac{1}{\mathrm{x}-1}\mathrm{dx}-\frac{1}{2}\int \frac{1}{{\mathrm{x}}^2+1}\mathrm{dx}\\ =-\frac{1}{4}\log \left|(\mathrm{x}+1)\right|+\frac{1}{4}\log \left|(\mathrm{x}-1)\right|-\frac{1}{2}{\tan }^{-1}\mathrm{x}+\mathrm{C}\\ =\frac{1}{4}\log \left|\frac{\mathrm{x}-1}{\mathrm{x}+1}\right|-\frac{1}{2}{\tan }^{-1}\mathrm{x}+\mathrm{C}\end{array}$

Q.52

$\text{Integrate the rational function\hspace{0.17em}}\frac{1}{\mathrm{x}({\mathrm{x}}^{\mathrm{n}}-1)}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\frac{1}{\mathrm{x}({\mathrm{x}}^{\mathrm{n}}-1)}=\frac{1}{\mathrm{x}({\mathrm{x}}^{\mathrm{n}}-1)}\times \frac{{\mathrm{x}}^{\mathrm{n}-1}}{{\mathrm{x}}^{\mathrm{n}-1}}\\ =\frac{{\mathrm{x}}^{\mathrm{n}-1}}{{\mathrm{x}}^{\mathrm{n}}({\mathrm{x}}^{\mathrm{n}}-1)}\\ \mathrm{Let}\mathrm{t}={\mathrm{x}}^{\mathrm{n}}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}={\mathrm{nx}}^{\mathrm{n}-1}\Rightarrow \frac{\mathrm{dt}}{{\mathrm{nx}}^{\mathrm{n}-1}}=\mathrm{dx}\\ \therefore \int \frac{1}{\mathrm{x}({\mathrm{x}}^{\mathrm{n}}-1)}\mathrm{dx}=\int \frac{{\mathrm{x}}^{\mathrm{n}-1}}{{\mathrm{x}}^{\mathrm{n}}({\mathrm{x}}^{\mathrm{n}}-1)}\mathrm{dx}\\ =\frac{1}{\mathrm{n}}\int \frac{1}{\mathrm{t}(\mathrm{t}-1)}\mathrm{dt}\\ \mathrm{Let}\frac{1}{\mathrm{t}(\mathrm{t}-1)}=\frac{\mathrm{A}}{\mathrm{t}}+\frac{\mathrm{B}}{\mathrm{t}-1}\\ 1=\mathrm{A}(\mathrm{t}-1)+\mathrm{Bt}\\ =\mathrm{t}(\mathrm{A}+\mathrm{B})-\mathrm{A}\\ \mathrm{Substituting}\text{ t}=\text{0 and 1 respectively, we get}\\ \mathrm{A}=1\mathrm{and}\mathrm{B}=-1\\ \therefore \frac{1}{\mathrm{t}(\mathrm{t}-1)}=\frac{1}{\mathrm{t}}+\frac{-1}{\mathrm{t}-1}\\ \Rightarrow \int \frac{1}{\mathrm{x}({\mathrm{x}}^{\mathrm{n}}-1)}\mathrm{dx}=\frac{1}{\mathrm{n}}\int (\frac{1}{\mathrm{t}}-\frac{1}{\mathrm{t}-1})\mathrm{dt}\\ =\frac{1}{\mathrm{n}}\int \frac{1}{\mathrm{t}}\mathrm{dt}-\frac{1}{\mathrm{n}}\int \frac{1}{\mathrm{t}-1}\mathrm{dt}\\ =\frac{1}{\mathrm{n}}\log \left|\mathrm{t}\right|-\frac{1}{\mathrm{n}}\log |\mathrm{t}-1|+\mathrm{C}\\ =\frac{1}{\mathrm{n}}\log \left|{\mathrm{x}}^{\mathrm{n}}\right|-\frac{1}{\mathrm{n}}\log |{\mathrm{x}}^{\mathrm{n}}-1|+\mathrm{C}\\ =\frac{1}{\mathrm{n}}\log \left|\frac{{\mathrm{x}}^{\mathrm{n}}}{{\mathrm{x}}^{\mathrm{n}}-1}\right|+\mathrm{C}\end{array}$

Q.53

$\text{Integrate the rational function\hspace{0.17em}}\frac{\mathrm{cosx}}{(1-\mathrm{sinx})(2-\mathrm{sinx})}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\frac{\mathrm{cosx}}{(1-\mathrm{sinx})(2-\mathrm{sinx})}\\ \mathrm{Let}\mathrm{t}=\mathrm{sinx}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=\mathrm{cosx}\Rightarrow \frac{\mathrm{dt}}{\mathrm{cosx}}=\mathrm{dx}\\ \therefore \int \frac{\mathrm{cosx}}{(1-\mathrm{sinx})(2-\mathrm{sinx})}\mathrm{dx}=\int \frac{\mathrm{cosx}}{(1-\mathrm{t})(2-\mathrm{t})}\frac{\mathrm{dt}}{\mathrm{cosx}}\\ =\int \frac{1}{(1-\mathrm{t})(2-\mathrm{t})}\mathrm{dt}\\ \mathrm{Let}\frac{1}{(1-\mathrm{t})(2-\mathrm{t})}=\frac{\mathrm{A}}{1-\mathrm{t}}+\frac{\mathrm{B}}{2-\mathrm{t}}\\ 1=\mathrm{A}(2-\mathrm{t})+\mathrm{B}(1-\mathrm{t})...\left(1\right)\\ \mathrm{Substituting}\text{t}=1\text{and 2 respectively in eqution}\left(1\right),\mathrm{we}\text{get}\\ \text{A}=\text{1 and B}=-1\\ \therefore \frac{1}{(1-\mathrm{t})(2-\mathrm{t})}=\frac{1}{1-\mathrm{t}}-\frac{1}{2-\mathrm{t}}\\ \mathrm{So},\\ \int \frac{\mathrm{cosx}}{(1-\mathrm{sinx})(2-\mathrm{sinx})}\mathrm{dx}=\int \frac{1}{(1-\mathrm{t})(2-\mathrm{t})}\mathrm{dt}\\ =\int \frac{1}{1-\mathrm{t}}\mathrm{dt}-\int \frac{1}{2-\mathrm{t}}\mathrm{dt}\\ =-\log |1-\mathrm{t}|+\log |2-\mathrm{t}|+\mathrm{C}\\ =\log \left|\frac{2-\mathrm{t}}{1-\mathrm{t}}\right|+\mathrm{C}\\ =\log \left|\frac{2-\mathrm{sinx}}{1-\mathrm{sinx}}\right|+\mathrm{C}\end{array}$

Q.54

$\text{Integrate the rational function\hspace{0.17em}}\frac{({\mathrm{x}}^2+1)({\mathrm{x}}^2+2)}{({\mathrm{x}}^2+3)({\mathrm{x}}^2+4)}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \frac{({\mathrm{x}}^2+1)({\mathrm{x}}^2+2)}{({\mathrm{x}}^2+3)({\mathrm{x}}^2+4)}=1+\frac{(4{\mathrm{x}}^2+10)}{({\mathrm{x}}^2+3)({\mathrm{x}}^2+4)}\\ \mathrm{Let},\\ \frac{(4{\mathrm{x}}^2+10)}{({\mathrm{x}}^2+3)({\mathrm{x}}^2+4)}=\frac{\mathrm{Ax}+\mathrm{B}}{({\mathrm{x}}^2+3)}+\frac{\mathrm{Cx}+\mathrm{D}}{({\mathrm{x}}^2+4)}\\ 4{\mathrm{x}}^2+10=(\mathrm{Ax}+\mathrm{B})({\mathrm{x}}^2+4)+(\mathrm{Cx}+\mathrm{D})({\mathrm{x}}^2+3)\\ ={\mathrm{x}}^3(\mathrm{A}+\mathrm{C})+{\mathrm{x}}^2(\mathrm{B}+\mathrm{D})+\mathrm{x}(4\mathrm{A}+3\mathrm{C})+(4\mathrm{B}+3\mathrm{D})\\ \mathrm{Comparing}{\text{the coefficients of x}}^{\text{3}}{\text{, x}}^{\text{2}}\text{, x and constant terms,}\\ \text{we get\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \mathrm{A}+\mathrm{C}=0\\ \mathrm{B}+\mathrm{D}=4\\ 4\mathrm{A}+3\mathrm{C}=0\\ 4\mathrm{B}+3\mathrm{D}=10\\ \mathrm{On}\text{ solving these equations, we get}\\ \text{A}=0,\mathrm{B}=-2,\mathrm{C}=0\text{and}\mathrm{D}=6\\ \therefore \frac{(4{\mathrm{x}}^2+10)}{({\mathrm{x}}^2+3)({\mathrm{x}}^2+4)}=\frac{-2}{({\mathrm{x}}^2+3)}+\frac{6}{({\mathrm{x}}^2+4)}\\ \mathrm{Then},\\ \int \frac{({\mathrm{x}}^2+1)({\mathrm{x}}^2+2)}{({\mathrm{x}}^2+3)({\mathrm{x}}^2+4)}\mathrm{dx}=\int 1\mathrm{dx}-(\int \frac{-2}{({\mathrm{x}}^2+3)}\mathrm{dx}+6\int \frac{1}{({\mathrm{x}}^2+4)}\mathrm{dx})\\ =\mathrm{x}+\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{\mathrm{x}}{\sqrt{3}}\right)-\frac{6}{2}{\tan }^{-1}\left(\frac{\mathrm{x}}{2}\right)+\mathrm{C}\\ =\mathrm{x}+\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{\mathrm{x}}{\sqrt{3}}\right)-3{\tan }^{-1}\left(\frac{\mathrm{x}}{2}\right)+\mathrm{C}\end{array}$

Q.55

$\text{Integrate the rational function\hspace{0.17em}}\frac{2\mathrm{x}}{({\mathrm{x}}^2+1)({\mathrm{x}}^2+3)}$

Ans.

$\begin{array}{l}\mathrm{We}\mathrm{have},\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\\ \mathrm{\hspace{0.17em}}\frac{2\mathrm{x}}{({\mathrm{x}}^2+1)({\mathrm{x}}^2+3)}\\ \mathrm{Let}\hspace{0.17em}\hspace{0.17em}\mathrm{t}={\mathrm{x}}^2\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=2\mathrm{x}\Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{2\mathrm{x}}\\ \int \frac{2\mathrm{x}}{({\mathrm{x}}^2+1)({\mathrm{x}}^2+3)}\mathrm{\hspace{0.17em}}\mathrm{dx}=\int \frac{2\mathrm{x}}{(\mathrm{t}+1)(\mathrm{t}+3)}\mathrm{\hspace{0.17em}}\frac{\mathrm{dt}}{2\mathrm{x}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\int \frac{1}{(\mathrm{t}+1)(\mathrm{t}+3)}\mathrm{\hspace{0.17em}}\mathrm{dt}\\ \mathrm{Let}\hspace{0.17em}\hspace{0.17em}\frac{1}{(\mathrm{t}+1)(\mathrm{t}+3)}=\frac{\mathrm{A}}{\mathrm{t}+1}+\frac{\mathrm{B}}{\mathrm{t}+3}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}1=\mathrm{A}(\mathrm{t}+3)+\mathrm{B}(\mathrm{t}+1)\\ \mathrm{Substituting}\mathrm{t}=-1\mathrm{and}-3\mathrm{respectively},\mathrm{we}\mathrm{get}\\ \mathrm{A}=\frac{1}{2}\hspace{0.17em}\hspace{0.17em}\mathrm{and}\hspace{0.17em}\hspace{0.17em}\mathrm{B}=-\frac{1}{2}\\ \therefore \frac{1}{(\mathrm{t}+1)(\mathrm{t}+3)}=\frac{1}{2(\mathrm{t}+1)}-\frac{1}{2(\mathrm{t}+3)}\\ \Rightarrow \int \frac{1}{(\mathrm{t}+1)(\mathrm{t}+3)}\mathrm{\hspace{0.17em}}\mathrm{dx}=\int \frac{1}{2(\mathrm{t}+1)}\mathrm{\hspace{0.17em}}\mathrm{dt}-\int \frac{1}{2(\mathrm{t}+3)}\mathrm{\hspace{0.17em}}\mathrm{dx}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{1}{2}\log |\mathrm{t}+1|-\frac{1}{2}\log |\mathrm{t}+3|+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{1}{2}\log \left|\frac{\mathrm{t}+1}{\mathrm{t}+3}\right|+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{1}{2}\log \left|\frac{{\mathrm{x}}^2+1}{{\mathrm{x}}^2+3}\right|+\mathrm{C}\end{array}$

Q.56

$\text{Integrate the rational function\hspace{0.17em}}\frac{1}{\mathrm{x}({\mathrm{x}}^4-1)}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\frac{1}{\mathrm{x}({\mathrm{x}}^4-1)}=\frac{1}{\mathrm{x}({\mathrm{x}}^4-1)}\times \frac{{\mathrm{x}}^3}{{\mathrm{x}}^3}\\ =\frac{{\mathrm{x}}^3}{{\mathrm{x}}^4({\mathrm{x}}^4-1)}\\ \mathrm{Let}\mathrm{t}={\mathrm{x}}^4\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=4{\mathrm{x}}^{4-1}\Rightarrow \frac{\mathrm{dt}}{4{\mathrm{x}}^3}=\mathrm{dx}\\ \therefore \int \frac{1}{\mathrm{x}({\mathrm{x}}^4-1)}\mathrm{dx}=\int \frac{{\mathrm{x}}^3}{{\mathrm{x}}^4({\mathrm{x}}^4-1)}\mathrm{dx}\\ =\int \frac{{\mathrm{x}}^3}{\mathrm{t}(\mathrm{t}-1)}\frac{\mathrm{dt}}{4{\mathrm{x}}^3}\\ =\frac{1}{4}\int \frac{1}{\mathrm{t}(\mathrm{t}-1)}\mathrm{dt}\\ \mathrm{Let}\frac{1}{\mathrm{t}(\mathrm{t}-1)}=\frac{\mathrm{A}}{\mathrm{t}}+\frac{\mathrm{B}}{\mathrm{t}-1}\\ 1=\mathrm{A}(\mathrm{t}-1)+\mathrm{Bt}\\ \mathrm{Substituting}\text{ t}=\text{0 and 1 respectively, we get}\\ \mathrm{A}=-1\mathrm{and}\mathrm{B}=1\\ \therefore \frac{1}{\mathrm{t}(\mathrm{t}-1)}=\frac{-1}{\mathrm{t}}+\frac{1}{\mathrm{t}-1}\\ \Rightarrow \int \frac{1}{\mathrm{x}({\mathrm{x}}^4-1)}\mathrm{dx}=\frac{1}{4}\int (-\frac{1}{\mathrm{t}}+\frac{1}{\mathrm{t}-1})\mathrm{dt}\\ =-\frac{1}{4}\int \frac{1}{\mathrm{t}}\mathrm{dt}+\frac{1}{4}\int \frac{1}{\mathrm{t}-1}\mathrm{dt}\\ =-\frac{1}{4}\log \left|\mathrm{t}\right|+\frac{1}{4}\log |\mathrm{t}-1|+\mathrm{C}\\ =-\frac{1}{4}\log \left|{\mathrm{x}}^4\right|+\frac{1}{4}\log |{\mathrm{x}}^4-1|+\mathrm{C}\\ =\frac{1}{4}\log \left|\frac{{\mathrm{x}}^4-1}{{\mathrm{x}}^4}\right|+\mathrm{C}\end{array}$

Q.57

$\text{Integrate the rational function\hspace{0.17em}}\frac{1}{({\mathrm{e}}^{\mathrm{x}}-1)}$

Ans.

$\begin{array}{l}\int \frac{1}{({\mathrm{e}}^{\mathrm{x}}-1)}\mathrm{dx}\\ \mathrm{Let}\mathrm{t}={\mathrm{e}}^{\mathrm{x}}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}={\mathrm{e}}^{\mathrm{x}}\\ \mathrm{dx}=\frac{\mathrm{dt}}{{\mathrm{e}}^{\mathrm{x}}}=\frac{\mathrm{dt}}{\mathrm{t}}\\ \therefore \int \frac{1}{({\mathrm{e}}^{\mathrm{x}}-1)}\mathrm{dx}=\int \frac{1}{(\mathrm{t}-1)}\frac{\mathrm{dt}}{\mathrm{t}}\\ =\int \frac{1}{\mathrm{t}(\mathrm{t}-1)}\mathrm{dt}\\ \mathrm{Let}\frac{1}{\mathrm{t}(\mathrm{t}-1)}=\frac{\mathrm{A}}{\mathrm{t}}+\frac{\mathrm{B}}{\mathrm{t}-1}\\ =\frac{\mathrm{A}(\mathrm{t}-1)+\mathrm{Bt}}{\mathrm{t}(\mathrm{t}-1)}\\ \Rightarrow 1=\mathrm{A}(\mathrm{t}-1)+\mathrm{Bt}...\left(\mathrm{i}\right)\\ \mathrm{Putting}\text{t}=0\text{ and 1 respectively in equation}\left(\mathrm{i}\right),\mathrm{we}\text{get}\\ \text{A}=-1\text{and B}=1\\ \therefore \frac{1}{\mathrm{t}(\mathrm{t}-1)}=\frac{-1}{\mathrm{t}}+\frac{1}{\mathrm{t}-1}\\ \Rightarrow \int \frac{1}{\mathrm{t}(\mathrm{t}-1)}\mathrm{dt}=\int \frac{-1}{\mathrm{t}}\mathrm{dt}+\int \frac{1}{\mathrm{t}-1}\mathrm{dt}\\ =-\log \left|\mathrm{t}\right|+\log |\mathrm{t}-1|+\mathrm{C}\\ =\log \left|\frac{\mathrm{t}-1}{\mathrm{t}}\right|+\mathrm{C}\\ \therefore \int \frac{1}{({\mathrm{e}}^{\mathrm{x}}-1)}\mathrm{dx}=\log \left|\frac{{\mathrm{e}}^{\mathrm{x}}-1}{{\mathrm{e}}^{\mathrm{x}}}\right|+\mathrm{C}\end{array}$

Q.58

$\begin{array}{l}\mathrm{Choose}\mathrm{the}\mathrm{correct}\mathrm{answer}\\ \int \frac{\mathrm{xdx}}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)}\mathrm{dx}\mathrm{equals}\\ \left(\mathrm{A}\right)\text{\hspace{0.33em}}\log \left|\frac{{\left(\mathrm{x}-1\right)}^2}{\mathrm{x}-2}\right|+\mathrm{C}\\ \left(\mathrm{B}\right)\log \left|\frac{{\left(\mathrm{x}-2\right)}^2}{\mathrm{x}-1}\right|+\mathrm{C}\\ \left(\mathrm{C}\right)\log \left|{\left(\frac{\mathrm{x}-1}{\mathrm{x}-2}\right)}^2\right|+\mathrm{C}\\ \left(\mathrm{D}\right)\log \left|\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)\right|+\mathrm{C}\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\int \frac{\mathrm{xdx}}{(\mathrm{x}-1)(\mathrm{x}-2)}\mathrm{dx}\\ \mathrm{Let}\frac{\mathrm{x}}{(\mathrm{x}-1)(\mathrm{x}-2)}=\frac{\mathrm{A}}{\mathrm{x}-1}+\frac{\mathrm{B}}{\mathrm{x}-2}\\ =\frac{\mathrm{A}(\mathrm{x}-2)+\mathrm{B}(\mathrm{x}-1)}{(\mathrm{x}-1)(\mathrm{x}-2)}\\ \Rightarrow \mathrm{x}=\mathrm{A}(\mathrm{x}-2)+\mathrm{B}(\mathrm{x}-1)...\left(\mathrm{i}\right)\\ \mathrm{Substituting}\text{x}=1\mathrm{and}2,\text{in equation}\left(\mathrm{i}\right),\text{\hspace{0.17em}we get}\\ \text{A}=-\text{1 and B}=2\\ \Rightarrow \frac{\mathrm{x}}{(\mathrm{x}-1)(\mathrm{x}-2)}=\frac{-1}{\mathrm{x}-1}+\frac{2}{\mathrm{x}-2}\\ \therefore \int \frac{\mathrm{x}}{(\mathrm{x}-1)(\mathrm{x}-2)}\mathrm{dx}=\int \frac{-1}{\mathrm{x}-1}\mathrm{dx}+\int \frac{2}{\mathrm{x}-2}\mathrm{dx}\\ =-\text{log}|\mathrm{x}-1|+2\log |\mathrm{x}-2|+\mathrm{C}\\ =\log \left|\frac{{(\mathrm{x}-2)}^2}{(\mathrm{x}-1)}\right|+\mathrm{C}\\ \mathrm{Hence},\text{the option}\left(\text{B}\right)\text{is correct.}\end{array}$

Q.59

$\begin{array}{l}\mathrm{Choose}\mathrm{the}\mathrm{correct}\mathrm{answer}\\ \int \frac{\mathrm{dx}}{\mathrm{x}\left({\mathrm{x}}^2+1\right)}\mathrm{equals}\\ \left(\mathrm{A}\right)\text{\hspace{0.33em}}\log \left|\mathrm{x}\right|-\frac{1}{2}\log \left({\mathrm{x}}^2+1\right)+\mathrm{C}\\ \left(\mathrm{B}\right)\log \left|\mathrm{x}\right|+\frac{1}{2}\log \left({\mathrm{x}}^2+1\right)+\mathrm{C}\\ \left(\mathrm{C}\right)-\log \left|\mathrm{x}\right|+\frac{1}{2}\log \left({\mathrm{x}}^2+1\right)+\mathrm{C}\\ \left(\mathrm{D}\right)\frac{1}{2}\log \left|\mathrm{x}\right|+\log \left({\mathrm{x}}^2+1\right)+\mathrm{C}\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\int \frac{\mathrm{dx}}{\mathrm{x}({\mathrm{x}}^2+1)}\\ \mathrm{Let}\frac{1}{\mathrm{x}({\mathrm{x}}^2+1)}=\frac{\mathrm{A}}{\mathrm{x}}+\frac{\mathrm{Bx}+\mathrm{C}}{({\mathrm{x}}^2+1)}\\ =\frac{\mathrm{A}({\mathrm{x}}^2+1)+(\mathrm{Bx}+\mathrm{C})\mathrm{x}}{\mathrm{x}({\mathrm{x}}^2+1)}\\ \Rightarrow 1=\mathrm{A}({\mathrm{x}}^2+1)+({\mathrm{Bx}}^2+\mathrm{Cx})\\ \Rightarrow 1={\mathrm{x}}^2(\mathrm{A}+\mathrm{B})+\mathrm{Cx}+\mathrm{A}\\ \mathrm{Equating}{\text{the coefficients of x}}^{\text{2}}\text{, x and constant term, we get}\\ \mathrm{A}+\mathrm{B}=0,\mathrm{C}=0\text{\hspace{0.17em}and A}=1\\ \mathrm{On}\text{solving these equations, we get}\\ \text{A}=\text{1,\hspace{0.17em} B}=-\text{1 and C}=0\\ \Rightarrow \frac{1}{\mathrm{x}({\mathrm{x}}^2+1)}=\frac{1}{\mathrm{x}}+\frac{-\mathrm{x}}{({\mathrm{x}}^2+1)}\\ \therefore \int \frac{1}{\mathrm{x}({\mathrm{x}}^2+1)}\mathrm{dx}=\int \frac{1}{\mathrm{x}}\mathrm{dx}-\int \frac{\mathrm{x}}{({\mathrm{x}}^2+1)}\mathrm{dx}\\ =\text{log}\left|\mathrm{x}\right|-2\log |{\mathrm{x}}^2+1|+\mathrm{C}\\ \mathrm{Thus},\text{the correct option is}\left(\mathrm{A}\right).\end{array}$

Q.60

Integrate the function x sin x.

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}=\int \mathrm{xsinx}\mathrm{dx}\\ \mathrm{Taking}\text{algebraic function i.e. x as a first function and sinx as}\\ \text{second function and integrating by parts, we get}\\ =\mathrm{x}\int \mathrm{sinx}\mathrm{dx}-\int \{\left(\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{x}\right)\int \mathrm{sinx}\mathrm{dx}\}\mathrm{dx}\\ =\mathrm{x}(-\mathrm{cosx})-\int 1.(-\mathrm{cosx})\mathrm{dx}\\ =-\mathrm{xcosx}+\mathrm{sinx}+\mathrm{C}\end{array}$

Q.61

Integrate the functions
x sin 3x

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}=\int \mathrm{xsin}3\mathrm{x}\mathrm{dx}\\ \mathrm{Taking}\text{algebraic function i.e. x as a first function and sin3x as}\\ \text{second function and integrating by parts, we get}\\ =\mathrm{x}\int \sin 3\mathrm{x}\mathrm{dx}-\int \{\left(\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{x}\right)\int \sin 3\mathrm{x}\mathrm{dx}\}\mathrm{dx}\\ =\mathrm{x}(-\frac{1}{3}\cos 3\mathrm{x})-\int 1.(-\frac{1}{3}\mathrm{cosx})\mathrm{dx}\\ =-\frac{1}{3}\mathrm{xcos}3\mathrm{x}+\frac{1}{9}\sin 3\mathrm{x}+\mathrm{C}\end{array}$

Q.62

Integrate the functions
x² e^x

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}=\int {\mathrm{x}}^2{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}\\ \mathrm{Taking}{\text{algebraic function i.e. x}}^{\text{2}}{\text{as a first function and e}}^{\text{x}}\text{as}\\ \text{second function and integrating by parts, we get}\\ ={\mathrm{x}}^2\int {\mathrm{e}}^{\mathrm{x}}\mathrm{dx}-\int \{\left(\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{x}}^2\right)\int {\mathrm{e}}^{\mathrm{x}}\mathrm{dx}\}\mathrm{dx}\\ ={\mathrm{x}}^2\left({\mathrm{e}}^{\mathrm{x}}\right)-\int 2\mathrm{x}.\left({\mathrm{e}}^{\mathrm{x}}\right)\mathrm{dx}\\ ={\mathrm{x}}^2{\mathrm{e}}^{\mathrm{x}}-2[\mathrm{x}\int {\mathrm{e}}^{\mathrm{x}}\mathrm{dx}-\int \{\left(\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{x}\right)\int {\mathrm{e}}^{\mathrm{x}}\mathrm{dx}\}\mathrm{dx}]+\mathrm{C}\\ ={\mathrm{x}}^2{\mathrm{e}}^{\mathrm{x}}-2[{\mathrm{xe}}^{\mathrm{x}}-\int 1.{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}]+\mathrm{C}\\ ={\mathrm{x}}^2{\mathrm{e}}^{\mathrm{x}}-2{\mathrm{xe}}^{\mathrm{x}}+2{\mathrm{e}}^{\mathrm{x}}+\mathrm{C}\\ ={\mathrm{e}}^{\mathrm{x}}({\mathrm{x}}^2-2\mathrm{x}+2)+\mathrm{C}\end{array}$

Q.63

Integrate the functions
x logx

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}=\int \mathrm{xlogx}\mathrm{dx}\\ \mathrm{Taking}\text{logx as a first function and}\mathrm{x}\text{as}\\ \text{second function and integrating by parts, we get}\\ =\mathrm{logx}\int \mathrm{x}\mathrm{dx}-\int \{\left(\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{logx}\right)\int \mathrm{x}\mathrm{dx}\}\mathrm{dx}\\ =\mathrm{logx}\left(\frac{{\mathrm{x}}^2}{2}\right)-\int \frac{1}{\mathrm{x}}.\left(\frac{{\mathrm{x}}^2}{2}\right)\mathrm{dx}\\ =\frac{{\mathrm{x}}^2}{2}\mathrm{logx}-\int \frac{\mathrm{x}}{2}\mathrm{dx}+\mathrm{C}\\ =\frac{{\mathrm{x}}^2}{2}\mathrm{logx}-\frac{{\mathrm{x}}^2}{4}+\mathrm{C}\end{array}$

Q.64

Integrate the functions
x log 2x

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}=\int \mathrm{xlog}2\mathrm{x}\mathrm{dx}\\ \mathrm{Taking}\text{log2x as a first function and}\mathrm{x}\text{as}\\ \text{second function and integrating by parts, we get}\\ =\log 2\mathrm{x}\int \mathrm{x}\mathrm{dx}-\int \{\left(\frac{\mathrm{d}}{\mathrm{dx}}\log 2\mathrm{x}\right)\int \mathrm{x}\mathrm{dx}\}\mathrm{dx}\\ =\log 2\mathrm{x}\left(\frac{{\mathrm{x}}^2}{2}\right)-\int \frac{1}{2\mathrm{x}}\times 2.\left(\frac{{\mathrm{x}}^2}{2}\right)\mathrm{dx}\\ =\frac{{\mathrm{x}}^2}{2}\log 2\mathrm{x}-\int \frac{\mathrm{x}}{2}\mathrm{dx}+\mathrm{C}\\ =\frac{{\mathrm{x}}^2}{2}\log 2\mathrm{x}-\frac{{\mathrm{x}}^2}{4}+\mathrm{C}\end{array}$

Q.65

Integrate the functions
x sin^-1x

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}=\int {\mathrm{xsin}}^{-1}\mathrm{x}\mathrm{dx}\\ \mathrm{Let}\text{t}={\text{sin}}^{-\text{1}}\text{x}\Rightarrow \mathrm{x}=\mathrm{sint}\\ \mathrm{cost}=\sqrt{1-{\sin }^2\mathrm{t}}=\sqrt{1-{\mathrm{x}}^2}\\ \mathrm{Then},\frac{\mathrm{dt}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}{\text{sin}}^{-\text{1}}\text{x}\\ =\frac{1}{\sqrt{1-{\mathrm{x}}^2}}\\ =\frac{1}{\mathrm{cost}}\\ \Rightarrow \mathrm{dx}=\mathrm{cost}\mathrm{dt}\\ \mathrm{So},\mathrm{I}=\int \mathrm{tsint}\mathrm{cost}\mathrm{dt}\\ =\frac{1}{2}\int \mathrm{tsin}2\mathrm{t}\mathrm{dt}\\ \mathrm{Taking}\text{algebraic function i.e. t as a first function and sin2t as}\\ \text{second function and integrating by parts, we get}\\ =\frac{1}{2}[\mathrm{t}\int \sin 2\mathrm{t}\mathrm{dt}-\int \{\left(\frac{\mathrm{d}}{\mathrm{dt}}\mathrm{t}\right)\int \sin 2\mathrm{t}\mathrm{dt}\}\mathrm{dt}]\\ =\frac{1}{2}[\mathrm{t}(-\frac{1}{2}\cos 2\mathrm{t})-\int 1.(-\frac{1}{2}\mathrm{co}2\mathrm{t})\mathrm{dx}]\\ =\frac{1}{2}[-\frac{1}{2}\mathrm{tcos}2\mathrm{t}+\frac{1}{4}\sin 2\mathrm{t}+\mathrm{C}]\\ =-\frac{1}{4}\mathrm{t}(1-2{\sin }^2\mathrm{t})+\frac{1}{8}\times 2\mathrm{sintcost}+\mathrm{C}\\ =\frac{1}{4}\mathrm{t}(2{\sin }^2\mathrm{t}-1)+\frac{1}{4}\mathrm{sintcost}+\mathrm{C}\\ =\frac{1}{4}(2{\mathrm{x}}^2-1){\sin }^{-1}\mathrm{x}+\frac{1}{4}\mathrm{x}\sqrt{1-{\mathrm{x}}^2}+\mathrm{C}\end{array}$

Q.66

Integrate the functions
x tan^-1x

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}=\int {\mathrm{xtan}}^{-1}\mathrm{x}\mathrm{dx}\\ \mathrm{Taking}{\tan }^{-1}\mathrm{x}\text{as a first function and x as second function and}\\ \text{integrating by parts, we get}\\ ={\tan }^{-1}\mathrm{x}\int \mathrm{x}\mathrm{dx}-\int \{\left(\frac{\mathrm{d}}{\mathrm{dx}}{\tan }^{-1}\mathrm{x}\right)\int \mathrm{x}\mathrm{dx}\}\mathrm{dx}\\ ={\tan }^{-1}\mathrm{x}\left(\frac{{\mathrm{x}}^2}{2}\right)-\int \frac{1}{1+{\mathrm{x}}^2}.\left(\frac{{\mathrm{x}}^2}{2}\right)\mathrm{dx}\\ =\frac{{\mathrm{x}}^2}{2}{\tan }^{-1}\mathrm{x}-\frac{1}{2}\int \frac{{\mathrm{x}}^2}{1+{\mathrm{x}}^2}\mathrm{dx}+\mathrm{C}\\ =\frac{{\mathrm{x}}^2}{2}{\tan }^{-1}\mathrm{x}-\frac{1}{2}\int \frac{1+{\mathrm{x}}^2-1}{1+{\mathrm{x}}^2}\mathrm{dx}+\mathrm{C}\\ =\frac{{\mathrm{x}}^2}{2}{\tan }^{-1}\mathrm{x}-\frac{1}{2}\int \frac{1+{\mathrm{x}}^2}{1+{\mathrm{x}}^2}\mathrm{dx}+\frac{1}{2}\int \frac{1}{1+{\mathrm{x}}^2}\mathrm{dx}+\mathrm{C}\\ =\frac{{\mathrm{x}}^2}{2}{\tan }^{-1}\mathrm{x}-\frac{1}{2}\int 1.\mathrm{dx}+\frac{1}{2}{\tan }^{-1}\mathrm{x}+\mathrm{C}\\ =\frac{{\mathrm{x}}^2}{2}{\tan }^{-1}\mathrm{x}-\frac{1}{2}\mathrm{x}\mathrm{dx}+\frac{1}{2}{\tan }^{-1}\mathrm{x}+\mathrm{C}\end{array}$

Q.67

Integrate the functions
x cos^-1x

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}=\int {\mathrm{xcos}}^{-1}\mathrm{x}\mathrm{dx}\\ \mathrm{Let}\text{t}={\cos }^{-\text{1}}\text{x}\Rightarrow \mathrm{x}=\mathrm{cost}\\ \mathrm{sint}=\sqrt{1-{\cos }^2\mathrm{t}}=\sqrt{1-{\mathrm{x}}^2}\\ \mathrm{Then},\frac{\mathrm{dt}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}{\cos }^{-\text{1}}\text{x}\\ =\frac{-1}{\sqrt{1-{\mathrm{x}}^2}}\\ =\frac{-1}{\mathrm{sint}}\\ \Rightarrow \mathrm{dx}=-\mathrm{sint}\mathrm{dt}\\ \mathrm{So},\mathrm{I}=\int \mathrm{tcost}(-\mathrm{sint})\mathrm{dt}\\ =-\frac{1}{2}\int \mathrm{tsin}2\mathrm{t}\mathrm{dt}\\ \mathrm{Taking}\text{algebraic function i.e. t as a first function and sin2t as}\\ \text{second function and integrating by parts, we get}\\ =-\frac{1}{2}[\mathrm{t}\int \sin 2\mathrm{t}\mathrm{dt}-\int \{\left(\frac{\mathrm{d}}{\mathrm{dt}}\mathrm{t}\right)\int \sin 2\mathrm{t}\mathrm{dt}\}\mathrm{dt}]\\ =-\frac{1}{2}[\mathrm{t}(-\frac{1}{2}\cos 2\mathrm{t})-\int 1.(-\frac{1}{2}\mathrm{co}2\mathrm{t})\mathrm{dx}]\\ =-\frac{1}{2}[-\frac{1}{2}\mathrm{tcos}2\mathrm{t}+\frac{1}{4}\sin 2\mathrm{t}+\mathrm{C}]\\ =\frac{1}{4}\mathrm{t}(1-2{\sin }^2\mathrm{t})-\frac{1}{8}\times 2\mathrm{sintcost}+\mathrm{C}\\ =\frac{1}{4}\mathrm{t}(1-2{\sin }^2\mathrm{t})-\frac{1}{4}\mathrm{sintcost}+\mathrm{C}\\ =\frac{1}{4}(1-2{\mathrm{x}}^2){\sin }^{-1}\mathrm{x}-\frac{1}{4}\mathrm{x}\sqrt{1-{\mathrm{x}}^2}+\mathrm{C}\end{array}$

Q.68

Integrate the functions
(sin^–1x)²

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}=\int {\left({\sin }^{-1}\mathrm{x}\right)}^2\mathrm{dx}\\ =\int {\left({\sin }^{-1}\mathrm{x}\right)}^2.1\mathrm{dx}\\ \mathrm{Taking}{\left({\sin }^{-1}\mathrm{x}\right)}^2\text{as first function and 1 as second function and}\\ \text{integrating by parts, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}={\left({\sin }^{-1}\mathrm{x}\right)}^2\int 1\mathrm{dx}-\int \{\frac{\mathrm{d}}{\mathrm{dx}}{\left({\sin }^{-1}\mathrm{x}\right)}^2\int 1\mathrm{dx}\}\mathrm{dx}\\ ={\left({\sin }^{-1}\mathrm{x}\right)}^2\mathrm{x}-\int \{2{\sin }^{-1}\mathrm{x}\frac{\mathrm{d}}{\mathrm{dx}}\left({\sin }^{-1}\mathrm{x}\right).\mathrm{x}\}\mathrm{dx}\\ ={\left({\sin }^{-1}\mathrm{x}\right)}^2\mathrm{x}-\int \frac{2{\mathrm{xsin}}^{-1}\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}\\ ={\left({\sin }^{-1}\mathrm{x}\right)}^2\mathrm{x}+\int {\sin }^{-1}\mathrm{x}\frac{-2\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}\\ ={\left({\sin }^{-1}\mathrm{x}\right)}^2\mathrm{x}+{\sin }^{-1}\mathrm{x}\int \frac{-2\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}-(\int \frac{\mathrm{d}}{\mathrm{dx}}{\sin }^{-1}\mathrm{x}\int \frac{-2\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx})\mathrm{dx}\\ ={\left({\sin }^{-1}\mathrm{x}\right)}^2\mathrm{x}+{\sin }^{-1}\mathrm{x}\int \frac{-2\mathrm{x}}{\sqrt{\mathrm{t}}}\frac{\mathrm{dt}}{-2\mathrm{x}}-\{\int \frac{\mathrm{d}}{\mathrm{dx}}{\sin }^{-1}\mathrm{x}\int \frac{-2\mathrm{x}}{\sqrt{\mathrm{t}}}\frac{\mathrm{dt}}{-2\mathrm{x}}\}\mathrm{dx}\\ \left[\mathrm{Let}\text{t}=1-{\mathrm{x}}^2\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=-2\mathrm{x}\Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{-2\mathrm{x}}\right]\\ ={\left({\sin }^{-1}\mathrm{x}\right)}^2\mathrm{x}+{\sin }^{-1}\mathrm{x}.2\sqrt{\mathrm{t}}-\{\int \frac{1}{\sqrt{1-{\mathrm{x}}^2}}2\sqrt{\mathrm{t}}\}\mathrm{dx}\\ ={\left({\sin }^{-1}\mathrm{x}\right)}^2\mathrm{x}+2{\sin }^{-1}\mathrm{x}.\sqrt{1-{\mathrm{x}}^2}-\int \frac{1}{\sqrt{1-{\mathrm{x}}^2}}2\sqrt{1-{\mathrm{x}}^2}\mathrm{dx}\\ ={\left({\sin }^{-1}\mathrm{x}\right)}^2\mathrm{x}+2{\sin }^{-1}\mathrm{x}.\sqrt{1-{\mathrm{x}}^2}-2\int 1\mathrm{dx}\\ =\mathrm{x}{\left({\sin }^{-1}\mathrm{x}\right)}^2+2\sqrt{1-{\mathrm{x}}^2}{\sin }^{-1}\mathrm{x}-2\mathrm{x}+\mathrm{C}\end{array}$

Q.69

Integrate the functions
x² logx

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}=\int {\mathrm{x}}^2\mathrm{logx}\mathrm{dx}\\ \mathrm{Taking}\text{logx as a first function and}{\mathrm{x}}^2\text{as}\\ \text{second function and integrating by parts, we get}\\ =\mathrm{logx}\int {\mathrm{x}}^2\mathrm{dx}-\int \{\left(\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{logx}\right)\int {\mathrm{x}}^2\mathrm{dx}\}\mathrm{dx}\\ =\mathrm{logx}\left(\frac{{\mathrm{x}}^3}{3}\right)-\int \frac{1}{\mathrm{x}}.\left(\frac{{\mathrm{x}}^3}{3}\right)\mathrm{dx}\\ =\frac{{\mathrm{x}}^3}{3}\mathrm{logx}-\int \frac{{\mathrm{x}}^2}{3}\mathrm{dx}+\mathrm{C}\\ =\frac{{\mathrm{x}}^3}{3}\mathrm{logx}-\frac{{\mathrm{x}}^3}{9}+\mathrm{C}\end{array}$

Q.70

$\begin{array}{l}\text{Integrate the functions}\\ \frac{{\mathrm{xcos}}^{–1}\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}=\int \frac{{\mathrm{xcos}}^{–1}\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}\\ =\frac{-1}{2}\int {\cos }^{-1}\mathrm{x}\frac{-2\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}\\ \mathrm{Taking}{\cos }^{-1}\mathrm{x}\text{as first function and}\frac{-2\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}}\text{as second function and}\\ \text{integrating by parts, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\frac{-1}{2}\{{\cos }^{-1}\mathrm{x}\int \frac{-2\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}-(\int \frac{\mathrm{d}}{\mathrm{dx}}{\cos }^{-1}\mathrm{x}\int \frac{-2\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx})\mathrm{dx}\}\\ =\frac{-1}{2}\{{\cos }^{-1}\mathrm{x}\int \frac{-2\mathrm{x}}{\sqrt{\mathrm{t}}}\frac{\mathrm{dt}}{-2\mathrm{x}}-(\int \frac{\mathrm{d}}{\mathrm{dx}}{\cos }^{-1}\mathrm{x}\int \frac{-2\mathrm{x}}{\sqrt{\mathrm{t}}}\frac{\mathrm{dt}}{-2\mathrm{x}})\mathrm{dx}\}\\ \left[\mathrm{Let}\text{t}=1-{\mathrm{x}}^2\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=-2\mathrm{x}\Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{-2\mathrm{x}}\right]\\ =\frac{-1}{2}[{\cos }^{-1}\mathrm{x}.2\sqrt{\mathrm{t}}-(\int \frac{-1}{\sqrt{1-{\mathrm{x}}^2}}2\sqrt{\mathrm{t}})\mathrm{dx}]\\ =\frac{-1}{2}[{\cos }^{-1}\mathrm{x}.2\sqrt{1-{\mathrm{x}}^2}-(\int \frac{-1}{\sqrt{1-{\mathrm{x}}^2}}2\sqrt{1-{\mathrm{x}}^2})\mathrm{dx}]\\ =\frac{-1}{2}(2{\cos }^{-1}\mathrm{x}\sqrt{1-{\mathrm{x}}^2}+2\int 1\mathrm{dx})\\ =\frac{-1}{2}(2{\cos }^{-1}\mathrm{x}\sqrt{1-{\mathrm{x}}^2}+2\mathrm{x})+\mathrm{C}\\ =-(\sqrt{1-{\mathrm{x}}^2}{\cos }^{-1}\mathrm{x}+\mathrm{x})+\mathrm{C}\end{array}$

Q.71

Integrate the functions x sec²x

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}=\int {\mathrm{xsec}}^2\mathrm{x}\mathrm{dx}\\ \mathrm{Taking}\text{algebraic function i.e. x as a first function and}{\sec }^2\mathrm{x}\text{as}\\ \text{second function and integrating by parts, we get}\\ =\mathrm{x}\int {\sec }^2\mathrm{x}\mathrm{dx}-\int \{\left(\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{x}\right)\int {\sec }^2\mathrm{x}\mathrm{dx}\}\mathrm{dx}\\ =\mathrm{x}\left(\mathrm{tanx}\right)-\int 1.\mathrm{tanx}\mathrm{dx}\\ =\mathrm{xtanx}-\int \mathrm{tanx}\mathrm{dx}+\mathrm{C}\\ =\mathrm{xtanx}+\log \left|\mathrm{cosx}\right|+\mathrm{C}\end{array}$

Q.72

Integrate the functions tan^-1x

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}=\int {\tan }^{-1}\mathrm{x}\mathrm{dx}=\int {\tan }^{-1}\mathrm{x}.1\mathrm{dx}\\ \mathrm{Taking}{\tan }^{-1}\mathrm{x}\text{as a first function and 1 as second function and}\\ \text{integrating by parts, we get}\\ ={\tan }^{-1}\mathrm{x}\int 1\mathrm{dx}-\int \{\left(\frac{\mathrm{d}}{\mathrm{dx}}{\tan }^{-1}\mathrm{x}\right)\int 1\mathrm{dx}\}\mathrm{dx}\\ ={\tan }^{-1}\mathrm{x}\left(\mathrm{x}\right)-\int \frac{1}{1+{\mathrm{x}}^2}.\left(\mathrm{x}\right)\mathrm{dx}\\ =\mathrm{x}{\tan }^{-1}\mathrm{x}-\int \frac{\mathrm{x}}{1+{\mathrm{x}}^2}\mathrm{dx}+\mathrm{C}\\ =\mathrm{x}{\tan }^{-1}\mathrm{x}-\frac{1}{2}\log |1+{\mathrm{x}}^2|+\mathrm{C}\end{array}$

Q.73

Integrate the functions x (logx)²

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}=\int \mathrm{x}{\left(\mathrm{logx}\right)}^2\mathrm{dx}\\ \mathrm{Taking}{\left(\text{logx}\right)}^{\text{2}}\text{as a first function and}\mathrm{x}\text{as}\\ \text{second function and integrating by parts, we get}\\ ={\left(\text{logx}\right)}^{\text{2}}\int \mathrm{x}\mathrm{dx}-\int \{\left(\frac{\mathrm{d}}{\mathrm{dx}}{\left(\text{logx}\right)}^{\text{2}}\right)\int \mathrm{x}\mathrm{dx}\}\mathrm{dx}\\ ={\left(\text{logx}\right)}^{\text{2}}\left(\frac{{\mathrm{x}}^2}{2}\right)-\int \frac{2\mathrm{logx}}{\mathrm{x}}.\left(\frac{{\mathrm{x}}^2}{2}\right)\mathrm{dx}\\ =\frac{{\mathrm{x}}^2}{2}{\left(\text{logx}\right)}^{\text{2}}-\int \mathrm{xlogx}\mathrm{dx}+\mathrm{C}\\ =\frac{{\mathrm{x}}^2}{2}{\left(\text{logx}\right)}^{\text{2}}-[\mathrm{logx}\int \mathrm{x}\mathrm{dx}-\int \{\left(\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{logx}\right)\int \mathrm{x}\mathrm{dx}\}\mathrm{dx}]\\ =\frac{{\mathrm{x}}^2}{2}{\left(\text{logx}\right)}^{\text{2}}-\{\mathrm{logx}\left(\frac{{\mathrm{x}}^2}{2}\right)-\int \frac{1}{\mathrm{x}}.\left(\frac{{\mathrm{x}}^2}{2}\right)\mathrm{dx}\}\\ =\frac{{\mathrm{x}}^2}{2}{\left(\text{logx}\right)}^{\text{2}}-\{\frac{{\mathrm{x}}^2}{2}\mathrm{logx}-\int \frac{\mathrm{x}}{2}\mathrm{dx}\}\\ =\frac{{\mathrm{x}}^2}{2}{\left(\text{logx}\right)}^{\text{2}}-\frac{{\mathrm{x}}^2}{2}\mathrm{logx}+\frac{{\mathrm{x}}^2}{4}+\mathrm{C}\end{array}$

Q.74

Integrate the functions (x²+1) logx

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}=\int ({\mathrm{x}}^2+1)\mathrm{logx}\mathrm{dx}\\ \mathrm{Taking}\text{logx as a first function and}\left({\mathrm{x}}^2\text{+1}\right)\text{as second function}\\ \text{and integrating by parts, we get}\\ =\mathrm{logx}\int ({\mathrm{x}}^2+1)\mathrm{dx}-\int \{\left(\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{logx}\right)\int ({\mathrm{x}}^2+1)\mathrm{dx}\}\mathrm{dx}\\ =\mathrm{logx}(\frac{{\mathrm{x}}^3}{3}+\mathrm{x})-\int \frac{1}{\mathrm{x}}.(\frac{{\mathrm{x}}^3}{3}+\mathrm{x})\mathrm{dx}\\ =(\frac{{\mathrm{x}}^3}{3}+\mathrm{x})\mathrm{logx}-\int (\frac{{\mathrm{x}}^2}{3}+1)\mathrm{dx}+\mathrm{C}\\ =(\frac{{\mathrm{x}}^3}{3}+\mathrm{x})\mathrm{logx}-\frac{{\mathrm{x}}^3}{9}-\mathrm{x}+\mathrm{C}\end{array}$

Q.75

Integrate the functions e^x (sinx + cosx)

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}=\int {\mathrm{e}}^{\mathrm{x}}(\mathrm{sinx}+\mathrm{cosx})\mathrm{dx}\\ \mathrm{Here},\text{f}\left(\mathrm{x}\right)=\mathrm{sinx}\text{and f’}\left(\mathrm{x}\right)=\mathrm{cosx}\\ \mathrm{Thus},{\text{the given integrand is of the form e}}^{\text{x}}[\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}‘\left(\mathrm{x}\right)].\\ \mathrm{Therefore},\\ \mathrm{I}=\int {\mathrm{e}}^{\mathrm{x}}(\mathrm{sinx}+\mathrm{cosx})\mathrm{dx}\\ ={\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}+\mathrm{C}[\because \int {\mathrm{e}}^{\mathrm{x}}\{\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}‘\left(\mathrm{x}\right)\}\mathrm{dx}={\mathrm{e}}^{\mathrm{x}}\mathrm{f}\left(\mathrm{x}\right)+\mathrm{C}]\end{array}$

Q.76

$\mathrm{Integrate}\mathrm{the}\mathrm{functions}\frac{{\mathrm{xe}}^{\mathrm{x}}}{{(1+\mathrm{x})}^2}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}=\int \frac{{\mathrm{xe}}^{\mathrm{x}}}{{(1+\mathrm{x})}^2}\mathrm{dx}\\ =\int {\mathrm{e}}^{\mathrm{x}}\left\{\frac{\mathrm{x}}{{(1+\mathrm{x})}^2}\right\}\mathrm{dx}\\ =\int {\mathrm{e}}^{\mathrm{x}}\left\{\frac{1+\mathrm{x}-1}{{(1+\mathrm{x})}^2}\right\}\mathrm{dx}\\ =\int {\mathrm{e}}^{\mathrm{x}}\{\frac{1+\mathrm{x}}{{(1+\mathrm{x})}^2}-\frac{1}{{(1+\mathrm{x})}^2}\}\mathrm{dx}\\ =\int {\mathrm{e}}^{\mathrm{x}}\{\frac{1}{(1+\mathrm{x})}-\frac{1}{{(1+\mathrm{x})}^2}\}\mathrm{dx}\\ \mathrm{Let}\text{ f}\left(\mathrm{x}\right)=\frac{1}{(1+\mathrm{x})}\mathrm{and}\text{f’}\left(\mathrm{x}\right)=-\frac{1}{{(1+\mathrm{x})}^2}\\ \mathrm{Thus},{\text{the given integrand is of the form e}}^{\text{x}}[\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}‘\left(\mathrm{x}\right)].\\ \mathrm{Therefore},\\ \mathrm{I}=\int \frac{{\mathrm{xe}}^{\mathrm{x}}}{{(1+\mathrm{x})}^2}\mathrm{dx}\\ =\int {\mathrm{e}}^{\mathrm{x}}\{\frac{1}{(1+\mathrm{x})}-\frac{1}{{(1+\mathrm{x})}^2}\}\mathrm{dx}\\ =\frac{{\mathrm{e}}^{\mathrm{x}}}{1+\mathrm{x}}+\mathrm{C}[\because {\text{e}}^{\text{x}}\{\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}‘\left(\mathrm{x}\right)\}={\mathrm{e}}^{\mathrm{x}}\mathrm{f}\left(\mathrm{x}\right)]\end{array}$

Q.77

$\mathrm{Integrate}\mathrm{the}\mathrm{functions}{\mathrm{e}}^{\mathrm{x}}\left(\frac{1+\mathrm{sinx}}{1+\mathrm{cosx}}\right)$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\\ {\mathrm{e}}^{\mathrm{x}}\left(\frac{1+\mathrm{sinx}}{1+\mathrm{cosx}}\right)={\mathrm{e}}^{\mathrm{x}}\left(\frac{{\sin }^2\frac{\mathrm{x}}{2}+{\cos }^2\frac{\mathrm{x}}{2}+2\sin \frac{\mathrm{x}}{2}\cos \frac{\mathrm{x}}{2}}{2{\cos }^2\frac{\mathrm{x}}{2}}\right)\\ =\frac{1}{2}{\mathrm{e}}^{\mathrm{x}}{\left(\frac{\sin \frac{\mathrm{x}}{2}+\cos \frac{\mathrm{x}}{2}}{\cos \frac{\mathrm{x}}{2}}\right)}^2\\ =\frac{1}{2}{\mathrm{e}}^{\mathrm{x}}{(\tan \frac{\mathrm{x}}{2}+1)}^2\\ =\frac{1}{2}{\mathrm{e}}^{\mathrm{x}}(1+{\tan }^2\frac{\mathrm{x}}{2}+2\tan \frac{\mathrm{x}}{2})\\ =\frac{1}{2}{\mathrm{e}}^{\mathrm{x}}({\sec }^2\frac{\mathrm{x}}{2}+2\tan \frac{\mathrm{x}}{2})\\ ={\mathrm{e}}^{\mathrm{x}}(\frac{1}{2}{\sec }^2\frac{\mathrm{x}}{2}+\tan \frac{\mathrm{x}}{2})\\ ={\mathrm{e}}^{\mathrm{x}}(\tan \frac{\mathrm{x}}{2}+\frac{1}{2}{\sec }^2\frac{\mathrm{x}}{2})\\ \mathrm{Let}\text{f}\left(\mathrm{x}\right)=\tan \frac{\mathrm{x}}{2}\text{\hspace{0.17em} and f’}\left(\mathrm{x}\right)=\frac{1}{2}{\sec }^2\frac{\mathrm{x}}{2}\\ \mathrm{Thus},{\text{the given integrand is of the form e}}^{\text{x}}[\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}‘\left(\mathrm{x}\right)].\\ \mathrm{Therefore},\\ \int {\mathrm{e}}^{\mathrm{x}}\left(\frac{1+\mathrm{sinx}}{1+\mathrm{cosx}}\right)\mathrm{dx}=\int {\mathrm{e}}^{\mathrm{x}}(\tan \frac{\mathrm{x}}{2}+\frac{1}{2}{\sec }^2\frac{\mathrm{x}}{2})\mathrm{dx}\\ ={\mathrm{e}}^{\mathrm{x}}\tan \frac{\mathrm{x}}{2}+\mathrm{C}[\because {\text{e}}^{\text{x}}\{\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}‘\left(\mathrm{x}\right)\}={\mathrm{e}}^{\mathrm{x}}\mathrm{f}\left(\mathrm{x}\right)]\end{array}$

Q.78

$\mathrm{Integrate}\mathrm{the}\mathrm{functionse}{}^{\mathrm{x}}(\frac{1}{\mathrm{x}}-\frac{1}{{\mathrm{x}}^2})$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}=\int {\mathrm{e}}^{\mathrm{x}}(\frac{1}{\mathrm{x}}-\frac{1}{{\mathrm{x}}^2})\mathrm{dx}\\ \mathrm{Let}\text{ f}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}}\mathrm{and}\text{f’}\left(\mathrm{x}\right)=-\frac{1}{{\mathrm{x}}^2}\\ \mathrm{Thus},{\text{the given integrand is of the form e}}^{\text{x}}[\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}‘\left(\mathrm{x}\right)].\\ \mathrm{Therefore},\mathrm{I}=\int {\mathrm{e}}^{\mathrm{x}}(\frac{1}{\mathrm{x}}-\frac{1}{{\mathrm{x}}^2})\mathrm{dx}\\ =\frac{{\mathrm{e}}^{\mathrm{x}}}{\mathrm{x}}+\mathrm{C}[\because {\text{e}}^{\text{x}}\{\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}‘\left(\mathrm{x}\right)\}={\mathrm{e}}^{\mathrm{x}}\mathrm{f}\left(\mathrm{x}\right)]\end{array}$

Q.79

$\begin{array}{l}\text{Integrate the functions}\\ \frac{(\mathrm{x}-3){\mathrm{e}}^{\mathrm{x}}}{{(\mathrm{x}-1)}^3}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}=\int \frac{(\mathrm{x}-3){\mathrm{e}}^{\mathrm{x}}}{{(\mathrm{x}-1)}^3}\mathrm{dx}\\ =\int {\mathrm{e}}^{\mathrm{x}}\left\{\frac{\mathrm{x}-1-2}{{(1-\mathrm{x})}^3}\right\}\mathrm{dx}\\ =\int {\mathrm{e}}^{\mathrm{x}}\{\frac{\mathrm{x}-1}{{(1-\mathrm{x})}^3}-\frac{2}{{(1-\mathrm{x})}^3}\}\mathrm{dx}\\ =\int {\mathrm{e}}^{\mathrm{x}}\{\frac{1}{{(1-\mathrm{x})}^2}-\frac{2}{{(1-\mathrm{x})}^3}\}\mathrm{dx}\\ \mathrm{Let}\text{ f}\left(\mathrm{x}\right)=\frac{1}{{(1-\mathrm{x})}^2}\mathrm{and}\text{f’}\left(\mathrm{x}\right)=-\frac{2}{{(1-\mathrm{x})}^3}\\ \mathrm{Thus},{\text{the given integrand is of the form e}}^{\text{x}}[\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}‘\left(\mathrm{x}\right)].\\ \mathrm{Therefore},\\ \mathrm{I}=\int \frac{(\mathrm{x}-3){\mathrm{e}}^{\mathrm{x}}}{{(\mathrm{x}-1)}^3}\mathrm{dx}\\ =\int {\mathrm{e}}^{\mathrm{x}}\{\frac{1}{{(1-\mathrm{x})}^2}-\frac{2}{{(1-\mathrm{x})}^3}\}\mathrm{dx}\\ =\frac{{\mathrm{e}}^{\mathrm{x}}}{{(1-\mathrm{x})}^2}+\mathrm{C}[\because {\text{e}}^{\text{x}}\{\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}‘\left(\mathrm{x}\right)\}={\mathrm{e}}^{\mathrm{x}}\mathrm{f}\left(\mathrm{x}\right)]\end{array}$

Q.80

Integrate the functions e^2x sinx

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}{\mathrm{e}}^{2\mathrm{x}}\mathrm{sinx}\\ \mathrm{Let}\mathrm{I}=\int {\mathrm{e}}^{2\mathrm{x}}\mathrm{sinx}\mathrm{dx}\\ \mathrm{Taking}{\text{sinx as first function and e}}^{\text{2x}}\text{as second function, then}\\ \mathrm{Integrating}\text{by parts, we get}\\ \mathrm{I}=\int {\mathrm{e}}^{2\mathrm{x}}\mathrm{sinx}\mathrm{dx}\\ =\mathrm{sinx}\int {\mathrm{e}}^{2\mathrm{x}}\mathrm{dx}-\int (\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{sinx}\int {\mathrm{e}}^{2\mathrm{x}}\mathrm{dx})\mathrm{dx}\\ =\mathrm{sinx}\left(\frac{{\mathrm{e}}^{2\mathrm{x}}}{2}\right)-\int \left\{\mathrm{cosx}\left(\frac{{\mathrm{e}}^{2\mathrm{x}}}{2}\right)\right\}\mathrm{dx}\\ =\frac{{\mathrm{e}}^{2\mathrm{x}}}{2}\mathrm{sinx}-\frac{1}{2}\int \mathrm{cosx}.{\mathrm{e}}^{2\mathrm{x}}\mathrm{dx}\\ =\frac{{\mathrm{e}}^{2\mathrm{x}}}{2}\mathrm{sinx}-\frac{1}{2}\{\mathrm{cosx}\int {\mathrm{e}}^{2\mathrm{x}}\mathrm{dx}-\int (\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{cosx}\int {\mathrm{e}}^{2\mathrm{x}}\mathrm{dx})\mathrm{dx}\}\\ =\frac{{\mathrm{e}}^{2\mathrm{x}}}{2}\mathrm{sinx}-\frac{1}{2}\{\mathrm{cosx}\left(\frac{{\mathrm{e}}^{2\mathrm{x}}}{2}\right)-\int \{-\mathrm{sinx}\left(\frac{{\mathrm{e}}^{2\mathrm{x}}}{2}\right)\}\mathrm{dx}\}\\ =\frac{{\mathrm{e}}^{2\mathrm{x}}}{2}\mathrm{sinx}-\frac{1}{2}\mathrm{cosx}\left(\frac{{\mathrm{e}}^{2\mathrm{x}}}{2}\right)-\frac{1}{4}\int {\mathrm{e}}^{2\mathrm{x}}\mathrm{sinx}\mathrm{dx}+\mathrm{C}\\ \mathrm{I}=\frac{{\mathrm{e}}^{2\mathrm{x}}\mathrm{sinx}}{2}-\frac{1}{4}\mathrm{cosx}{\mathrm{e}}^{2\mathrm{x}}-\frac{1}{4}\mathrm{I}+\mathrm{C}\\ \mathrm{I}+\frac{1}{4}\mathrm{I}=\frac{{\mathrm{e}}^{2\mathrm{x}}\mathrm{sinx}}{2}-\frac{1}{4}\mathrm{cosx}{\mathrm{e}}^{2\mathrm{x}}+\mathrm{C}\\ \frac{5}{4}\mathrm{I}=\frac{{\mathrm{e}}^{2\mathrm{x}}\mathrm{sinx}}{2}-\frac{1}{4}\mathrm{cosx}{\mathrm{e}}^{2\mathrm{x}}+\mathrm{C}\\ \mathrm{I}=\frac{4}{5}\frac{{\mathrm{e}}^{2\mathrm{x}}}{4}(2\mathrm{sinx}-\mathrm{cosx})+\mathrm{C}\end{array}$

Q.81

$\begin{array}{l}\text{Integrate the functions}\\ {\sin }^{–1}\left(\frac{2\mathrm{x}}{1+{\mathrm{x}}^2}\right)\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}{\sin }^{–1}\left(\frac{2\mathrm{x}}{1+{\mathrm{x}}^2}\right)\\ \mathrm{Let}\mathrm{I}=\int {\sin }^{–1}\left(\frac{2\mathrm{x}}{1+{\mathrm{x}}^2}\right)\mathrm{dx}\\ \mathrm{Let}\mathrm{x}=\mathrm{tan\theta }\Rightarrow \frac{\mathrm{dx}}{\mathrm{d\theta }}={\sec }^2\mathrm{\theta }\\ \therefore \mathrm{I}=\int {\sin }^{–1}\left(\frac{2\mathrm{x}}{1+{\mathrm{x}}^2}\right)\mathrm{dx}\\ =\int {\sin }^{–1}\left(\frac{2\mathrm{tan\theta }}{1+{\tan }^2\mathrm{\theta }}\right){\sec }^2\mathrm{\theta }\mathrm{d\theta }\\ =\int {\sin }^{–1}\left(\sin 2\mathrm{\theta }\right){\sec }^2\mathrm{\theta }\mathrm{d\theta }\\ =\int 2\mathrm{\theta }{\sec }^2\mathrm{\theta }\mathrm{d\theta }\\ =2[\mathrm{\theta }\int {\sec }^2\mathrm{\theta }\mathrm{d\theta }-(\int \frac{\mathrm{d}}{\mathrm{d\theta }}\mathrm{\theta }\int {\sec }^2\mathrm{\theta }\mathrm{d\theta })\mathrm{d\theta }]\\ =2[\mathrm{\theta tan\theta }-\int 1\mathrm{tan\theta }\mathrm{d\theta }]\\ =2\mathrm{\theta tan\theta }-2\log \left|\mathrm{sec\theta }\right|+\mathrm{C}\\ =2\mathrm{\theta tan\theta }-2\log \left|\sqrt{1+{\tan }^2\mathrm{\theta }}\right|+\mathrm{C}\\ \mathrm{I}=2{\mathrm{xtan}}^{-1}\mathrm{x}-2\log \left|\sqrt{1+{\mathrm{x}}^2}\right|+\mathrm{C}.\end{array}$

Q.82

$\begin{array}{l}\mathrm{Choose}\mathrm{the}\mathrm{correct}\mathrm{answer}\\ \int {\mathrm{x}}^2{\mathrm{e}}^{{\mathrm{x}}^3}\mathrm{dx}\mathrm{equals}\\ \left(\mathrm{A}\right)\text{\hspace{0.33em}}\frac{1}{3}{\mathrm{e}}^{{\mathrm{x}}^3}+\mathrm{C}\\ \left(\mathrm{B}\right)\frac{1}{3}{\mathrm{e}}^{{\mathrm{x}}^2}+\mathrm{C}\\ \left(\mathrm{C}\right)\frac{1}{2}{\mathrm{e}}^{{\mathrm{x}}^3}+\mathrm{C}\\ \left(\mathrm{D}\right)\frac{1}{2}{\mathrm{e}}^{{\mathrm{x}}^2}+\mathrm{C}\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\mathrm{have}\int {\mathrm{x}}^2{\mathrm{e}}^{{\mathrm{x}}^3}\mathrm{dx}\\ \mathrm{Let}\text{ t}={\text{x}}^{\text{3}}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=3{\mathrm{x}}^2\\ \therefore \int {\mathrm{x}}^2{\mathrm{e}}^{{\mathrm{x}}^3}\mathrm{dx}=\int {\mathrm{x}}^2{\mathrm{e}}^{\mathrm{t}}\frac{\mathrm{dt}}{3{\mathrm{x}}^2}\\ =\frac{1}{3}\int {\mathrm{e}}^{\mathrm{t}}\mathrm{dt}\\ =\frac{1}{3}{\mathrm{e}}^{{\mathrm{x}}^3}+\mathrm{C}\\ \mathrm{Hence},\text{the correct option is A.}\end{array}$

Q.83

$\begin{array}{l}\mathrm{Choose}\mathrm{the}\mathrm{correct}\mathrm{answer}\\ \int {\mathrm{e}}^{\mathrm{x}}\mathrm{secx}\left(1+\mathrm{tanx}\right)\mathrm{dx}\mathrm{equals}\\ \left(\mathrm{A}\right)\text{\hspace{0.33em}}{\mathrm{e}}^{\mathrm{x}}\mathrm{cosx}+\mathrm{C}\\ \left(\mathrm{B}\right){\mathrm{e}}^{\mathrm{x}}\mathrm{secx}+\mathrm{C}\\ \left(\mathrm{C}\right){\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}+\mathrm{C}\\ \left(\mathrm{D}\right){\mathrm{e}}^{\mathrm{x}}\mathrm{tanx}+\mathrm{C}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}=\int {\mathrm{e}}^{\mathrm{x}}\mathrm{secx}(1+\mathrm{tanx})\mathrm{dx}\\ =\int {\mathrm{e}}^{\mathrm{x}}(\mathrm{secx}+\mathrm{secxtanx})\mathrm{dx}\\ \mathrm{Let}\text{ f}\left(\mathrm{x}\right)=\mathrm{secx}\mathrm{and}\text{f’}\left(\mathrm{x}\right)=\mathrm{secxtanx}\\ \mathrm{Thus},{\text{the given integrand is of the form e}}^{\text{x}}[\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}‘\left(\mathrm{x}\right)].\\ \mathrm{Therefore},\\ \mathrm{I}=\int {\mathrm{e}}^{\mathrm{x}}(\mathrm{secx}+\mathrm{secxtanx})\mathrm{dx}\\ ={\mathrm{e}}^{\mathrm{x}}\mathrm{secx}+\mathrm{C}[\because {\text{e}}^{\text{x}}\{\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}‘\left(\mathrm{x}\right)\}={\mathrm{e}}^{\mathrm{x}}\mathrm{f}\left(\mathrm{x}\right)]\\ \mathrm{Hence},\text{the correct option is B.}\end{array}$

Q.84

$\int \sqrt{4-{\mathrm{x}}^2}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\sqrt{4-{\mathrm{x}}^2}\\ \mathrm{Let}\text{I}=\int \sqrt{4-{\mathrm{x}}^2}\mathrm{dx}\\ =\int \sqrt{2^2-{\mathrm{x}}^2}\mathrm{dx}\\ [\because \int \sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}\mathrm{dx}=\frac{1}{2}\mathrm{x}\sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}+\frac{{\mathrm{a}}^2}{2}{\sin }^{-1}\frac{\mathrm{x}}{\mathrm{a}}+\mathrm{C}]\\ =\frac{1}{2}\mathrm{x}\sqrt{2^2-{\mathrm{x}}^2}+\frac{2^2}{2}{\sin }^{-1}\frac{\mathrm{x}}{2}+\mathrm{C}\\ =\frac{\mathrm{x}}{2}\sqrt{4-{\mathrm{x}}^2}+2{\sin }^{-1}\frac{\mathrm{x}}{2}+\mathrm{C}\end{array}$

Q.85

$\int \sqrt{1-4{\mathrm{x}}^2}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\sqrt{1-4{\mathrm{x}}^2}\\ \mathrm{Let}\text{I}=\int \sqrt{1-4{\mathrm{x}}^2}\mathrm{dx}\\ =\int \sqrt{1^2-{\left(2\mathrm{x}\right)}^2}\mathrm{dx}\\ \left[\mathrm{Let}\text{t}=2\mathrm{x}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=2\Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{2}\right]\\ =\int \sqrt{1^2-{\mathrm{t}}^2}\frac{\mathrm{dt}}{2}\\ [\because \int \sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}\mathrm{dx}=\frac{1}{2}\mathrm{x}\sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}+\frac{{\mathrm{a}}^2}{2}{\sin }^{-1}\frac{\mathrm{x}}{\mathrm{a}}+\mathrm{C}]\\ =\frac{1}{2}(\frac{1}{2}\mathrm{t}\sqrt{1^2-{\mathrm{t}}^2}+\frac{{\mathrm{a}}^2}{2}{\sin }^{-1}\frac{\mathrm{t}}{1})+\mathrm{C}\\ =\frac{1}{2}(\frac{2\mathrm{x}}{2}\sqrt{1-4{\mathrm{x}}^2}+\frac{1}{2}{\sin }^{-1}\frac{2\mathrm{x}}{1})+\mathrm{C}\\ =\frac{\mathrm{x}}{2}\sqrt{1-4{\mathrm{x}}^2}+\frac{1}{4}{\sin }^{-1}2\mathrm{x}+\mathrm{C}\end{array}$

Q.86

$\int \sqrt{{\mathrm{x}}^2+4\mathrm{x}+6}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\sqrt{{\mathrm{x}}^2+4\mathrm{x}+6}=\sqrt{{(\mathrm{x}+2)}^2+2}\\ \text{Let I}=\int \sqrt{{(\mathrm{x}+2)}^2+{\left(\sqrt{2}\right)}^2}\mathrm{dx}\\ =\int \sqrt{{\mathrm{t}}^2+{\left(\sqrt{2}\right)}^2}\mathrm{dt}\left[\mathrm{Let}\text{\hspace{0.17em}t}=\mathrm{x}+2\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=1\right]\\ [\because \int \sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}\mathrm{dx}=\frac{\mathrm{x}}{2}\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}+\frac{{\mathrm{a}}^2}{2}\log |\mathrm{x}+\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}|]\\ =\frac{\mathrm{t}}{2}\sqrt{{\mathrm{t}}^2+{\left(\sqrt{2}\right)}^2}+\frac{{\left(\sqrt{2}\right)}^2}{2}\log |\mathrm{t}+\sqrt{{\mathrm{t}}^2+{\left(\sqrt{2}\right)}^2}|+\mathrm{C}\\ \text{Putting t}=\mathrm{x}+2,\text{\hspace{0.17em}we\hspace{0.17em}get}\\ =\frac{(\mathrm{x}+2)}{2}\sqrt{{(\mathrm{x}+2)}^2+{\left(\sqrt{2}\right)}^2}+\frac{2}{2}\log |(\mathrm{x}+2)+\sqrt{{(\mathrm{x}+2)}^2+{\left(\sqrt{2}\right)}^2}|+\mathrm{C}\\ =\frac{(\mathrm{x}+2)}{2}\sqrt{{\mathrm{x}}^2+4\mathrm{x}+6}+\log |(\mathrm{x}+2)+\sqrt{{\mathrm{x}}^2+4\mathrm{x}+6}|+\mathrm{C}\end{array}$

Q.87

$\int \sqrt{{\mathrm{x}}^2+4\mathrm{x}+1}\mathrm{dx}$

Ans.

$\begin{array}{l}\text{We have,}\sqrt{{\mathrm{x}}^2+4\mathrm{x}+1}=\sqrt{{(\mathrm{x}+2)}^2-3}\\ \text{Let I}=\int \sqrt{{(\mathrm{x}+2)}^2-{\left(\sqrt{3}\right)}^2}\mathrm{dx}\\ =\int \sqrt{{\mathrm{t}}^2-{\left(\sqrt{3}\right)}^2}\mathrm{dt}\left[\mathrm{Let}\text{\hspace{0.17em}t}=\mathrm{x}+2\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=1\right]\\ [\because \int \sqrt{{\mathrm{x}}^2-{\mathrm{a}}^2}\mathrm{dx}=\frac{\mathrm{x}}{2}\sqrt{{\mathrm{x}}^2-{\mathrm{a}}^2}-\frac{{\mathrm{a}}^2}{2}\log |\mathrm{x}+\sqrt{{\mathrm{x}}^2-{\mathrm{a}}^2}|]\\ =\frac{\mathrm{t}}{2}\sqrt{{\mathrm{t}}^2-{\left(\sqrt{3}\right)}^2}-\frac{{\left(\sqrt{3}\right)}^2}{2}\log |\mathrm{t}+\sqrt{{\mathrm{t}}^2-{\left(\sqrt{3}\right)}^2}|+\mathrm{C}\\ \text{Putting t}=\mathrm{x}+2,\text{\hspace{0.17em}we\hspace{0.17em}get}\\ =\frac{(\mathrm{x}+2)}{2}\sqrt{{(\mathrm{x}+2)}^2-{\left(\sqrt{3}\right)}^2}-\frac{3}{2}\log |(\mathrm{x}+2)+\sqrt{{(\mathrm{x}+2)}^2-{\left(\sqrt{3}\right)}^2}|+\mathrm{C}\\ =\frac{(\mathrm{x}+2)}{2}\sqrt{{\mathrm{x}}^2+4\mathrm{x}+1}-\frac{3}{2}\log |(\mathrm{x}+2)+\sqrt{{\mathrm{x}}^2+4\mathrm{x}+1}|+\mathrm{C}\end{array}$

Q.88

$\int \sqrt{1-4\mathrm{x}-{\mathrm{x}}^2}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\sqrt{1-4\mathrm{x}-{\mathrm{x}}^2}=\sqrt{5-({\mathrm{x}}^2+4\mathrm{x}+4)}\\ =\sqrt{{\left(\sqrt{5}\right)}^2-{(\mathrm{x}+2)}^2}\\ \mathrm{Let}\text{I}=\int \sqrt{{\left(\sqrt{5}\right)}^2-{(\mathrm{x}+2)}^2}\mathrm{dx}\\ =\int \sqrt{{\left(\sqrt{5}\right)}^2-{\mathrm{t}}^2}\mathrm{dt}\left[\mathrm{Let}\text{\hspace{0.17em}t}=\mathrm{x}+2\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=1\right]\\ [\because \int \sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}\mathrm{dx}=\frac{\mathrm{x}}{2}\sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}+\frac{{\mathrm{a}}^2}{2}{\sin }^{-1}\frac{\mathrm{x}}{\mathrm{a}}]\\ =\frac{\mathrm{t}}{2}\sqrt{{\left(\sqrt{5}\right)}^2-{\mathrm{t}}^2}+\frac{{\left(\sqrt{5}\right)}^2}{2}{\sin }^{-1}\frac{\mathrm{t}}{\left(\sqrt{5}\right)}+\mathrm{C}\\ \mathrm{Putting}\text{t}=\mathrm{x}+2,\mathrm{we}\mathrm{get}\\ =\frac{(\mathrm{x}+2)}{2}\sqrt{{\left(\sqrt{5}\right)}^2-{(\mathrm{x}+2)}^2}+\frac{{\left(\sqrt{5}\right)}^2}{2}{\sin }^{-1}\frac{(\mathrm{x}+2)}{\left(\sqrt{5}\right)}+\mathrm{C}\\ =\frac{(\mathrm{x}+2)}{2}\sqrt{1-4\mathrm{x}-{\mathrm{x}}^2}+\frac{5}{2}{\sin }^{-1}\frac{(\mathrm{x}+2)}{\left(\sqrt{5}\right)}+\mathrm{C}\end{array}$

Q.89

$\int \sqrt{{\mathrm{x}}^2+4\mathrm{x}-5}\mathrm{dx}$

Ans.

$\begin{array}{l}\text{We have,}\sqrt{{\mathrm{x}}^2+4\mathrm{x}-5}=\sqrt{{(\mathrm{x}+2)}^2-9}\\ \text{Let I}=\int \sqrt{{(\mathrm{x}+2)}^2-{\left(3\right)}^2}\mathrm{dx}\\ =\int \sqrt{{\mathrm{t}}^2-{\left(3\right)}^2}\mathrm{dt}\left[\mathrm{Let}\text{\hspace{0.17em}t}=\mathrm{x}+2\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=1\right]\\ [\because \int \sqrt{{\mathrm{x}}^2-{\mathrm{a}}^2}\mathrm{dx}=\frac{\mathrm{x}}{2}\sqrt{{\mathrm{x}}^2-{\mathrm{a}}^2}-\frac{{\mathrm{a}}^2}{2}\log |\mathrm{x}+\sqrt{{\mathrm{x}}^2-{\mathrm{a}}^2}|]\\ =\frac{\mathrm{t}}{2}\sqrt{{\mathrm{t}}^2-{\left(3\right)}^2}-\frac{{\left(3\right)}^2}{2}\log |\mathrm{t}+\sqrt{{\mathrm{t}}^2-{\left(3\right)}^2}|+\mathrm{C}\\ \text{Putting t}=\mathrm{x}+2,\text{\hspace{0.17em}we\hspace{0.17em}get}\\ =\frac{(\mathrm{x}+2)}{2}\sqrt{{(\mathrm{x}+2)}^2-{\left(3\right)}^2}-\frac{9}{2}\log |(\mathrm{x}+2)+\sqrt{{(\mathrm{x}+2)}^2-{\left(3\right)}^2}|+\mathrm{C}\\ =\frac{(\mathrm{x}+2)}{2}\sqrt{{\mathrm{x}}^2+4\mathrm{x}-5}-\frac{9}{2}\log |(\mathrm{x}+2)+\sqrt{{\mathrm{x}}^2+4\mathrm{x}-5}|+\mathrm{C}\end{array}$

Q.90

$\int \sqrt{1+3\mathrm{x}-{\mathrm{x}}^2}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\sqrt{1+3\mathrm{x}-{\mathrm{x}}^2}=\sqrt{1-({\mathrm{x}}^2-3\mathrm{x}+\frac{9}{4}-\frac{9}{4})}\\ =\sqrt{{\left(\frac{\sqrt{13}}{2}\right)}^2-{(\mathrm{x}-\frac{3}{2})}^2}\\ \mathrm{Let}\text{I}=\int \sqrt{{\left(\frac{\sqrt{13}}{2}\right)}^2-{(\mathrm{x}-\frac{3}{2})}^2}\mathrm{dx}\\ =\int \sqrt{{\left(\frac{\sqrt{13}}{2}\right)}^2-{\mathrm{t}}^2}\mathrm{dt}\left[\mathrm{Let}\text{\hspace{0.17em}t}=\mathrm{x}-\frac{3}{2}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=1\right]\\ [\because \int \sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}\mathrm{dx}=\frac{\mathrm{x}}{2}\sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}+\frac{{\mathrm{a}}^2}{2}{\sin }^{-1}\frac{\mathrm{x}}{\mathrm{a}}]\\ =\frac{\mathrm{t}}{2}\sqrt{{\left(\frac{\sqrt{13}}{2}\right)}^2-{\mathrm{t}}^2}+\frac{{\left(\frac{\sqrt{13}}{2}\right)}^2}{2}{\sin }^{-1}\frac{\mathrm{t}}{\left(\frac{\sqrt{13}}{2}\right)}+\mathrm{C}\\ \text{Putting t}=\mathrm{x}-\frac{3}{2},\text{\hspace{0.17em}we\hspace{0.17em}get}\\ =\frac{(\mathrm{x}-\frac{3}{2})}{2}\sqrt{{\left(\frac{\sqrt{13}}{2}\right)}^2-{(\mathrm{x}-\frac{3}{2})}^2}+\frac{{\left(\frac{\sqrt{13}}{2}\right)}^2}{2}{\sin }^{-1}\frac{(\mathrm{x}-\frac{3}{2})}{\left(\frac{\sqrt{13}}{2}\right)}+\mathrm{C}\\ =\frac{(2\mathrm{x}-3)}{4}\sqrt{1+3\mathrm{x}-{\mathrm{x}}^2}+\frac{13}{8}{\sin }^{-1}\frac{(2\mathrm{x}-3)}{\left(\sqrt{13}\right)}+\mathrm{C}\end{array}$

Q.91

$\int \sqrt{{\mathrm{x}}^2+3\mathrm{x}}\mathrm{dx}$

Ans.

$\begin{array}{l}\text{We have,}\sqrt{{\mathrm{x}}^2+3\mathrm{x}}=\sqrt{({\mathrm{x}}^2+3\mathrm{x}+\frac{9}{4})-\frac{9}{4}}\\ =\sqrt{{(\mathrm{x}+\frac{3}{2})}^2-{\left(\frac{3}{2}\right)}^2}\\ \text{Let I}=\int \sqrt{{(\mathrm{x}+\frac{3}{2})}^2-{\left(\frac{3}{2}\right)}^2}\mathrm{dx}\\ =\int \sqrt{{\mathrm{t}}^2-{\left(\frac{3}{2}\right)}^2}\mathrm{dt}\left[\text{Let\hspace{0.17em}t}=\mathrm{x}+\frac{3}{2}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=1\right]\\ [\because \int \sqrt{{\mathrm{x}}^2-{\mathrm{a}}^2}\mathrm{dx}=\frac{\mathrm{x}}{2}\sqrt{{\mathrm{x}}^2-{\mathrm{a}}^2}-\frac{{\mathrm{a}}^2}{2}\log |\mathrm{x}+\sqrt{{\mathrm{x}}^2-{\mathrm{a}}^2}|]\\ =\frac{\mathrm{t}}{2}\sqrt{{\mathrm{t}}^2-{\left(\frac{3}{2}\right)}^2}-\frac{{\left(\frac{3}{2}\right)}^2}{2}\log |\mathrm{t}+\sqrt{{\mathrm{t}}^2-{\left(\frac{3}{2}\right)}^2}|+\mathrm{C}\\ \text{Putting t}=\mathrm{x}+\frac{3}{2},\text{\hspace{0.17em}we\hspace{0.17em}get}\\ =\frac{(\mathrm{x}+\frac{3}{2})}{2}\sqrt{{(\mathrm{x}+\frac{3}{2})}^2-{\left(\frac{3}{2}\right)}^2}-\frac{9}{8}\log |(\mathrm{x}+\frac{3}{2})+\sqrt{{(\mathrm{x}+\frac{3}{2})}^2-{\left(\frac{3}{2}\right)}^2}|+\mathrm{C}\\ =\frac{(2\mathrm{x}+3)}{4}\sqrt{{\mathrm{x}}^2+3\mathrm{x}}-\frac{9}{8}\log |(\mathrm{x}+\frac{3}{2})+\sqrt{{\mathrm{x}}^2+3\mathrm{x}}|+\mathrm{C}\end{array}$

Q.92

$\int \sqrt{1+\frac{{\mathrm{x}}^2}{9}}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{We}\mathrm{have},\sqrt{1+\frac{{\mathrm{x}}^2}{9}}=\frac{1}{3}\sqrt{9+{\mathrm{x}}^2}\\ \mathrm{Let}\int \frac{1}{3}\sqrt{9+{\mathrm{x}}^2}\mathrm{dx}=\frac{1}{3}\int \sqrt{3^2+{\mathrm{x}}^2}\mathrm{dx}\\ =\frac{1}{3}\int \sqrt{{\mathrm{x}}^2+3^2}\mathrm{dx}\\ [\because \int \sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}\mathrm{dx}=\frac{\mathrm{x}}{2}\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}+\frac{{\mathrm{a}}^2}{2}\log |\mathrm{x}+\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}|]\\ =\frac{1}{3}(\frac{\mathrm{x}}{2}\sqrt{{\mathrm{x}}^2+3^2}+\frac{3^2}{2}\log |\mathrm{x}+\sqrt{{\mathrm{x}}^2+3^2}|)+\mathrm{C}\\ =\frac{\mathrm{x}}{6}\sqrt{{\mathrm{x}}^2+9}+\frac{3}{2}\log |\mathrm{x}+\sqrt{{\mathrm{x}}^2+9}|+\mathrm{C}\end{array}$

Q.93

$\begin{array}{l}\begin{array}{l}\mathrm{Choose}\mathrm{the}\mathrm{correct}\mathrm{answer}\\ \int \sqrt{1+{\text{x}}^2}\mathrm{dx}\mathrm{is}\mathrm{equal}\mathrm{to}\end{array}\\ \left(\text{A}\right)\frac{\text{x}}{2}\sqrt{1+{\text{x}}^2}+\frac{1}{2}\log \left|\left(\text{x}+\sqrt{1+{\text{x}}^2}\right)\right|+\text{C}\\ \left(\text{B}\right)\frac{2}{3}{\left(1+{\text{x}}^2\right)}^{\frac{3}{2}}+\text{C}\\ \left(\text{C}\right)\frac{2}{3}\text{x}{\left(1+{\text{x}}^2\right)}^{\frac{3}{2}}+\text{C}\\ \left(\text{D}\right)\frac{{\text{x}}^2}{2}\sqrt{1+{\text{x}}^2}+\frac{1}{2}{\text{\hspace{0.33em}x}}^2\log \left|\left(\text{x}+\sqrt{1+{\text{x}}^2}\right)\right|+\text{C}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}=\int \sqrt{1+{\mathrm{x}}^2}\mathrm{dx}\\ =\int \sqrt{{\mathrm{x}}^2+1^2}\mathrm{dx}\\ [\because \int \sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}\mathrm{dx}=\frac{\mathrm{x}}{2}\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}+\frac{{\mathrm{a}}^2}{2}\log |\mathrm{x}+\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}|]\\ =(\frac{\mathrm{x}}{2}\sqrt{{\mathrm{x}}^2+1^2}+\frac{1^2}{2}\log |\mathrm{x}+\sqrt{{\mathrm{x}}^2+1^2}|)+\mathrm{C}\\ =\frac{\mathrm{x}}{2}\sqrt{{\mathrm{x}}^2+1}+\frac{1}{2}\log |\mathrm{x}+\sqrt{{\mathrm{x}}^2+1}|+\mathrm{C}\\ =\frac{\mathrm{x}}{2}\sqrt{1+{\mathrm{x}}^2}+\frac{1}{2}\log |\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}|+\mathrm{C}\\ \mathrm{Hence},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{the}\text{correct option is A.}\end{array}$

Q.94

$\begin{array}{l}\begin{array}{l}\mathrm{Choose}\mathrm{the}\mathrm{correct}\mathrm{answer}\\ \int \sqrt{{\mathrm{x}}^2-8\mathrm{x}+7}\mathrm{dx}\mathrm{is}\mathrm{equal}\mathrm{to}\end{array}\\ \left(\mathrm{A}\right)\frac{1}{2}\left(\mathrm{x}-4\right)\sqrt{{\mathrm{x}}^2-8\mathrm{x}+7}+9\log \left|\mathrm{x}-4+\sqrt{{\mathrm{x}}^2-8\mathrm{x}+7}\right|+\mathrm{C}\\ \left(\mathrm{B}\right)\frac{1}{4}\left(\mathrm{x}+4\right)\sqrt{{\mathrm{x}}^2-8\mathrm{x}+7}+9\log \left|\mathrm{x}+4+\sqrt{{\mathrm{x}}^2-8\mathrm{x}+7}\right|+\mathrm{C}\\ \left(\mathrm{C}\right)\frac{1}{2}\left(\mathrm{x}-4\right)\sqrt{{\mathrm{x}}^2-8\mathrm{x}+7}-3\sqrt{2}\log \left|\mathrm{x}-4+\sqrt{{\mathrm{x}}^2-8\mathrm{x}+7}\right|+\mathrm{C}\\ \left(\mathrm{D}\right)\frac{1}{2}\left(\mathrm{x}-4\right)\sqrt{{\mathrm{x}}^2-8\mathrm{x}+7}-\frac{9}{2}\log \left|\mathrm{x}-4+\sqrt{{\mathrm{x}}^2-8\mathrm{x}+7}\right|+\mathrm{C}\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}\sqrt{{\mathrm{x}}^2-8\mathrm{x}+7}=\sqrt{{(\mathrm{x}-4)}^2-9}\\ \mathrm{Let}\text{I}=\int \sqrt{{(\mathrm{x}-4)}^2-{\left(3\right)}^2}\mathrm{dx}\\ =\int \sqrt{{\mathrm{t}}^2-{\left(3\right)}^2}\mathrm{dt}\left[\mathrm{Let}\text{\hspace{0.17em}t}=\mathrm{x}-4\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=1\right]\\ [\because \int \sqrt{{\mathrm{x}}^2-{\mathrm{a}}^2}\mathrm{dx}=\frac{\mathrm{x}}{2}\sqrt{{\mathrm{x}}^2-{\mathrm{a}}^2}-\frac{{\mathrm{a}}^2}{2}\log |\mathrm{x}+\sqrt{{\mathrm{x}}^2-{\mathrm{a}}^2}|]\\ =\frac{\mathrm{t}}{2}\sqrt{{\mathrm{t}}^2-{\left(3\right)}^2}-\frac{{\left(3\right)}^2}{2}\log |\mathrm{t}+\sqrt{{\mathrm{t}}^2-{\left(3\right)}^2}|+\mathrm{C}\\ \mathrm{Putting}\text{t}=\mathrm{x}-4,\mathrm{we}\mathrm{get}\\ =\frac{(\mathrm{x}-4)}{2}\sqrt{{(\mathrm{x}-4)}^2-{\left(3\right)}^2}-\frac{9}{2}\log |(\mathrm{x}-4)+\sqrt{{(\mathrm{x}-4)}^2-{\left(3\right)}^2}|+\mathrm{C}\\ =\frac{(\mathrm{x}-4)}{2}\sqrt{{\mathrm{x}}^2-8\mathrm{x}+7}-\frac{9}{2}\log |(\mathrm{x}-4)+\sqrt{{\mathrm{x}}^2-8\mathrm{x}+7}|+\mathrm{C}\\ \mathrm{Hence},\text{the correct option is D.}\end{array}$

Q.95

$\int \mathrm{x}\sqrt{\mathrm{x}+{\mathrm{x}}^2}\mathrm{dx}$

Ans.

$\begin{array}{l}\int \mathrm{x}\sqrt{\mathrm{x}+{\mathrm{x}}^2}\mathrm{dx}\\ \mathrm{Let}\text{x}=\mathrm{A}\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}+{\mathrm{x}}^2)+\mathrm{B}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{A}(1+2\mathrm{x})+\mathrm{B}\\ \Rightarrow 1=2\mathrm{A}\text{\& A}+\text{B}=\text{0}\\ \Rightarrow \mathrm{A}=\frac{1}{2}\text{\&\hspace{0.17em}B}=-\frac{1}{2}\\ \therefore \int \mathrm{x}\sqrt{\mathrm{x}+{\mathrm{x}}^2}\mathrm{dx}=\frac{1}{2}\int (1+\mathrm{x})\sqrt{\mathrm{x}+{\mathrm{x}}^2}\mathrm{dx}-\frac{1}{2}\int \sqrt{\mathrm{x}+{\mathrm{x}}^2}\mathrm{dx}\\ =\frac{1}{2}\times \frac{{(\mathrm{x}+{\mathrm{x}}^2)}^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}-\frac{1}{2}\int \sqrt{{(\mathrm{x}+\frac{1}{2})}^2-{\left(\frac{1}{2}\right)}^2}\mathrm{dx}\\ =\frac{1}{3}\times {(\mathrm{x}+{\mathrm{x}}^2)}^{\frac{3}{2}}-\frac{1}{2}\times \frac{(\mathrm{x}+\frac{1}{2})}{2}\sqrt{{(\mathrm{x}+\frac{1}{2})}^2-{\left(\frac{1}{2}\right)}^2}\\ +\frac{1}{2}\times \frac{{\left(\frac{1}{2}\right)}^2}{2}\log |\mathrm{x}+\sqrt{{(\mathrm{x}+\frac{1}{2})}^2-{\left(\frac{1}{2}\right)}^2}|+\mathrm{C}\\ =\frac{{(\mathrm{x}+{\mathrm{x}}^2)}^{\frac{3}{2}}}{3}-\frac{1}{8}(2\mathrm{x}+1)\sqrt{\mathrm{x}+{\mathrm{x}}^2}+\frac{1}{16}\log |\mathrm{x}+\sqrt{\mathrm{x}+{\mathrm{x}}^2}|+\mathrm{C}\end{array}$

Q.96

$\int \left(\mathrm{x}+1\right)\sqrt{2{\mathrm{x}}^2+3}\mathrm{dx}$

Ans.

$\begin{array}{l}\int (\mathrm{x}+1)\sqrt{2{\mathrm{x}}^2+3}\mathrm{dx}\\ \mathrm{Let}\text{x}+1=\mathrm{A}\frac{\mathrm{d}}{\mathrm{dx}}(2{\mathrm{x}}^2+3)+\mathrm{B}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}+1=\mathrm{A}(4\mathrm{x}+0)+\mathrm{B}\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}+1=4\mathrm{Ax}\text{\hspace{0.17em}\hspace{0.17em} \& B}=\text{1}\\ \Rightarrow \mathrm{A}=\frac{1}{4}\text{\&\hspace{0.17em}B}=1\\ \therefore \int (\mathrm{x}+1)\sqrt{2{\mathrm{x}}^2+3}\mathrm{dx}=\frac{1}{4}\int 4\mathrm{x}\sqrt{2{\mathrm{x}}^2+3}\mathrm{dx}+1\int \sqrt{2{\mathrm{x}}^2+3}\mathrm{dx}\\ =\frac{1}{4}\times \frac{{(2{\mathrm{x}}^2+3)}^{\frac{3}{2}}}{\frac{3}{2}}+\sqrt{2}\int \sqrt{{\mathrm{x}}^2+{\left(\sqrt{\frac{3}{2}}\right)}^2}\mathrm{dx}\\ =\frac{1}{6}\times {(2{\mathrm{x}}^2+3)}^{\frac{3}{2}}+\sqrt{2}\int \sqrt{{\mathrm{x}}^2+{\left(\sqrt{\frac{3}{2}}\right)}^2}\mathrm{dx}\\ =\frac{1}{6}\times {(2{\mathrm{x}}^2+3)}^{\frac{3}{2}}+\sqrt{2}\left(\begin{array}{l}\frac{1}{2}\mathrm{x}\sqrt{{\mathrm{x}}^2+{\left(\sqrt{\frac{3}{2}}\right)}^2}\\ +\frac{{\left(\sqrt{\frac{3}{2}}\right)}^2}{2}\log |\mathrm{x}+{\left(\sqrt{\frac{3}{2}}\right)}^2|\end{array}\right)+\mathrm{C}\\ =\frac{1}{6}\times {(2{\mathrm{x}}^2+3)}^{\frac{3}{2}}+\frac{\sqrt{2}}{2}\mathrm{x}\sqrt{{\mathrm{x}}^2+\frac{3}{2}}\\ +\frac{3\sqrt{2}}{4}\log |\mathrm{x}+\sqrt{{\mathrm{x}}^2+{\left(\sqrt{\frac{3}{2}}\right)}^2}|+\mathrm{C}\\ =\frac{{(2{\mathrm{x}}^2+3)}^{\frac{3}{2}}}{6}-\frac{\mathrm{x}}{2}\sqrt{2{\mathrm{x}}^2+3}-\frac{3\sqrt{2}}{4}\log |\mathrm{x}+\sqrt{{\mathrm{x}}^2+\frac{3}{2}}|+\mathrm{C}\end{array}$

Q.97

$(\mathrm{x}+3)\sqrt{3–4\mathrm{x}–{\mathrm{x}}^2}$

Ans.

$\begin{array}{l}\int (\mathrm{x}+3)\sqrt{3-4\mathrm{x}-{\mathrm{x}}^2}\mathrm{dx}\\ \mathrm{Let}\text{x}+3=\mathrm{A}\frac{\mathrm{d}}{\mathrm{dx}}(3-4\mathrm{x}-{\mathrm{x}}^2)+\mathrm{B}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}+3=\mathrm{A}(-4-2\mathrm{x})+\mathrm{B}\\ \Rightarrow 3=-4\mathrm{A}+\mathrm{B}\text{\hspace{0.17em}\hspace{0.17em} \&}-2\text{A}=\text{1}\\ \Rightarrow \mathrm{A}=\frac{1}{-2}\\ \text{\&\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}B}=3+4\mathrm{A}\\ =3+4\left(\frac{1}{-2}\right)\\ =3-2=1\\ \therefore \int (\mathrm{x}+3)\sqrt{3-4\mathrm{x}-{\mathrm{x}}^2}\mathrm{dx}=\frac{1}{-2}\int (-4-2\mathrm{x})\sqrt{3-4\mathrm{x}-{\mathrm{x}}^2}\mathrm{dx}\\ +1\int \sqrt{3-4\mathrm{x}-{\mathrm{x}}^2}\mathrm{dx}\\ =\frac{1}{-2}\frac{{(3-4\mathrm{x}-{\mathrm{x}}^2)}^{\frac{3}{2}}}{\frac{3}{2}}+\int \sqrt{7-{(\mathrm{x}+2)}^2}\mathrm{dx}\\ =\frac{{(3-4\mathrm{x}-{\mathrm{x}}^2)}^{\frac{3}{2}}}{-3}+\int \sqrt{{\left(\sqrt{7}\right)}^2-{(\mathrm{x}+2)}^2}\mathrm{dx}\\ =\frac{{(3-4\mathrm{x}-{\mathrm{x}}^2)}^{\frac{3}{2}}}{-3}+\frac{1}{2}(\mathrm{x}+2)\sqrt{{\left(\sqrt{7}\right)}^2-{(\mathrm{x}+2)}^2}+\frac{7}{2}{\sin }^{-1}\left(\frac{\mathrm{x}+2}{\sqrt{7}}\right)+\mathrm{C}\\ +\frac{{\left(\sqrt{7}\right)}^2}{2}{\sin }^{-1}\left(\frac{\mathrm{x}+2}{\sqrt{7}}\right)+\mathrm{C}\\ =\frac{{(3-4\mathrm{x}-{\mathrm{x}}^2)}^{\frac{3}{2}}}{-3}+\frac{1}{2}(\mathrm{x}+2)\sqrt{3-4\mathrm{x}-{\mathrm{x}}^2}+\frac{7}{2}{\sin }^{-1}\left(\frac{\mathrm{x}+2}{\sqrt{7}}\right)+\mathrm{C}\end{array}$

Q.98

$$\begin{gathered} \mathrm{Evaluate}\mathrm{the}\mathrm{following}\mathrm{definite}\mathrm{integrals}\mathrm{as}\mathrm{limit}\mathrm{of}\mathrm{sums}. \\ {\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{x}\mathrm{dx} \end{gathered}$$

Ans.

$\begin{array}{l}\text{By definition}\\ {\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=(\mathrm{b}-\mathrm{a})\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[\mathrm{f}\left(\mathrm{a}\right)+\mathrm{f}(\mathrm{a}+\mathrm{h})+...+\mathrm{f}(\mathrm{a}+(\mathrm{n}-1)\mathrm{h})],\\ \text{where, h}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}\\ \text{Here},\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x},\mathrm{h}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}\\ \therefore {\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{x}\mathrm{dx}=(\mathrm{b}-\mathrm{a})\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[\mathrm{f}\left(\mathrm{a}\right)+\mathrm{f}(\mathrm{a}+\mathrm{h})+...+\mathrm{f}(\mathrm{a}+(\mathrm{n}-1)\mathrm{h})]\\ =(\mathrm{b}-\mathrm{a})\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[\mathrm{a}+\mathrm{a}+\mathrm{h}+...+\mathrm{a}+(\mathrm{n}-1)\mathrm{h}]\\ =(\mathrm{b}-\mathrm{a})\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[\mathrm{na}+\mathrm{h}+2\mathrm{h}+...+(\mathrm{n}-1)\mathrm{h}]\\ =(\mathrm{b}-\mathrm{a})\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[\mathrm{na}+\{1+2+...+(\mathrm{n}-1)\}\mathrm{h}]\\ =(\mathrm{b}-\mathrm{a})\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[\mathrm{na}+\frac{(\mathrm{n}-1)\mathrm{n}}{2}\mathrm{h}][\because \sum \mathrm{n}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}]\\ =(\mathrm{b}-\mathrm{a})\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }[\mathrm{a}+\frac{(\mathrm{n}-1)}{2}\left(\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}\right)]\\ =(\mathrm{b}-\mathrm{a})\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }[\mathrm{a}+\frac{(\mathrm{b}-\mathrm{a})}{2}(1-\frac{1}{\mathrm{n}})]\\ =(\mathrm{b}-\mathrm{a})[\mathrm{a}+\frac{(\mathrm{b}-\mathrm{a})}{2}(1-0)]\\ =(\mathrm{b}-\mathrm{a})\left(\frac{2\mathrm{a}+\mathrm{b}-\mathrm{a}}{2}\right)\\ =(\mathrm{b}-\mathrm{a})\frac{(\mathrm{b}+\mathrm{a})}{2}=\frac{{\mathrm{b}}^2-{\mathrm{a}}^2}{2}\end{array}$

Q.99

$$\begin{gathered} \mathrm{Evaluate}\mathrm{the}\mathrm{following}\mathrm{definite}\mathrm{integrals}\mathrm{as}\mathrm{limit}\mathrm{of}\mathrm{sums}. \\ {\int }_0^5(\mathrm{x}+1)\mathrm{dx} \end{gathered}$$

Ans.

$\begin{array}{l}\mathrm{We}\mathrm{have},{\int }_0^5(\mathrm{x}+1)\mathrm{dx}\\ \mathrm{By}\text{definition}\\ {\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=(\mathrm{b}-\mathrm{a})\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[\mathrm{f}\left(\mathrm{a}\right)+\mathrm{f}(\mathrm{a}+\mathrm{h})+...+\mathrm{f}(\mathrm{a}+(\mathrm{n}-1)\mathrm{h})],\\ \mathrm{where},\text{h}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}\\ \mathrm{Here},\mathrm{a}=0,\mathrm{b}=5,\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}+1,\mathrm{h}=\frac{5-0}{\mathrm{n}}=\frac{5}{\mathrm{n}}\\ \therefore {\int }_0^5(\mathrm{x}+1)\mathrm{dx}=(\mathrm{b}-\mathrm{a})\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[\mathrm{f}\left(0\right)+\mathrm{f}(0+\mathrm{h})+...+\mathrm{f}(0+(\mathrm{n}-1)\mathrm{h})]\\ =5\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[(0+1)+(\mathrm{h}+1)+...+\{(\mathrm{n}-1)\mathrm{h}+1\}]\\ =5\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[\mathrm{n}+\mathrm{h}+2\mathrm{h}+...+(\mathrm{n}-1)\mathrm{h}]\\ =5\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[\mathrm{n}+\{1+2+...+(\mathrm{n}-1)\}\mathrm{h}]\\ =5\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[\mathrm{n}+\frac{(\mathrm{n}-1)\mathrm{n}}{2}\mathrm{h}][\because \sum \mathrm{n}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}]\\ =5\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }[1+\frac{(\mathrm{n}-1)}{2}\left(\frac{5}{\mathrm{n}}\right)]\\ =5\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }[1+\frac{5}{2}(1-\frac{1}{\mathrm{n}})]\\ =5[1+\frac{5}{2}(1-0)]\\ =5\times \frac{7}{2}=\frac{35}{2}\end{array}$

Q.100

$$\begin{gathered} \mathrm{Evaluate}\mathrm{the}\mathrm{following}\mathrm{definite}\mathrm{integrals}\mathrm{as}\mathrm{limit}\mathrm{of}\mathrm{sums}. \\ {\int }_2^3{\mathrm{x}}^2\mathrm{dx} \end{gathered}$$

Ans.

$\begin{array}{l}\text{We have,}{\int }_2^3{\mathrm{x}}^2\mathrm{dx}\\ {\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=(\mathrm{b}-\mathrm{a})\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[\mathrm{f}\left(\mathrm{a}\right)+\mathrm{f}(\mathrm{a}+\mathrm{h})+...+\mathrm{f}(\mathrm{a}+(\mathrm{n}-1)\mathrm{h})],\\ \text{where},\text{h}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}\\ \text{Here},\mathrm{a}=2,\mathrm{b}=3,\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2,\mathrm{h}=\frac{3-2}{\mathrm{n}}=\frac{1}{\mathrm{n}}\\ \therefore {\int }_2^3{\mathrm{x}}^2\mathrm{dx}\\ =(3-2)\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[\mathrm{f}\left(2\right)+\mathrm{f}(2+\mathrm{h})+...+\mathrm{f}(2+(\mathrm{n}-1)\mathrm{h})]\\ =\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[{\left(2\right)}^2+{(2+\mathrm{h})}^2+...+{\{2+(\mathrm{n}-1)\mathrm{h}\}}^2]\\ =\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[4+(4+4\mathrm{h}+{\mathrm{h}}^2)+...+\{4+4(\mathrm{n}-1)\mathrm{h}+{(\mathrm{n}-1)}^2{\mathrm{h}}^2\}]\\ =\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[4\mathrm{n}+4\{1+2+...+(\mathrm{n}-1)\}\mathrm{h}+\{1+2^2+3^2+...+{(\mathrm{n}-1)}^2\}{\mathrm{h}}^2]\\ =\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[4\mathrm{n}+4\frac{(\mathrm{n}-1)\mathrm{n}}{2}\mathrm{h}+\frac{(\mathrm{n}-1)\mathrm{n}(2\mathrm{n}-2+1)}{6}{\mathrm{h}}^2]\\ =\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[4\mathrm{n}+4\frac{(\mathrm{n}-1)\mathrm{n}}{2}.\frac{1}{\mathrm{n}}+\frac{(\mathrm{n}-1)\mathrm{n}(2\mathrm{n}-2+1)}{6}.\frac{1}{{\mathrm{n}}^2}]\\ [\because \sum \mathrm{n}=\frac{\mathrm{n}(\mathrm{n}+1)}{2},\sum {\mathrm{n}}^2=\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}]\\ =\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }[4+4\frac{(1-\frac{1}{\mathrm{n}})}{2}+\frac{(1-\frac{1}{\mathrm{n}})(2-\frac{1}{\mathrm{n}})}{6}]\\ =[4+4\frac{(1-0)}{2}+\frac{(1-0)(2-0)}{6}]\\ =4+2+\frac{1}{3}\\ =\frac{19}{3}\end{array}$

Q.101

$$\begin{gathered} \mathrm{Evaluate}\mathrm{the}\mathrm{following}\mathrm{definite}\mathrm{integrals}\mathrm{as}\mathrm{limit}\mathrm{of}\mathrm{sums}. \\ {\int }_1^4({\mathrm{x}}^2-\mathrm{x})\mathrm{dx} \end{gathered}$$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}{\int }_1^4({\mathrm{x}}^2-\mathrm{x})\mathrm{dx}={\int }_1^4{\mathrm{x}}^2\mathrm{dx}-{\int }_1^4\mathrm{x}\mathrm{dx}\\ ={\mathrm{I}}_1-{\mathrm{I}}_2\left(\mathrm{Let}\right)\\ {\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=(\mathrm{b}-\mathrm{a})\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[\mathrm{f}\left(\mathrm{a}\right)+\mathrm{f}(\mathrm{a}+\mathrm{h})+...+\mathrm{f}(\mathrm{a}+(\mathrm{n}-1)\mathrm{h})]\\ \mathrm{where},\text{h}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}\\ \mathrm{Here},\mathrm{a}=1,\mathrm{b}=4,\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2,\mathrm{h}=\frac{4-1}{\mathrm{n}}=\frac{3}{\mathrm{n}}\\ \therefore {\mathrm{I}}_1={\int }_1^4{\mathrm{x}}^2\mathrm{dx}\\ =(4-1)\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[\mathrm{f}\left(1\right)+\mathrm{f}(1+\mathrm{h})+...+\mathrm{f}(1+(\mathrm{n}-1)\mathrm{h})]\\ =3\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[{\left(1\right)}^2+{(1+\mathrm{h})}^2+...+{\{1+(\mathrm{n}-1)\mathrm{h}\}}^2]\\ =3\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[1+(1+2\mathrm{h}+{\mathrm{h}}^2)+...+\{1+2(\mathrm{n}-1)\mathrm{h}+{(\mathrm{n}-1)}^2{\mathrm{h}}^2\}]\\ =3\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[\mathrm{n}+\{1+2+...+(\mathrm{n}-1)\}2\mathrm{h}+\{1+2^2+3^2+...+{(\mathrm{n}-1)}^2\}{\mathrm{h}}^2]\\ =3\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[\mathrm{n}+\frac{(\mathrm{n}-1)\mathrm{n}}{2}2\mathrm{h}+\frac{(\mathrm{n}-1)\mathrm{n}(2\mathrm{n}-2+1)}{6}{\mathrm{h}}^2]\\ =3\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[\mathrm{n}+(\mathrm{n}-1)\mathrm{n}.\frac{3}{\mathrm{n}}+\frac{(\mathrm{n}-1)\mathrm{n}(2\mathrm{n}-2+1)}{6}.\frac{9}{{\mathrm{n}}^2}]\\ \left[\because \sum \text{n}=\frac{\text{n}\left(\text{n}+1\right)}{2},\sum {\text{n}}^2=\frac{\text{n}\left(\text{n}+1\right)\left(2\text{n}+1\right)}{6}\right]\\ =3\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }[1+(1-\frac{1}{\mathrm{n}})\times 3+\frac{(1-\frac{1}{\mathrm{n}})(2-\frac{1}{\mathrm{n}})}{6}\times 9]\end{array}$

$\begin{array}{l}=3[1+3(1-0)+\frac{3(1-0)(2-0)}{2}]\\ =3(1+3+3)\\ =21\\ \mathrm{For}{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}}_{\text{2}}={\int }_1^4\mathrm{x}\mathrm{dx}\\ \therefore {\int }_1^4\mathrm{x}\mathrm{dx}=(4-1)\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[\mathrm{f}\left(1\right)+\mathrm{f}(1+\mathrm{h})+...+\mathrm{f}(1+(\mathrm{n}-1)\mathrm{h})]\\ =3\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[1+1+\mathrm{h}+...+1+(\mathrm{n}-1)\mathrm{h}]\\ =3\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[\mathrm{n}+\mathrm{h}+2\mathrm{h}+...+(\mathrm{n}-1)\mathrm{h}]\\ =3\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[\mathrm{n}+\{1+2+...+(\mathrm{n}-1)\}\mathrm{h}]\\ =3\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[\mathrm{n}+\frac{(\mathrm{n}-1)\mathrm{n}}{2}\mathrm{h}][\because \sum \mathrm{n}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}]\\ =3\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }[1+\frac{(\mathrm{n}-1)}{2}\left(\frac{3}{\mathrm{n}}\right)]\\ =3\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }[1+\frac{3}{2}(1-\frac{1}{\mathrm{n}})]\\ =3[1+\frac{3}{2}(1-0)]\\ =3\left(\frac{5}{2}\right)=\frac{15}{2}\\ \mathrm{So},{\int }_1^4({\mathrm{x}}^2-\mathrm{x})\mathrm{dx}\\ ={\mathrm{I}}_1-{\mathrm{I}}_2=21-\frac{15}{2}=\frac{27}{2}\end{array}$

Q.102

$$\begin{gathered} \mathrm{Evaluate}\mathrm{the}\mathrm{following}\mathrm{definite}\mathrm{integrals}\mathrm{as}\mathrm{limit}\mathrm{of}\mathrm{sums}. \\ {\int }_{–1}^1{\mathrm{e}}^{\mathrm{x}}\mathrm{dx} \end{gathered}$$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em} have,}{\int }_{–1}^1{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}\\ \because {\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=(\mathrm{b}-\mathrm{a})\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[\mathrm{f}\left(\mathrm{a}\right)+\mathrm{f}(\mathrm{a}+\mathrm{h})+...+\mathrm{f}(\mathrm{a}+(\mathrm{n}-1)\mathrm{h})],\\ \mathrm{where},\text{h}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}\\ \mathrm{Here},\mathrm{a}=-1,\mathrm{b}=1,\mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{x}},\mathrm{h}=\frac{1-(-1)}{\mathrm{n}}=\frac{2}{\mathrm{n}}\\ \therefore {\int }_{–1}^1{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}=(1+1)\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[\mathrm{f}(-1)+\mathrm{f}(-1+\mathrm{h})+...+\mathrm{f}(-1+(\mathrm{n}-1)\mathrm{h})]\\ =2\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[{\mathrm{e}}^{-1}+{\mathrm{e}}^{-1+\mathrm{h}}+...+{\mathrm{e}}^{-1+(\mathrm{n}-1)\mathrm{h}}]\\ \text{Using the sum to n terms of a G.P., where a}={\mathrm{e}}^{-1},\text{r}={\mathrm{e}}^{\mathrm{h}},\\ \mathrm{we}\text{have}\\ \therefore {\int }_{–1}^1{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}=2\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}\left[\frac{{\mathrm{e}}^{-1}({\mathrm{e}}^{\mathrm{nh}}-1)}{({\mathrm{e}}^{\mathrm{h}}-1)}\right]\\ =2\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}\left[\frac{{\mathrm{e}}^{-1}({\mathrm{e}}^{\mathrm{n}.\frac{2}{\mathrm{n}}}-1)}{({\mathrm{e}}^{\frac{2}{\mathrm{n}}}-1)}\right]\\ =2\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}\left[\frac{{\mathrm{e}}^{-1}({\mathrm{e}}^2-1)}{({\mathrm{e}}^{\frac{2}{\mathrm{n}}}-1)}\right]\\ =\frac{2{\mathrm{e}}^{-1}({\mathrm{e}}^2-1)}{\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\left[\frac{{\mathrm{e}}^{\frac{2}{\mathrm{n}}}-1}{\frac{2}{\mathrm{n}}}\right].2}\\ =\frac{2{\mathrm{e}}^{-1}({\mathrm{e}}^2-1)}{2}\\ =\mathrm{e}-\frac{1}{\mathrm{e}}\left[\because \underset{\text{x}\to 0}{\lim }\frac{{\text{e}}^{\text{h}}-1}{\text{h}}=1\right]\end{array}$

Q.103

$$\begin{gathered} \mathrm{Evaluate}\mathrm{the}\mathrm{following}\mathrm{definite}\mathrm{integrals}\mathrm{as}\mathrm{limit}\mathrm{of}\mathrm{sums}. \\ {\int }_0^4(\mathrm{x}+{\mathrm{e}}^{2\mathrm{x}})\mathrm{dx} \end{gathered}$$

Ans.

$\begin{array}{l}\mathrm{We}\text{have,}{\int }_0^4(\mathrm{x}+{\mathrm{e}}^{2\mathrm{x}})\mathrm{dx}={\int }_0^4\mathrm{x}\mathrm{dx}+{\int }_0^4{\mathrm{e}}^{2\mathrm{x}}\mathrm{dx}\\ ={\mathrm{I}}_1+{\mathrm{I}}_2\left(\mathrm{Let}\right)\\ {\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=(\mathrm{b}-\mathrm{a})\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[\mathrm{f}\left(\mathrm{a}\right)+\mathrm{f}(\mathrm{a}+\mathrm{h})+...+\mathrm{f}(\mathrm{a}+(\mathrm{n}-1)\mathrm{h})]\\ \mathrm{where},\text{h}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}\\ \mathrm{Here},\mathrm{a}=0,\mathrm{b}=4,\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x},\mathrm{h}=\frac{4-0}{\mathrm{n}}=\frac{4}{\mathrm{n}}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}}_{\text{1}}={\int }_0^4\mathrm{x}\mathrm{dx}\\ \therefore {\int }_0^4\mathrm{x}\mathrm{dx}=(4-0)\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[\mathrm{f}\left(0\right)+\mathrm{f}(0+\mathrm{h})+...+\mathrm{f}(0+(\mathrm{n}-1)\mathrm{h})]\\ =4\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[0+\mathrm{h}+...+(\mathrm{n}-1)\mathrm{h}]\\ =4\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[\mathrm{h}+2\mathrm{h}+...+(\mathrm{n}-1)\mathrm{h}]\\ =4\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}\left[\{1+2+...+(\mathrm{n}-1)\}\mathrm{h}\right]\\ =4\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}\left[\frac{(\mathrm{n}-1)\mathrm{n}}{2}\mathrm{h}\right][\because \sum \mathrm{n}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}]\\ =4\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\left[\frac{(\mathrm{n}-1)}{2}\left(\frac{4}{\mathrm{n}}\right)\right]\\ =4\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\left[\frac{4}{2}(1-\frac{1}{\mathrm{n}})\right]\\ =4\left[2(1-0)\right]\\ =4\left(2\right)=8\\ {\mathrm{I}}_2={\int }_0^4{\mathrm{e}}^{2\mathrm{x}}\mathrm{dx}\\ =(4-0)\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[\mathrm{f}\left(0\right)+\mathrm{f}(0+\mathrm{h})+...+\mathrm{f}(0+(\mathrm{n}-1)\mathrm{h})]\\ =4\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[{\mathrm{e}}^0+{\mathrm{e}}^{2\mathrm{h}}+...+{\mathrm{e}}^{2(\mathrm{n}-1)\mathrm{h}}]\\ =4\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[1+{\mathrm{e}}^{2\mathrm{h}}+...+{\mathrm{e}}^{2(\mathrm{n}-1)\mathrm{h}}]\\ \text{Using the sum to n terms of a G.P., where a}=1,\text{r}={\mathrm{e}}^{2\mathrm{h}},\\ \mathrm{we}\text{have}\end{array}$

$\begin{array}{l}\therefore {\int }_0^4{\mathrm{e}}^{2\mathrm{x}}\mathrm{dx}=4\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}\left[\frac{1.({\mathrm{e}}^{2\mathrm{nh}}-1)}{({\mathrm{e}}^{2\mathrm{h}}-1)}\right]\\ =4\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}\left[\frac{({\mathrm{e}}^{2\mathrm{n}.\frac{4}{\mathrm{n}}}-1)}{({\mathrm{e}}^{\frac{8}{\mathrm{n}}}-1)}\right]\\ =4\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}\left[\frac{({\mathrm{e}}^8-1)}{({\mathrm{e}}^{\frac{8}{\mathrm{n}}}-1)}\right]\\ =\frac{4({\mathrm{e}}^8-1)}{\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\left[\frac{{\mathrm{e}}^{\frac{8}{\mathrm{n}}}-1}{\frac{8}{\mathrm{n}}}\right].8}\\ =\frac{4({\mathrm{e}}^8-1)}{8}[\because \underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{h}}-1}{\mathrm{h}}=1]\\ =\frac{{\mathrm{e}}^8-1}{2}\\ \mathrm{So},{\int }_0^4(\mathrm{x}+{\mathrm{e}}^{2\mathrm{x}})\mathrm{dx}\\ ={\mathrm{I}}_1+{\mathrm{I}}_2\\ =8+\frac{{\mathrm{e}}^8-1}{2}\\ =\frac{15+{\mathrm{e}}^8}{2}\end{array}$

Q.104

${\int }_{–1}^1(\mathrm{x}+1)\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}={\int }_{-1}^1(\mathrm{x}+1)\mathrm{dx}\\ \because \int (\mathrm{x}+1)\mathrm{dx}=\frac{{\mathrm{x}}^2}{2}+\mathrm{x}=\mathrm{F}\left(\mathrm{x}\right)\\ \mathrm{Therefore},\text{by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(1\right)-\mathrm{F}(-1)\\ =(\frac{1^2}{2}+1)-\{\frac{{(-1)}^2}{2}+(-1)\}\\ =\frac{3}{2}-(-\frac{1}{2})\\ =2\end{array}$

Q.105

${\int }_2^3\frac{1}{\mathrm{x}}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}={\int }_2^3\frac{1}{\mathrm{x}}\mathrm{dx}\\ \because \int \frac{1}{\mathrm{x}}\mathrm{dx}=\mathrm{logx}=\mathrm{F}\left(\mathrm{x}\right)\\ \mathrm{Therefore},\text{by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(3\right)-\mathrm{F}\left(2\right)\\ =\log 3-\log 2=\log \frac{3}{2}\end{array}$

Q.106

${\int }_1^2(4{\mathrm{x}}^3-5{\mathrm{x}}^2+6\mathrm{x}+9)\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}={\int }_1^2(4{\mathrm{x}}^3-5{\mathrm{x}}^2+6\mathrm{x}+9)\mathrm{dx}\\ \because \int (4{\mathrm{x}}^3-5{\mathrm{x}}^2+6\mathrm{x}+9)\mathrm{dx}\\ =\int 4{\mathrm{x}}^3\mathrm{dx}-\int 5{\mathrm{x}}^2\mathrm{dx}+\int 6\mathrm{x}\mathrm{dx}+\int 9\mathrm{dx}\\ ={\mathrm{x}}^4-\frac{5}{3}{\mathrm{x}}^3+3{\mathrm{x}}^2+9\mathrm{x}=\mathrm{F}\left(\mathrm{x}\right)\\ \mathrm{Therefore},\text{by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(2\right)-\mathrm{F}\left(1\right)\\ =\{2^4-\frac{5}{3}{\left(2\right)}^3+3{\left(2\right)}^2+9\left(2\right)\}-\{{\left(1\right)}^4-\frac{5}{3}{\left(1\right)}^3+3{\left(1\right)}^2+9\left(1\right)\}\\ =(16-\frac{40}{3}+12+18)-(1-\frac{5}{3}+3+9)\\ =(46-\frac{40}{3})-(13-\frac{5}{3})\\ =46-\frac{40}{3}-13+\frac{5}{3}\\ =33-\frac{35}{3}\\ =\frac{64}{3}\end{array}$

Q.107

${\int }_0^{\frac{\mathrm{\pi }}{4}}\sin 2\mathrm{x}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{4}}\sin 2\mathrm{x}\mathrm{dx}\\ \because \int \sin 2\mathrm{x}\mathrm{dx}=-\frac{1}{2}\cos 2\mathrm{x}=\mathrm{F}\left(\mathrm{x}\right)\\ \mathrm{Therefore},\text{by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(\frac{\mathrm{\pi }}{4}\right)-\mathrm{F}\left(0\right)\\ =-\frac{1}{2}\cos 2\left(\frac{\mathrm{\pi }}{4}\right)-(-\frac{1}{2}\cos 0)\\ =-\frac{1}{2}\cos \frac{\mathrm{\pi }}{2}+\frac{1}{2}\cos 0\\ =0+\frac{1}{2}\left(1\right)=\frac{1}{2}\end{array}$

Q.108

${\int }_0^{\frac{\mathrm{\pi }}{2}}\cos 2\mathrm{x}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}\cos 2\mathrm{x}\mathrm{dx}\\ \because \int \cos 2\mathrm{x}\mathrm{dx}=\frac{1}{2}\sin 2\mathrm{x}=\mathrm{F}\left(\mathrm{x}\right)\\ \mathrm{Therefore},\text{by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(\frac{\mathrm{\pi }}{2}\right)-\mathrm{F}\left(0\right)\\ =-\frac{1}{2}\sin 2\left(\frac{\mathrm{\pi }}{2}\right)-(-\frac{1}{2}\sin 0)\\ =-\frac{1}{2}\mathrm{sin\pi }+\frac{1}{2}\sin 0\\ =0+0=0\end{array}$

Q.109

${\int }_4^5{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}={\int }_4^5{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}\\ \because \int {\mathrm{e}}^{\mathrm{x}}\mathrm{dx}={\mathrm{e}}^{\mathrm{x}}=\mathrm{F}\left(\mathrm{x}\right)\\ \mathrm{Therefore},\text{by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(5\right)-\mathrm{F}\left(4\right)\\ ={\mathrm{e}}^5-{\mathrm{e}}^4\\ ={\mathrm{e}}^4(\mathrm{e}-1)\end{array}$

Q.110

${\int }_0^{\frac{\mathrm{\pi }}{4}}\tan \mathrm{x}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{4}}\tan \mathrm{x}\mathrm{dx}\\ \because \int \tan \mathrm{x}\mathrm{dx}=-\log \left|\mathrm{cosx}\right|=\mathrm{F}\left(\mathrm{x}\right)\\ \mathrm{Therefore},\text{by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(\frac{\mathrm{\pi }}{4}\right)-\mathrm{F}\left(0\right)\\ =-\log \left|\cos \frac{\mathrm{\pi }}{4}\right|-(-\log \left|\cos 0\right|)\\ =-\log \left|\cos \frac{\mathrm{\pi }}{4}\right|+\log \left|\cos 0\right|\\ =-\log \left|\frac{1}{\sqrt{2}}\right|+\log 1\\ =\log \sqrt{2}+0=\frac{1}{2}\log 2\end{array}$

Q.111

${\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{4}}\mathrm{cosec}\mathrm{x}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{4}}\mathrm{cosec}\mathrm{x}\mathrm{dx}\\ \because \int \mathrm{cosec}\mathrm{x}\mathrm{dx}=\log |\mathrm{cosecx}-\mathrm{cotx}|=\mathrm{F}\left(\mathrm{x}\right)\\ \mathrm{Therefore},\text{by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(\frac{\mathrm{\pi }}{4}\right)-\mathrm{F}\left(\frac{\mathrm{\pi }}{6}\right)\\ =\log |\mathrm{cosec}\frac{\mathrm{\pi }}{4}-\cot \frac{\mathrm{\pi }}{4}|-\log |\mathrm{cosec}\frac{\mathrm{\pi }}{6}-\cot \frac{\mathrm{\pi }}{6}|\\ =\log |\sqrt{2}-1|-\log |2-\sqrt{3}|\\ =\log \left(\frac{\sqrt{2}-1}{2-\sqrt{3}}\right)\end{array}$

Q.112

${\int }_0^1\frac{1}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}={\int }_0^1\frac{1}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}\\ \because \int \frac{1}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}={\sin }^{-1}\mathrm{x}=\mathrm{F}\left(\mathrm{x}\right)\\ \mathrm{Therefore},\text{by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(1\right)-\mathrm{F}\left(0\right)\\ ={\sin }^{-1}\left(1\right)-{\sin }^{-1}\left(0\right)\\ =\frac{\mathrm{\pi }}{2}-0=\frac{\mathrm{\pi }}{2}\end{array}$

Q.113

${\int }_0^1\frac{1}{(1+{\mathrm{x}}^2)}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}={\int }_0^1\frac{1}{1+{\mathrm{x}}^2}\mathrm{dx}\\ \because \int \frac{1}{1+{\mathrm{x}}^2}\mathrm{dx}={\tan }^{-1}\mathrm{x}=\mathrm{F}\left(\mathrm{x}\right)\\ \mathrm{Therefore},\text{by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(1\right)-\mathrm{F}\left(0\right)\\ ={\tan }^{-1}\left(1\right)-{\tan }^{-1}\left(0\right)\\ =\frac{\mathrm{\pi }}{4}-0=\frac{\mathrm{\pi }}{4}\end{array}$

Q.114

${\int }_2^3\frac{\mathrm{dx}}{({\mathrm{x}}^2-1)}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}={\int }_2^3\frac{\mathrm{dx}}{({\mathrm{x}}^2-1)}\\ \because \int \frac{1}{{\mathrm{x}}^2-1}\mathrm{dx}=\frac{1}{2}\log \left|\frac{\mathrm{x}-1}{\mathrm{x}+1}\right|=\mathrm{F}\left(\mathrm{x}\right)\\ \mathrm{Therefore},\text{by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(3\right)-\mathrm{F}\left(2\right)\\ =\frac{1}{2}\log \left|\frac{3-1}{3+1}\right|-\frac{1}{2}\log \left|\frac{2-1}{2+1}\right|\\ =\frac{1}{2}\log \left|\frac{2}{4}\right|-\frac{1}{2}\log \left|\frac{1}{3}\right|\\ =\frac{1}{2}\log |\frac{2}{4}\times \frac{3}{1}|\\ =\frac{1}{2}\log \left|\frac{3}{2}\right|\end{array}$

Q.115

${\int }_0^{\frac{\mathrm{\pi }}{2}}{\cos }^2\mathrm{x}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}{\cos }^2\mathrm{x}\mathrm{dx}={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1+\cos 2\mathrm{x}}{2}\mathrm{dx}\\ \because \int \frac{1+\cos 2\mathrm{x}}{2}\mathrm{dx}=\frac{1}{2}(\mathrm{x}+\frac{\sin 2\mathrm{x}}{2})=\mathrm{F}\left(\mathrm{x}\right)\\ \mathrm{Therefore},\text{by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(\frac{\mathrm{\pi }}{2}\right)-\mathrm{F}\left(0\right)\\ =\frac{1}{2}\{\frac{\mathrm{\pi }}{2}+\frac{\sin 2.\left(\frac{\mathrm{\pi }}{2}\right)}{2}\}-\frac{1}{2}\{0+\frac{\sin 2\left(0\right)}{2}\}\\ =\frac{1}{2}(\frac{\mathrm{\pi }}{2}+\frac{\mathrm{sin\pi }}{2})-\frac{1}{2}\{0+0\}\\ =\frac{1}{2}\left(\frac{\mathrm{\pi }}{2}\right)=\frac{\mathrm{\pi }}{4}\end{array}$

Q.116

${\int }_2^3\frac{\mathrm{x}\mathrm{dx}}{({\mathrm{x}}^2+1)}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}={\int }_2^3\frac{\mathrm{x}\mathrm{dx}}{({\mathrm{x}}^2+1)}\\ \because \int \frac{\mathrm{x}}{({\mathrm{x}}^2+1)}\mathrm{dx}=\frac{1}{2}\log |{\mathrm{x}}^2+1|=\mathrm{F}\left(\mathrm{x}\right)\\ \mathrm{Therefore},\text{by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(3\right)-\mathrm{F}\left(2\right)\\ =\frac{1}{2}\log |3^2+1|-\frac{1}{2}\log |2^2+1|\\ =\frac{1}{2}\log \left|10\right|-\frac{1}{2}\log \left|5\right|\\ =\frac{1}{2}\log \left|\frac{10}{5}\right|\\ =\frac{1}{2}\log \left|2\right|\end{array}$

Q.117

${\int }_0^1\frac{2\mathrm{x}+3}{(5{\mathrm{x}}^2+1)}\mathrm{dx}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_0^1\frac{2\mathrm{x}+3}{(5{\mathrm{x}}^2+1)}\mathrm{dx}\\ =\frac{1}{5}{\int }_0^1\frac{10\mathrm{x}}{(5{\mathrm{x}}^2+1)}\mathrm{dx}+{\int }_0^1\frac{3}{(5{\mathrm{x}}^2+1)}\mathrm{dx}\\ \int \frac{2\mathrm{x}+3}{(5{\mathrm{x}}^2+1)}\mathrm{dx}=\frac{1}{5}\int \frac{10\mathrm{x}}{(5{\mathrm{x}}^2+1)}\mathrm{dx}+3\int \frac{1}{(5{\mathrm{x}}^2+1)}\mathrm{dx}\\ =\frac{1}{5}\log |5{\mathrm{x}}^2+1|+3\int \frac{1}{{\left(\sqrt{5}\mathrm{x}\right)}^2+1}\mathrm{dx}\\ =\frac{1}{5}\log |5{\mathrm{x}}^2+1|+3{\tan }^{-1}\sqrt{5}\mathrm{x}.\frac{1}{\sqrt{5}}\\ =\frac{1}{5}\log |5{\mathrm{x}}^2+1|+\frac{3}{\sqrt{5}}{\tan }^{-1}\left(\sqrt{5}\mathrm{x}\right)=\mathrm{F}\left(\mathrm{x}\right)\\ \text{Therefore, by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(1\right)-\mathrm{F}\left(0\right)\\ =\frac{1}{5}\log |5{\left(1\right)}^2+1|+\frac{3}{\sqrt{5}}{\tan }^{-1}\left\{\sqrt{5}\left(1\right)\right\}-\left[\begin{array}{l}\frac{1}{5}\log |5{\left(0\right)}^2+1|\\ +\frac{3}{\sqrt{5}}{\tan }^{-1}\left\{\sqrt{5}\left(0\right)\right\}\end{array}\right]\\ =\frac{1}{5}\log \left(6\right)+\frac{3}{\sqrt{5}}{\tan }^{-1}\left(\sqrt{5}\right)-\frac{1}{5}\log \left|1\right|-\frac{3}{\sqrt{5}}{\tan }^{-1}\left(0\right)\\ =\frac{1}{5}\log \left(6\right)+\frac{3}{\sqrt{5}}{\tan }^{-1}\left(\sqrt{5}\right)\end{array}$

Q.118

${\int }_0^1\mathrm{x}{\mathrm{e}}^{{\mathrm{x}}^2}\mathrm{dx}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_0^1\mathrm{x}{\mathrm{e}}^{{\mathrm{x}}^2}\mathrm{dx}\\ \because \int \mathrm{x}{\mathrm{e}}^{{\mathrm{x}}^2}\mathrm{dx}=\int \mathrm{x}{\mathrm{e}}^{\mathrm{t}}\frac{\mathrm{dt}}{2\mathrm{x}}\left[\text{Let t}={\mathrm{x}}^2\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=2\mathrm{x}\Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{2\mathrm{x}}\right]\\ =\frac{1}{2}{\mathrm{e}}^{\mathrm{t}}\\ =\frac{1}{2}{\mathrm{e}}^{{\mathrm{x}}^2}=\mathrm{F}\left(\mathrm{x}\right)\\ \text{Therefore, by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(1\right)-\mathrm{F}\left(0\right)\\ =\frac{1}{2}{\mathrm{e}}^{1^2}-\frac{1}{2}{\mathrm{e}}^0=\frac{1}{2}(\mathrm{e}-1)\end{array}$

Q.119

${\int }_1^2\frac{5{\mathrm{x}}^2}{{\mathrm{x}}^2+4\mathrm{x}+3}\mathrm{dx}$

Ans.

$\begin{array}{l}\text{Let \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}={\int }_1^2\frac{5{\mathrm{x}}^2}{{\mathrm{x}}^2+4\mathrm{x}+3}\mathrm{dx}\\ \int \frac{5{\mathrm{x}}^2}{{\mathrm{x}}^2+4\mathrm{x}+3}\mathrm{dx}=\int (5-\frac{20\mathrm{x}+15}{{\mathrm{x}}^2+4\mathrm{x}+3})\mathrm{dx}\\ =\int 5\mathrm{dx}-\int \frac{20\mathrm{x}+15}{{\mathrm{x}}^2+4\mathrm{x}+3}\mathrm{dx}...\left(\mathrm{i}\right)\\ \text{Let\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}20\mathrm{x}+15=\mathrm{A}\frac{\mathrm{d}}{\mathrm{dx}}({\mathrm{x}}^2+4\mathrm{x}+3)+\mathrm{B}\\ =\mathrm{A}(2\mathrm{x}+4)+\mathrm{B}\\ =2\mathrm{Ax}+4\mathrm{A}+\mathrm{B}\\ \text{Equating the coefficients of x and constant term, we get}\\ \Rightarrow 2\mathrm{A}=20\text{\hspace{0.17em}and\hspace{0.17em}4A}+\text{B}=\text{15}\\ \Rightarrow \mathrm{A}=10\mathrm{and}\mathrm{B}=-25\\ \therefore 20\mathrm{x}+15=10(2\mathrm{x}+4)-25\\ \text{From equation}\left(\text{i}\right)\text{, we have}\\ \int \frac{5{\mathrm{x}}^2}{{\mathrm{x}}^2+4\mathrm{x}+3}\mathrm{dx}=\int 5\mathrm{dx}-\int \frac{10(2\mathrm{x}+4)-25}{{\mathrm{x}}^2+4\mathrm{x}+3}\mathrm{dx}\\ =\int 5\mathrm{dx}-10\int \frac{(2\mathrm{x}+4)}{{\mathrm{x}}^2+4\mathrm{x}+3}\mathrm{dx}+25\int \frac{1}{{\mathrm{x}}^2+4\mathrm{x}+3}\mathrm{dx}\\ =5\mathrm{x}-10\log ({\mathrm{x}}^2+4\mathrm{x}+3)+25\int \frac{1}{{(\mathrm{x}+2)}^2-1}\mathrm{dx}\\ =5\mathrm{x}-10\log ({\mathrm{x}}^2+4\mathrm{x}+3)+25\times \frac{1}{2}\log \left|\frac{\mathrm{x}+2-1}{\mathrm{x}+2+1}\right|+\mathrm{C}\\ =5\mathrm{x}-10\log ({\mathrm{x}}^2+4\mathrm{x}+3)+\frac{25}{2}\log \left|\frac{\mathrm{x}+1}{\mathrm{x}+3}\right|\\ {\int }_1^2\frac{5{\mathrm{x}}^2}{{\mathrm{x}}^2+4\mathrm{x}+3}\mathrm{dx}={{[5\mathrm{x}-10\log ({\mathrm{x}}^2+4\mathrm{x}+3)+\frac{25}{2}\log \left|\frac{\mathrm{x}+1}{\mathrm{x}+3}\right|]}_1}^2\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\; \hspace{0.17em}=[5\left(2\right)-10\log (2^2+4\times 2+3)+\frac{25}{2}\log \left|\frac{2+1}{2+3}\right|]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\; \hspace{0.17em}-[5\left(1\right)-10\log (1^2+4\times 1+3)+\frac{25}{2}\log \left|\frac{1+1}{1+3}\right|]\\ {\int }_1^2\frac{5{\mathrm{x}}^2}{{\mathrm{x}}^2+4\mathrm{x}+3}\mathrm{dx}={{[5\mathrm{x}-10\log ({\mathrm{x}}^2+4\mathrm{x}+3)+\frac{25}{2}\log \left|\frac{\mathrm{x}+1}{\mathrm{x}+3}\right|]}_1}^2\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\; \hspace{0.17em}=[5\left(2\right)-10\log (2^2+4\times 2+3)+\frac{25}{2}\log \left|\frac{2+1}{2+3}\right|]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\; \hspace{0.17em}-[5\left(1\right)-10\log (1^2+4\times 1+3)+\frac{25}{2}\log \left|\frac{1+1}{1+3}\right|]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\; \hspace{0.17em}=(10-10\log 15+\frac{25}{2}\log \frac{3}{5})-(5-10\log 8+\frac{25}{2}\log \frac{2}{4})\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\; \hspace{0.17em}=5-10(\log 15-\log 8)+\frac{25}{2}(\log \frac{3}{5}-\log \frac{2}{4})\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\; \hspace{0.17em}=5-10(\log 5+\log 3-\log 4-\log 2)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}+\frac{25}{2}(\log 3-\log 5-\log 2+\log 4)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\; \hspace{0.17em}=5-(10+\frac{25}{2})\log 5+(10+\frac{25}{2})\log 4+(-10+\frac{25}{2})\log 3\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\; \hspace{0.17em}+(10-\frac{25}{2})\log 2\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\; \hspace{0.17em}=5-\frac{45}{2}\log 5+\frac{45}{2}\log 4+\frac{5}{2}\log 3-\frac{5}{2}\log 2\\ {\int }_1^2\frac{5{\mathrm{x}}^2}{{\mathrm{x}}^2+4\mathrm{x}+3}\mathrm{dx}=5-\frac{45}{2}\log \frac{5}{4}+\frac{5}{2}\log \frac{3}{2}=5-\frac{5}{2}(9\log \frac{5}{4}-\log \frac{3}{2})\end{array}$

Q.120

${\int }_0^{\frac{\mathrm{\pi }}{4}}(2{\sec }^2\mathrm{x}+{\mathrm{x}}^3+2)\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{4}}(2{\sec }^2\mathrm{x}+{\mathrm{x}}^3+2)\mathrm{dx}\\ \int (2{\sec }^2\mathrm{x}+{\mathrm{x}}^3+2)\mathrm{dx}\\ =2\int {\sec }^2\mathrm{x}\mathrm{dx}+\int {\mathrm{x}}^3\mathrm{dx}+\int 2\mathrm{dx}\\ =2\mathrm{tanx}+\frac{{\mathrm{x}}^4}{4}+2\mathrm{x}=\mathrm{F}\left(\mathrm{x}\right)\\ \text{Therefore, by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(\frac{\mathrm{\pi }}{4}\right)-\mathrm{F}\left(0\right)\\ =\{2\tan \left(\frac{\mathrm{\pi }}{4}\right)+\frac{{\left(\frac{\mathrm{\pi }}{4}\right)}^4}{4}+2\left(\frac{\mathrm{\pi }}{4}\right)\}-\{2\tan \left(0\right)+\frac{{\left(0\right)}^4}{4}+2\left(0\right)\}\\ =2[\tan \frac{\mathrm{\pi }}{4}-\tan 0]+\frac{1}{4}[{\left(\frac{\mathrm{\pi }}{4}\right)}^4-0]+2[\frac{\mathrm{\pi }}{4}-0]\\ =2[1-0]+\frac{1}{4}\left[{\left(\frac{\mathrm{\pi }}{4}\right)}^4\right]+2\left[\frac{\mathrm{\pi }}{4}\right]\\ =2+\frac{{\mathrm{\pi }}^4}{1024}+\frac{\mathrm{\pi }}{2}\end{array}$

Q.121

${\int }_0^{\mathrm{\pi }}({\sin }^2\frac{\mathrm{x}}{2}-{\cos }^2\frac{\mathrm{x}}{2})\mathrm{dx}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_0^{\mathrm{\pi }}({\sin }^2\frac{\mathrm{x}}{2}-{\cos }^2\frac{\mathrm{x}}{2})\mathrm{dx}\\ \int ({\sin }^2\frac{\mathrm{x}}{2}-{\cos }^2\frac{\mathrm{x}}{2})\mathrm{dx}=-\int ({\cos }^2\frac{\mathrm{x}}{2}-{\sin }^2\frac{\mathrm{x}}{2})\mathrm{dx}\\ =-\int \mathrm{cosx}\mathrm{dx}[\because {\cos }^2\mathrm{\theta }-{\sin }^2\mathrm{\theta }=\cos 2\mathrm{\theta }]\\ =-\mathrm{sinx}=\mathrm{F}\left(\mathrm{x}\right)\\ \text{Therefore},\text{by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(\mathrm{\pi }\right)-\mathrm{F}\left(0\right)\\ =-\mathrm{sin\pi }-(-\sin 0)=0\end{array}$

Q.122

${\int }_0^2\frac{6\mathrm{x}+3}{{\mathrm{x}}^2+4}\mathrm{dx}$

Ans.

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Let\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_0^2\frac{6\mathrm{x}+3}{{\mathrm{x}}^2+4}\mathrm{dx}\\ \int \frac{6\mathrm{x}+3}{{\mathrm{x}}^2+4}\mathrm{dx}=\int \frac{6\mathrm{x}}{{\mathrm{x}}^2+4}\mathrm{dx}+\int \frac{3}{{\mathrm{x}}^2+4}\mathrm{dx}\\ =3\int \frac{2\mathrm{x}}{{\mathrm{x}}^2+4}\mathrm{dx}+3\int \frac{1}{{\mathrm{x}}^2+2^2}\mathrm{dx}\\ =3\left[\log ({\mathrm{x}}^2+4)\right]+\frac{3}{2}\left[{\tan }^{-1}\frac{\mathrm{x}}{2}\right]=\mathrm{F}\left(\mathrm{x}\right)\\ \text{Therefore, by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(2\right)-\mathrm{F}\left(0\right)\\ =3[\log (2^2+4)-\frac{3}{2}{\tan }^{-1}\frac{2}{2}]-[\log (0^2+4)+\frac{3}{2}{\tan }^{-1}\frac{0}{2}]\\ =3[\log 8-\log 4]-\frac{3}{2}[{\tan }^{-1}\left(1\right)-{\tan }^{-1}\left(0\right)]\\ =3\log \frac{8}{4}-\frac{3}{2}[\frac{\mathrm{\pi }}{4}-0]\\ {\int }_0^2\frac{6\mathrm{x}+3}{{\mathrm{x}}^2+4}\mathrm{dx}=3\log 2-\frac{3\mathrm{\pi }}{8}\end{array}$

Q.123

${\int }_0^1(\mathrm{x}{\mathrm{e}}^{\mathrm{x}}+\sin \frac{\mathrm{\pi x}}{4})\mathrm{dx}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_0^1(\mathrm{x}{\mathrm{e}}^{\mathrm{x}}+\sin \frac{\mathrm{\pi x}}{4})\mathrm{dx}\\ \int (\mathrm{x}{\mathrm{e}}^{\mathrm{x}}+\sin \frac{\mathrm{\pi x}}{4})\mathrm{dx}=\int \mathrm{x}{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}+\int \sin \frac{\mathrm{\pi x}}{4}\mathrm{dx}\\ =\mathrm{x}\int {\mathrm{e}}^{\mathrm{x}}\mathrm{dx}-\int (\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{x}\int {\mathrm{e}}^{\mathrm{x}}\mathrm{dx})\mathrm{dx}-\frac{1}{\left(\frac{\mathrm{\pi }}{4}\right)}\cos \frac{\mathrm{\pi x}}{4}+\mathrm{C}\\ ={\mathrm{xe}}^{\mathrm{x}}-\int {\mathrm{e}}^{\mathrm{x}}\mathrm{dx}-\frac{4}{\mathrm{\pi }}\cos \frac{\mathrm{\pi x}}{4}\\ ={\mathrm{xe}}^{\mathrm{x}}-{\mathrm{e}}^{\mathrm{x}}-\frac{4}{\mathrm{\pi }}\cos \frac{\mathrm{\pi x}}{4}\\ ={\mathrm{e}}^{\mathrm{x}}(\mathrm{x}-1)-\frac{4}{\mathrm{\pi }}\cos \frac{\mathrm{\pi x}}{4}=\mathrm{F}\left(\mathrm{x}\right)\\ \text{Therefore, by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(1\right)-\mathrm{F}\left(0\right)\\ =\{{\mathrm{e}}^1(1-1)-\frac{4}{\mathrm{\pi }}\cos \frac{\mathrm{\pi }\left(1\right)}{4}\}-\{{\mathrm{e}}^0(0-1)-\frac{4}{\mathrm{\pi }}\cos \frac{\mathrm{\pi }\left(0\right)}{4}\}\\ =\{-\frac{4}{\mathrm{\pi }}\cos \frac{\mathrm{\pi }}{4}\}-\{-1-\frac{4}{\mathrm{\pi }}\times 1\}\\ =\{-\frac{4}{\mathrm{\pi }}\times \frac{1}{\sqrt{2}}\}-\{-1-\frac{4}{\mathrm{\pi }}\}\\ =-\frac{4}{\sqrt{2}\mathrm{\pi }}+1+\frac{4}{\mathrm{\pi }}\\ =1+\frac{4}{\mathrm{\pi }}-\frac{2\sqrt{2}}{\mathrm{\pi }}\end{array}$

Q.124

$\begin{array}{l}\mathrm{Choose}\mathrm{the}\mathrm{correct}\mathrm{answer}\\ {\int }_1^{\sqrt{3}}\frac{\mathrm{dx}}{1+{\mathrm{x}}^2}\text{equals}\\ \left(\text{A}\right)\frac{\mathrm{\pi }}{3}\\ \left(\text{B}\right)\frac{2\mathrm{\pi }}{3}\\ \left(\text{C}\right)\frac{\mathrm{\pi }}{6}\\ \left(\text{D}\right)\frac{\mathrm{\pi }}{12}\end{array}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_1^{\sqrt{3}}\frac{\mathrm{dx}}{1+{\mathrm{x}}^2}\\ \int \frac{1}{1+{\mathrm{x}}^2}\mathrm{dx}={\tan }^{-1}\mathrm{x}\\ \text{Therefore, by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(\sqrt{3}\right)-\mathrm{F}\left(1\right)\\ ={\text{tan}}^{-1}\left(\sqrt{3}\right)-{\text{tan}}^{-1}\left(1\right)\\ =\frac{\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{4}\\ =\frac{4\mathrm{\pi }-3\mathrm{\pi }}{12}\\ =\frac{\mathrm{\pi }}{12}\\ \text{Hence, the correct option is D.}\end{array}$

Q.125

$\begin{array}{l}\mathrm{Choose}\mathrm{the}\mathrm{correct}\mathrm{answer}\\ {\int }_0^{\frac{2}{3}}\frac{\mathrm{dx}}{4+9{\mathrm{x}}^2}\text{\hspace{0.33em}equals}\\ \left(\text{A}\right)\frac{\mathrm{\pi }}{6}\\ \left(\text{B}\right)\frac{\mathrm{\pi }}{12}\\ \left(\text{C}\right)\frac{\mathrm{\pi }}{24}\\ \left(\text{D}\right)\frac{\mathrm{\pi }}{4}\end{array}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_0^{\frac{2}{3}}\frac{\mathrm{dx}}{4+9{\mathrm{x}}^2}\\ \int \frac{1}{4+9{\mathrm{x}}^2}\mathrm{dx}=\int \frac{1}{2^2+{\left(3\mathrm{x}\right)}^2}\mathrm{dx}\\ =\frac{1}{2}{\tan }^{-1}\left(\frac{3\mathrm{x}}{2}\right)\times \frac{1}{3}\\ =\frac{1}{6}{\tan }^{-1}\left(\frac{3\mathrm{x}}{2}\right)=\mathrm{F}\left(\mathrm{x}\right)\\ \text{Therefore, by the second fundamental theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}=\mathrm{F}\left(\frac{2}{3}\right)-\mathrm{F}\left(0\right)\\ =\frac{1}{6}{\tan }^{-1}(\frac{3}{2}\times \frac{2}{3})-\frac{1}{6}{\tan }^{-1}(\frac{3}{2}\times 0)\\ =\frac{1}{6}{\tan }^{-1}\left(1\right)-\frac{1}{6}{\tan }^{-1}\left(0\right)\\ =\frac{1}{6}\times \frac{\mathrm{\pi }}{4}=\frac{\mathrm{\pi }}{24}\\ \text{Hence, the correct option is C.}\end{array}$

Q.126

$\mathrm{Evaluate}\mathrm{the}\mathrm{integrals}{\int }_0^1\frac{\mathrm{x}}{{\mathrm{x}}^2+1}\mathrm{dx}$

Ans.

$\begin{array}{l}{\int }_0^1\frac{\mathrm{x}}{{\mathrm{x}}^2+1}\mathrm{dx}\\ \text{Let\hspace{0.17em}\hspace{0.17em}}\mathrm{t}={\mathrm{x}}^2+1\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=2\mathrm{x}\\ \text{When x}=0,\text{\hspace{0.17em}t}=1\text{and when x}=\text{1, t}=\text{2}\\ \therefore {\int }_0^1\frac{\mathrm{x}}{{\mathrm{x}}^2+1}\mathrm{dx}=\frac{1}{2}{\int }_1^2\frac{1}{\mathrm{t}}\mathrm{dt}\\ =\frac{1}{2}{{\left[\log \left|\mathrm{t}\right|\right]}_1}^2\\ =\frac{1}{2}[\log 2-\log 1]\\ =\frac{1}{2}\log 2\end{array}$

Q.127

$\mathrm{Evaluate}\mathrm{the}\mathrm{integrals}{\int }_0^{\frac{\mathrm{\pi }}{2}}\sqrt{\mathrm{sin\varphi }}{\cos }^5\mathrm{\varphi }\mathrm{d\varphi }$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}\sqrt{\mathrm{sin\varphi }}{\cos }^5\mathrm{\varphi }\mathrm{d}\mathrm{\varphi }\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\sqrt{\mathrm{sin\varphi }}{\cos }^4\mathrm{\varphi }\mathrm{cos\varphi }\mathrm{d\varphi }\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\sqrt{\mathrm{sin\varphi }}{\left({\cos }^2\mathrm{\varphi }\right)}^2\mathrm{cos\varphi }\mathrm{d\varphi }\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\sqrt{\mathrm{sin\varphi }}{(1-{\sin }^2\mathrm{\varphi })}^2\mathrm{cos\varphi }\mathrm{d\varphi }\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\sqrt{\mathrm{sin\varphi }}(1-2{\sin }^2\mathrm{\varphi }+{\sin }^4\mathrm{\varphi })\mathrm{cos\varphi }\mathrm{d\varphi }\\ \text{Let t}=\mathrm{sin\varphi }\Rightarrow \frac{\mathrm{dt}}{\mathrm{d\varphi }}=\mathrm{cos\varphi }\\ \text{When\hspace{0.17em}}\mathrm{\varphi }=0,\mathrm{t}=0\text{and when}\mathrm{\varphi }=\frac{\mathrm{\pi }}{2},\mathrm{t}=1\\ \therefore \mathrm{I}={\int }_0^1\sqrt{\mathrm{t}}(1-2{\mathrm{t}}^2+{\mathrm{t}}^4)\mathrm{dt}\\ ={\int }_0^1({\mathrm{t}}^{\frac{1}{2}}-2{\mathrm{t}}^{\frac{5}{2}}+{\mathrm{t}}^{\frac{9}{2}})\mathrm{dt}\\ ={{[\frac{{\mathrm{t}}^{\frac{3}{2}}}{\frac{3}{2}}-2\frac{{\mathrm{t}}^{\frac{7}{2}}}{\frac{7}{2}}+\frac{{\mathrm{t}}^{\frac{11}{2}}}{\frac{11}{2}}]}_0}^1\\ =[\frac{1^{\frac{3}{2}}}{\frac{3}{2}}-2\frac{1^{\frac{7}{2}}}{\frac{7}{2}}+\frac{1^{\frac{11}{2}}}{\frac{11}{2}}]-[\frac{0^{\frac{3}{2}}}{\frac{3}{2}}-2\frac{0^{\frac{7}{2}}}{\frac{7}{2}}+\frac{0^{\frac{11}{2}}}{\frac{11}{2}}]\\ =[\frac{2}{3}-2.\frac{2}{7}+\frac{2}{11}]\\ =\frac{2}{3}-\frac{4}{7}+\frac{2}{11}\\ =\frac{154-132+42}{231}=\frac{64}{231}\end{array}$

Q.128

$\mathrm{Evaluate}\mathrm{the}\mathrm{integrals}{\int }_0^1{\sin }^{–1}\left(\frac{2\mathrm{x}}{1+{\mathrm{x}}^2}\right)\mathrm{dx}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_0^1{\sin }^{–1}\left(\frac{2\mathrm{x}}{1+{\mathrm{x}}^2}\right)\mathrm{dx}\\ \text{Also, let x}=\text{tan}\mathrm{\theta }\Rightarrow \frac{\mathrm{dx}}{\mathrm{d\theta }}={\text{sec}}^2\mathrm{\theta }\\ \text{When\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=0,\mathrm{\theta }=0\text{\hspace{0.17em}and when x}=\text{1,}\mathrm{\theta }=\frac{\mathrm{\pi }}{4}\\ \therefore \mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{4}}{\text{sin}}^{–1}\left(\frac{2\text{tan}\mathrm{\theta }}{1+{\text{tan}}^2\mathrm{\theta }}\right){\text{\hspace{0.17em}sec}}^2\mathrm{\theta }\mathrm{d}\mathrm{\theta }\\ ={\int }_0^{\frac{\mathrm{\pi }}{4}}{\text{sin}}^{–1}\left(\text{sin}2\mathrm{\theta }\right){\text{\hspace{0.17em}sec}}^2\mathrm{\theta d\theta }\\ ={\int }_0^{\frac{\mathrm{\pi }}{4}}2\mathrm{\theta }{\text{\hspace{0.17em}sec}}^2\mathrm{\theta }\mathrm{d\theta }\\ \text{Now},\\ \int 2\mathrm{\theta }{\text{\hspace{0.17em}sec}}^2\mathrm{\theta }\mathrm{d\theta }\\ =2[\mathrm{\theta }\int {\text{sec}}^2\mathrm{\theta }\mathrm{d\theta }-\int (\frac{\mathrm{d}}{\mathrm{d\theta }}\mathrm{\theta }\int {\sec }^2\mathrm{\theta }\mathrm{d\theta })\mathrm{d\theta }]\\ =2[\mathrm{\theta }\text{\hspace{0.17em}tan}\mathrm{\theta }-\int (1.\text{\hspace{0.17em}tan}\mathrm{\theta })\mathrm{d\theta }]\\ =2[\mathrm{\theta }\text{\hspace{0.17em}tan\hspace{0.17em}}\mathrm{\theta }+\text{log\hspace{0.17em}cos\hspace{0.17em}}\mathrm{\theta }]\\ \therefore \mathrm{I}=2{{[\mathrm{\theta }\text{\hspace{0.17em}tan\hspace{0.17em}}\mathrm{\theta }+\text{log\hspace{0.17em}cos\hspace{0.17em}}\mathrm{\theta }]}_0}^{\frac{\mathrm{\pi }}{4}}\\ =2[\left(\frac{\mathrm{\pi }}{4}\right)\text{tan\hspace{0.17em}}\frac{\mathrm{\pi }}{4}+\text{log\hspace{0.17em}cos\hspace{0.17em}}\frac{\mathrm{\pi }}{4}]-[2\left(0\right)\text{tan\hspace{0.17em}}0+\text{log\hspace{0.17em}cos\hspace{0.17em}}0]\\ =2[\frac{\mathrm{\pi }}{4}\times 1+\text{log\hspace{0.17em}}\frac{1}{\sqrt{2}}]-[0+\text{log\hspace{0.17em}}1]\\ =[\frac{\mathrm{\pi }}{2}-2\times \frac{1}{2}\text{\hspace{0.17em}log}2]-[0+0]\\ =(\frac{\mathrm{\pi }}{2}-\text{log\hspace{0.17em}}2)\end{array}$

Q.129

$\mathrm{Evaluate}\mathrm{the}\mathrm{integrals}{\int }_0^2\mathrm{x}\sqrt{\mathrm{x}+2}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{I}={\int }_0^2\mathrm{x}\sqrt{\mathrm{x}+2}\mathrm{dx}\\ {\text{Let t}}^{\text{2}}=\text{x}+\text{2}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=2\mathrm{t}\\ \text{When x}=\text{0, t}=\sqrt{\text{2}}\text{and when x}=\text{2, t}=\text{2}\\ \therefore \mathrm{I}={\int }_{\sqrt{2}}^2({\mathrm{t}}^2-2)\sqrt{{\mathrm{t}}^2}2\mathrm{t}\mathrm{dt}\\ =2{\int }_{\sqrt{2}}^2({\mathrm{t}}^4-2{\mathrm{t}}^2)\mathrm{dt}\\ =2{{[\frac{{\mathrm{t}}^5}{5}-2\times \frac{{\mathrm{t}}^3}{3}]}_{\sqrt{2}}}^2\\ =2[\frac{2^5}{5}-2\times \frac{2^3}{3}]-2[\frac{{\left(\sqrt{2}\right)}^5}{5}-2\times \frac{{\left(\sqrt{2}\right)}^3}{3}]\\ =2[\frac{2^5}{5}-\frac{2^4}{3}]-2[\frac{4\sqrt{2}}{5}-\frac{4\sqrt{2}}{3}]\\ =2^5[\frac{2}{5}-\frac{1}{3}]-2^3\sqrt{2}[\frac{1}{5}-\frac{1}{3}]\\ =2^5\left(\frac{6-5}{15}\right)-2^3\sqrt{2}\left(\frac{3-5}{15}\right)\\ =\frac{2^5}{15}-2^3\sqrt{2}\left(\frac{-2}{15}\right)\\ =2^4(\frac{2}{15}+\frac{\sqrt{2}}{15})\\ =16\sqrt{2}\left(\frac{\sqrt{2}+1}{15}\right)\end{array}$

Q.130

$\begin{array}{l}\mathrm{I}={\int }_0^2\mathrm{x}\sqrt{\mathrm{x}+2}\mathrm{dx}\\ {\text{Let t}}^{\text{2}}=\text{x}+\text{2}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=2\mathrm{t}\\ \text{When x}=\text{0, t}=\sqrt{\text{2}}\text{and when x}=\text{2, t}=\text{2}\\ \therefore \mathrm{I}={\int }_{\sqrt{2}}^2({\mathrm{t}}^2-2)\sqrt{{\mathrm{t}}^2}2\mathrm{t}\mathrm{dt}\\ =2{\int }_{\sqrt{2}}^2({\mathrm{t}}^4-2{\mathrm{t}}^2)\mathrm{dt}\\ =2{{[\frac{{\mathrm{t}}^5}{5}-2\times \frac{{\mathrm{t}}^3}{3}]}_{\sqrt{2}}}^2\\ =2[\frac{2^5}{5}-2\times \frac{2^3}{3}]-2[\frac{{\left(\sqrt{2}\right)}^5}{5}-2\times \frac{{\left(\sqrt{2}\right)}^3}{3}]\\ =2[\frac{2^5}{5}-\frac{2^4}{3}]-2[\frac{4\sqrt{2}}{5}-\frac{4\sqrt{2}}{3}]\\ =2^5[\frac{2}{5}-\frac{1}{3}]-2^3\sqrt{2}[\frac{1}{5}-\frac{1}{3}]\\ =2^5\left(\frac{6-5}{15}\right)-2^3\sqrt{2}\left(\frac{3-5}{15}\right)\\ =\frac{2^5}{15}-2^3\sqrt{2}\left(\frac{-2}{15}\right)\\ =2^4(\frac{2}{15}+\frac{\sqrt{2}}{15})\\ =16\sqrt{2}\left(\frac{\sqrt{2}+1}{15}\right)\end{array}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\mathrm{sinx}}{1+{\cos }^2\mathrm{x}}\mathrm{dx}\\ \text{Again, let t}=\text{cosx}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=-\text{sin}\mathrm{x}\\ \text{When x}=\text{0, t}=1\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}when\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\frac{\mathrm{\pi }}{2},\mathrm{t}=0\\ \therefore \mathrm{I}={\int }_1^0\frac{\mathrm{sinx}}{1+{\mathrm{t}}^2}\frac{\mathrm{dt}}{-\mathrm{sinx}}\\ =-{\int }_1^0\frac{1}{1+{\mathrm{t}}^2}\mathrm{dt}\\ =-{{\left[{\tan }^{-1}\mathrm{t}\right]}_1}^0\\ =-({\tan }^{-1}0-{\tan }^{-1}1)\\ =-(0-\frac{\mathrm{\pi }}{4})=\frac{\mathrm{\pi }}{4}\end{array}$

Q.131

$\mathrm{Evaluate}\mathrm{the}\mathrm{integrals}{\int }_0^2\frac{\mathrm{dx}}{\mathrm{x}+4-{\mathrm{x}}^2}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em} \hspace{0.17em}}\mathrm{I}={\int }_0^2\frac{\mathrm{dx}}{\mathrm{x}+4-{\mathrm{x}}^2}\\ ={\int }_0^2\frac{\mathrm{dx}}{-({\mathrm{x}}^2-\mathrm{x}-4)}\\ ={\int }_0^2\frac{\mathrm{dx}}{-({\mathrm{x}}^2-\mathrm{x}+\frac{1}{4}-\frac{1}{4}-4)}\\ =-{\int }_0^2\frac{\mathrm{dx}}{{(\mathrm{x}-\frac{1}{2})}^2-\frac{17}{4}}\\ =-{\int }_0^2\frac{\mathrm{dx}}{{(\mathrm{x}-\frac{1}{2})}^2-{\left(\frac{\sqrt{17}}{4}\right)}^2}\\ \text{Let t}=\mathrm{x}-\frac{1}{2}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=1\\ \text{When x}=0,\mathrm{t}=-\frac{1}{2}\text{and when x}=2,\mathrm{t}=\frac{3}{2}\\ \mathrm{I}=-{\int }_{-\frac{1}{2}}^{\frac{3}{2}}\frac{\mathrm{dt}}{{\left(\frac{\sqrt{17}}{4}\right)}^2-{\mathrm{t}}^2}\\ =\frac{1}{2\left(\frac{\sqrt{17}}{4}\right)}\log {{\left\{\frac{\left(\frac{\sqrt{17}}{4}\right)+\mathrm{t}}{\left(\frac{\sqrt{17}}{4}\right)-\mathrm{t}}\right\}}_{-\frac{1}{2}}}^{\frac{3}{2}}\\ =\frac{1}{2\left(\frac{\sqrt{17}}{4}\right)}[\log \left\{\frac{\left(\frac{\sqrt{17}}{4}\right)+\frac{3}{2}}{\left(\frac{\sqrt{17}}{4}\right)-\frac{3}{2}}\right\}-\log \left\{\frac{\left(\frac{\sqrt{17}}{4}\right)-\frac{1}{2}}{\left(\frac{\sqrt{17}}{4}\right)+\frac{1}{2}}\right\}]\\ =\frac{1}{2\left(\frac{\sqrt{17}}{4}\right)}[\log \left\{\frac{\sqrt{17}+3}{\sqrt{17}-3}\right\}-\log \left\{\frac{\sqrt{17}-1}{\sqrt{17}+1}\right\}]\\ =\frac{1}{2\left(\frac{\sqrt{17}}{4}\right)}\log \{\frac{\sqrt{17}+3}{\sqrt{17}-3}\times \frac{\sqrt{17}+1}{\sqrt{17}-1}\}\\ =\frac{1}{2\left(\frac{\sqrt{17}}{4}\right)}\log \left\{\frac{17+\sqrt{17}+3\sqrt{17}+3}{17-\sqrt{17}-3\sqrt{17}+3}\right\}\\ =\frac{2}{\sqrt{17}}\log \left\{\frac{20+4\sqrt{17}}{20-4\sqrt{17}}\right\}\\ =\frac{2}{\sqrt{17}}\log \left\{\frac{5+\sqrt{17}}{5-\sqrt{17}}\right\}\\ =\frac{2}{\sqrt{17}}\log \{\frac{5+\sqrt{17}}{5-\sqrt{17}}\times \frac{5+\sqrt{17}}{5+\sqrt{17}}\}\\ =\frac{2}{\sqrt{17}}\log \left\{\frac{25+17+10\sqrt{17}}{25-17}\right\}\\ =\frac{2}{\sqrt{17}}\log \left\{\frac{42+10\sqrt{17}}{8}\right\}\\ =\frac{2}{\sqrt{17}}\log \left\{\frac{21+5\sqrt{17}}{4}\right\}\end{array}$

Q.132

$\mathrm{Evaluate}\mathrm{the}\mathrm{integrals}{\int }_{–1}^1\frac{\mathrm{dx}}{{\mathrm{x}}^2+2\mathrm{x}+5}\mathrm{dx}$

Ans.

$\begin{array}{l}{\int }_{–1}^1\frac{\mathrm{dx}}{{\mathrm{x}}^2+2\mathrm{x}+5}\mathrm{dx}={\int }_{–1}^1\frac{\mathrm{dx}}{{(\mathrm{x}+1)}^2+2^2}\mathrm{dx}\\ \text{Let t}=\mathrm{x}+1\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=1\\ \text{When x}=-1,\mathrm{t}=0\text{\hspace{0.17em}\hspace{0.17em}and when x}=1,\mathrm{t}=2\\ \therefore {\int }_{–1}^1\frac{\mathrm{dx}}{{(\mathrm{x}+1)}^2+2^2}\mathrm{dx}={\int }_0^2\frac{\mathrm{dx}}{{\mathrm{t}}^2+2^2}\mathrm{dt}\\ ={{\left\{\frac{1}{2}{\tan }^{-1}\left(\frac{\mathrm{t}}{2}\right)\right\}}_0}^2\\ =\frac{1}{2}{\tan }^{-1}\left(\frac{2}{2}\right)-\frac{1}{2}{\tan }^{-1}\left(\frac{0}{2}\right)\\ =\frac{1}{2}{\tan }^{-1}\left(1\right)-\frac{1}{2}{\tan }^{-1}\left(0\right)\\ =\frac{1}{2}\times \frac{\mathrm{\pi }}{4}-\frac{1}{2}\times 0=\frac{\mathrm{\pi }}{8}\\ \text{Thus},{\int }_{–1}^1\frac{\mathrm{dx}}{{\mathrm{x}}^2+2\mathrm{x}+5}\mathrm{dx}=\frac{\mathrm{\pi }}{8}\end{array}$

Q.133

$\mathrm{Evaluate}\mathrm{the}\mathrm{integrals}{\int }_1^2(\frac{1}{\mathrm{x}}-\frac{1}{2{\mathrm{x}}^2}){\mathrm{e}}^{2\mathrm{x}}\mathrm{dx}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}={\int }_1^2(\frac{1}{\mathrm{x}}-\frac{1}{2{\mathrm{x}}^2}){\mathrm{e}}^{2\mathrm{x}}\mathrm{dx}\\ ={\int }_1^2(\frac{2}{2\mathrm{x}}-\frac{2}{{\left(2\mathrm{x}\right)}^2}){\mathrm{e}}^{2\mathrm{x}}\mathrm{dx}\\ \mathrm{Again}\mathrm{let}\text{ t}=2\text{x}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=2\\ \mathrm{When}\mathrm{x}=1,\mathrm{t}=2\mathrm{and}\mathrm{when}\mathrm{x}=2,\mathrm{t}=4\\ \mathrm{I}={\int }_2^4(\frac{2}{\mathrm{t}}-\frac{2}{{\mathrm{t}}^2}){\mathrm{e}}^{\mathrm{t}}\frac{\mathrm{dt}}{2}\\ ={\int }_2^4(\frac{1}{\mathrm{t}}-\frac{1}{{\mathrm{t}}^2}){\mathrm{e}}^{\mathrm{t}}\mathrm{dt}\\ \mathrm{Let}\mathrm{F}\left(\mathrm{t}\right)=\frac{1}{\mathrm{t}},\mathrm{Then}\mathrm{F}‘\left(\mathrm{t}\right)=-\frac{1}{{\mathrm{t}}^2}\\ \Rightarrow {\int }_2^4(\frac{1}{\mathrm{t}}-\frac{1}{{\mathrm{t}}^2}){\mathrm{e}}^{\mathrm{t}}\mathrm{dt}={\int }_2^4(\mathrm{F}\left(\mathrm{t}\right)-\mathrm{F}‘\left(\mathrm{t}\right)){\mathrm{e}}^{\mathrm{t}}\mathrm{dt}\\ ={{\left[{\mathrm{e}}^{\mathrm{t}}\mathrm{F}\left(\mathrm{t}\right)\right]}_2}^4\\ ={{[{\mathrm{e}}^{\mathrm{t}}.\frac{1}{\mathrm{t}}]}_2}^4\\ ={\mathrm{e}}^4.\frac{2}{4}-{\mathrm{e}}^2.\frac{1}{2}\\ =\frac{1}{4}{\mathrm{e}}^2({\mathrm{e}}^2-2)\\ \mathrm{Thus},\\ {\int }_1^2(\frac{1}{\mathrm{x}}-\frac{1}{2{\mathrm{x}}^2}){\mathrm{e}}^{2\mathrm{x}}\mathrm{dx}=\frac{1}{4}{\mathrm{e}}^2({\mathrm{e}}^2-2)\end{array}$

Q.134

$\begin{array}{l}\mathrm{Choose}\mathrm{the}\mathrm{correct}\mathrm{answer}\\ \text{The value of the integral\hspace{0.33em}}{\int }_{-\frac{1}{3}}^1\frac{{\left(\text{x}-{\text{x}}^3\right)}^{\frac{1}{3}}}{{\text{x}}^4}\mathrm{dx}\text{is}\\ \left(\mathrm{A}\right)6\\ \left(\mathrm{B}\right)0\\ \left(\mathrm{C}\right)3\\ \left(\mathrm{D}\right)4\end{array}$

Ans.

$\begin{array}{l}{\int }_{–\frac{1}{3}}^1\frac{{(\mathrm{x}-{\mathrm{x}}^3)}^{\frac{1}{3}}}{{\mathrm{x}}^4}\mathrm{dx}\\ \text{Let\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\mathrm{sin\theta }\Rightarrow \frac{\mathrm{dx}}{\mathrm{d\theta }}=\mathrm{cos\theta }\\ \text{When\hspace{0.17em}}\mathrm{x}=-\frac{1}{3},\mathrm{\theta }={\sin }^{-1}(-\frac{1}{3})\text{and when x}=\text{1,\hspace{0.17em}}\mathrm{\theta }\text{=}\frac{\mathrm{\pi }}{2}\\ {\int }_{–\frac{1}{3}}^1\frac{{(\mathrm{x}-{\mathrm{x}}^3)}^{\frac{1}{3}}}{{\mathrm{x}}^4}\mathrm{dx}={\int }_{{\sin }^{–1}(-\frac{1}{3})}^{\frac{\mathrm{\pi }}{2}}\frac{{(\mathrm{sin\theta }-{\sin }^3\mathrm{\theta })}^{\frac{1}{3}}}{{\sin }^4\mathrm{\theta }}.\mathrm{cos\theta }\mathrm{d\theta }\\ ={\int }_{{\sin }^{–1}(-\frac{1}{3})}^{\frac{\mathrm{\pi }}{2}}\frac{{\left(\mathrm{sin\theta }\right)}^{\frac{1}{3}}{(1-{\sin }^2\mathrm{\theta })}^{\frac{1}{3}}}{{\sin }^4\mathrm{\theta }}.\mathrm{cos\theta }\mathrm{d}\mathrm{\theta }\\ ={\int }_{{\sin }^{–1}(-\frac{1}{3})}^{\frac{\mathrm{\pi }}{2}}\frac{{\left(\mathrm{sin\theta }\right)}^{\frac{1}{3}}{\left({\cos }^2\mathrm{\theta }\right)}^{\frac{1}{3}}}{{\sin }^4\mathrm{\theta }}.\mathrm{cos\theta }\mathrm{d}\mathrm{\theta }\\ ={\int }_{{\sin }^{–1}(-\frac{1}{3})}^{\frac{\mathrm{\pi }}{2}}\frac{{\left(\mathrm{sin\theta }\right)}^{\frac{1}{3}}{\left(\mathrm{cos\theta }\right)}^{\frac{2}{3}}}{{\sin }^2\mathrm{\theta }.{\sin }^2\mathrm{\theta }}.\mathrm{cos\theta }\mathrm{d\theta }\\ ={\int }_{{\sin }^{–1}(-\frac{1}{3})}^{\frac{\mathrm{\pi }}{2}}\frac{{\left(\mathrm{cos\theta }\right)}^{\frac{5}{3}}}{{\sin }^{\frac{5}{3}}\mathrm{\theta }}.{\text{cosec}}^2\mathrm{\theta }\mathrm{d\theta }\\ ={\int }_{{\sin }^{–1}(-\frac{1}{3})}^{\frac{\mathrm{\pi }}{2}}{\left(\mathrm{cot\theta }\right)}^{\frac{5}{3}}.{\text{cosec}}^2\mathrm{\theta }\mathrm{d\theta }\\ \text{Let\hspace{0.17em}\hspace{0.17em}}\mathrm{t}=\mathrm{cot\theta }\Rightarrow \frac{\mathrm{dt}}{\mathrm{d\theta }}=-{\mathrm{cosec}}^2\mathrm{\theta }\\ \text{When}\mathrm{\theta }={\text{sin}}^{-\text{1}}(-\frac{1}{3}),\mathrm{t}=2\sqrt{2}\text{and when}\mathrm{\theta }=\frac{\mathrm{\pi }}{2},\mathrm{t}=0\\ =-{\int }_{2\sqrt{2}}^0{\left(\mathrm{t}\right)}^{\frac{5}{3}}.\mathrm{dt}=-{{\left[\frac{{\mathrm{t}}^{\frac{8}{3}}}{\left(\frac{8}{3}\right)}\right]}_{2\sqrt{2}}}^0\\ =-\frac{3}{8}[0-{\left(2\sqrt{2}\right)}^{\frac{8}{3}}]\\ =-\frac{3}{8}{(-2^{\frac{3}{2}})}^{\frac{8}{3}}=\\ =\frac{3}{8}\times 2^4\\ =\frac{3}{8}\times 16=6\\ \text{Thus, the correct option is A.}\end{array}$

Q.135

$\begin{array}{l}\mathrm{Choose}\mathrm{the}\mathrm{correct}\mathrm{answer}\\ \text{If}\mathrm{f}\left(\mathrm{x}\right)={\int }_0^{\mathrm{x}}\mathrm{t}\mathrm{sint}\mathrm{dt},\text{then}\mathrm{f}‘\left(\mathrm{x}\right)\text{is}\\ \left(\mathrm{A}\right)\mathrm{cosx}+\mathrm{x}\mathrm{sinx}\\ \left(\mathrm{B}\right)\mathrm{x}\mathrm{sinx}\\ \left(\mathrm{C}\right)\mathrm{x}\mathrm{cosx}\\ \left(\mathrm{D}\right)\mathrm{sinx}+\mathrm{x}\mathrm{cosx}\end{array}$

Ans.

$\begin{array}{l}\text{We have}\mathrm{f}\left(\mathrm{x}\right)={\int }_0^{\mathrm{x}}\mathrm{t}\mathrm{sint}\mathrm{dt}\\ ={{[\mathrm{t}\int \mathrm{sint}\mathrm{dt}-\int \frac{\mathrm{d}}{\mathrm{dt}}\mathrm{t}\int \mathrm{sint}\mathrm{dt}\mathrm{dt}]}_0}^{\mathrm{x}}\\ ={{[\mathrm{t}(-\mathrm{cost})-\int 1.(-\mathrm{cost})\mathrm{dt}]}_0}^{\mathrm{x}}\\ ={{[\mathrm{t}(-\mathrm{cost})+\int \mathrm{cost}\mathrm{dt}]}_0}^{\mathrm{x}}\\ ={{[\mathrm{t}(-\mathrm{cost})+\mathrm{sint}]}_0}^{\mathrm{x}}\\ =\mathrm{x}(-\mathrm{cosx})+\mathrm{sinx}-0(-\cos 0)-\sin 0\\ =-\mathrm{xcosx}+\mathrm{sinx}\\ \text{Differentiating w.r.t. x, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}f’}\left(\mathrm{x}\right)=-(\mathrm{x}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{cosx}+\mathrm{cosx}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{x})+\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{sinx}\\ =-(\mathrm{x}\times -\mathrm{sinx}+\mathrm{cosx}\times 1)+\mathrm{cosx}\\ =\mathrm{xsinx}-\mathrm{cosx}+\mathrm{cosx}\\ =\mathrm{xsinx}\\ \text{Hence,\hspace{0.17em}the correct option is B.}\end{array}$

Q.136

${\int }_0^{\frac{\mathrm{\pi }}{2}}{\cos }^2\mathrm{x}\mathrm{dx}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}{\cos }^2\mathrm{x}\mathrm{dx}...\left(\mathrm{i}\right)\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}{\cos }^2(\frac{\mathrm{\pi }}{2}-\mathrm{x})\mathrm{dx}[\because {\mathrm{P}}_4:{\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_0^{\mathrm{a}}\mathrm{f}(\mathrm{a}-\mathrm{x})\mathrm{dx}]\\ \mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}{\text{sin}}^2\mathrm{x}\mathrm{dx}...\left(\mathrm{ii}\right)\\ \text{Adding equation}\left(\text{i}\right)\text{ and\hspace{0.17em}\hspace{0.17em}}\left(\text{ii}\right)\text{, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2I}={\int }_0^{\frac{\mathrm{\pi }}{2}}({\cos }^2\mathrm{x}+{\text{sin}}^2\mathrm{x})\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2I}={\int }_0^{\frac{\mathrm{\pi }}{2}}1\mathrm{dx}[\because {\cos }^2\mathrm{x}+{\text{sin}}^2\mathrm{x}=1]\\ ={{\left[\mathrm{x}\right]}_0}^{\frac{\mathrm{\pi }}{2}}\\ =\frac{\mathrm{\pi }}{2}-0\\ =\frac{\mathrm{\pi }}{2}\\ \mathrm{I}=\frac{\mathrm{\pi }}{4}\end{array}$

Q.137

${\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{\sin \mathrm{x}}}{\sqrt{\sin \mathrm{x}}+\sqrt{\cos \mathrm{x}}}\mathrm{dx}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{\mathrm{sinx}}}{\sqrt{\mathrm{sinx}}+\sqrt{\mathrm{cosx}}}\mathrm{dx}...\left(\mathrm{i}\right)\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{\sin (\frac{\mathrm{\pi }}{2}-\mathrm{x})}}{\sqrt{\sin (\frac{\mathrm{\pi }}{2}-\mathrm{x})}+\sqrt{\cos (\frac{\mathrm{\pi }}{2}-\mathrm{x})}}\mathrm{dx}\\ [\because {\mathrm{P}}_4:{\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_0^{\mathrm{a}}\mathrm{f}(\mathrm{a}-\mathrm{x})\mathrm{dx}]\\ \mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{\mathrm{cosx}}}{\sqrt{\mathrm{cosx}}+\sqrt{\mathrm{sinx}}}\mathrm{dx}...\left(\mathrm{ii}\right)\\ \text{Adding\hspace{0.17em}\hspace{0.17em}equation}\left(\text{i}\right)\text{\hspace{0.17em}and equation}\left(\text{ii}\right)\text{, we get}\\ 2\mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{\mathrm{sinx}}}{\sqrt{\mathrm{sinx}}+\sqrt{\mathrm{cosx}}}\mathrm{dx}+{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{\mathrm{cosx}}}{\sqrt{\mathrm{cosx}}+\sqrt{\mathrm{sinx}}}\mathrm{dx}\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{\mathrm{sinx}}+\sqrt{\mathrm{cosx}}}{\sqrt{\mathrm{sinx}}+\sqrt{\mathrm{cosx}}}\mathrm{dx}\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}1.\mathrm{dx}\\ ={{\left[\mathrm{x}\right]}_0}^{\frac{\mathrm{\pi }}{2}}\\ =\frac{\mathrm{\pi }}{2}-0\\ =\frac{\mathrm{\pi }}{2}\\ \mathrm{I}=\frac{\mathrm{\pi }}{4}\end{array}$

Q.138

${\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\sin }^{\frac{3}{2}}\mathrm{x}}{{\sin }^{\frac{3}{2}}\mathrm{x}+{\cos }^{\frac{3}{2}}\mathrm{x}}\mathrm{dx}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\sin }^{\frac{3}{2}}\mathrm{x}}{{\sin }^{\frac{3}{2}}\mathrm{x}+{\cos }^{\frac{3}{2}}\mathrm{x}}\mathrm{dx}...\left(\mathrm{i}\right)\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\sin }^{\frac{3}{2}}(\frac{\mathrm{\pi }}{2}-\mathrm{x})}{{\sin }^{\frac{3}{2}}(\frac{\mathrm{\pi }}{2}-\mathrm{x})+{\cos }^{\frac{3}{2}}(\frac{\mathrm{\pi }}{2}-\mathrm{x})}\mathrm{dx}\\ [\because {\mathrm{P}}_4:{\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_0^{\mathrm{a}}\mathrm{f}(\mathrm{a}-\mathrm{x})\mathrm{dx}]\\ \mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\cos }^{\frac{3}{2}}\mathrm{x}}{{\cos }^{\frac{3}{2}}\mathrm{x}+{\sin }^{\frac{3}{2}}\mathrm{x}}\mathrm{dx}...\left(\mathrm{ii}\right)\\ \text{Adding\hspace{0.17em}\hspace{0.17em}equation}\left(\text{i}\right)\text{\hspace{0.17em}and equation}\left(\text{ii}\right)\text{, we get}\\ 2\mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\sin }^{\frac{3}{2}}\mathrm{x}}{{\sin }^{\frac{3}{2}}\mathrm{x}+{\cos }^{\frac{3}{2}}\mathrm{x}}\mathrm{dx}+{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\cos }^{\frac{3}{2}}\mathrm{x}}{{\cos }^{\frac{3}{2}}\mathrm{x}+{\sin }^{\frac{3}{2}}\mathrm{x}}\mathrm{dx}\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\sin }^{\frac{3}{2}}\mathrm{x}+{\cos }^{\frac{3}{2}}\mathrm{x}}{{\cos }^{\frac{3}{2}}\mathrm{x}+{\sin }^{\frac{3}{2}}\mathrm{x}}\mathrm{dx}\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}1.\mathrm{dx}\\ ={{\left[\mathrm{x}\right]}_0}^{\frac{\mathrm{\pi }}{2}}\\ =\frac{\mathrm{\pi }}{2}-0\\ =\frac{\mathrm{\pi }}{2}\end{array}$

Q.139

${\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\cos }^5\mathrm{x}}{{\sin }^5\mathrm{x}+{\cos }^5\mathrm{x}}\mathrm{dx}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\cos }^5\mathrm{x}}{{\sin }^5\mathrm{x}+{\cos }^5\mathrm{x}}\mathrm{dx}...\left(\mathrm{i}\right)\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\cos }^5(\frac{\mathrm{\pi }}{2}-\mathrm{x})}{{\sin }^5(\frac{\mathrm{\pi }}{2}-\mathrm{x})+{\cos }^5(\frac{\mathrm{\pi }}{2}-\mathrm{x})}\mathrm{dx}\left[\because {\text{P}}_4:{\int }_0^{\text{a}}\text{f}\left(\text{x}\right)\mathrm{dx}={\int }_0^{\text{a}}\text{f}\left(\text{a}-\text{x}\right)\mathrm{dx}\right]\\ \mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\sin }^5\mathrm{x}}{{\cos }^5\mathrm{x}+{\sin }^5\mathrm{x}}\mathrm{dx}...\left(\mathrm{ii}\right)\\ \text{Adding\hspace{0.17em}\hspace{0.17em}equation}\left(\text{i}\right)\text{\hspace{0.17em}and equation}\left(\text{ii}\right)\text{, we get}\\ 2\mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\cos }^5\mathrm{x}}{{\sin }^5\mathrm{x}+{\cos }^5\mathrm{x}}\mathrm{dx}+{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\sin }^5\mathrm{x}}{{\cos }^5\mathrm{x}+{\sin }^5\mathrm{x}}\mathrm{dx}\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\cos }^5\mathrm{x}+{\sin }^5\mathrm{x}}{{\sin }^5\mathrm{x}+{\cos }^5\mathrm{x}}\mathrm{dx}\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}1.\mathrm{dx}\\ ={{\left[\mathrm{x}\right]}_0}^{\frac{\mathrm{\pi }}{2}}\\ =\frac{\mathrm{\pi }}{2}-0\\ =\frac{\mathrm{\pi }}{2}\\ \mathrm{I}=\frac{\mathrm{\pi }}{4}\end{array}$

Q.140

${\int }_{–5}^5|\mathrm{x}+2|\mathrm{dx}$

Ans.

$\begin{array}{l}\text{Let \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} I}={\int }_{–5}^5|\mathrm{x}+2|\mathrm{dx}\\ \text{Since},\text{}(\mathrm{x}+2)\le 0\text{on\hspace{0.17em}}[-5,-2]\text{and}(\mathrm{x}+2)\ge 0\text{on\hspace{0.17em}}[-2,5].\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I}={\int }_{–5}^{-2}-(\mathrm{x}+2)\mathrm{dx}+{\int }_{–2}^5(\mathrm{x}+2)\mathrm{dx}\\ ={{[-\frac{{\mathrm{x}}^2}{2}-2\mathrm{x}]}_{-5}}^{-2}+{{[\frac{{\mathrm{x}}^2}{2}+2\mathrm{x}]}_{-2}}^5\\ =-[\frac{{(-2)}^2}{2}+2(-2)-\frac{{(-5)}^2}{2}-2(-5)]+[\frac{5^2}{2}+2\left(5\right)-\frac{{(-2)}^2}{2}-2(-2)]\\ =-[2-4-\frac{25}{2}+10]+[\frac{25}{2}+10-2+4]\\ =-8+\frac{25}{2}+\frac{25}{2}+12\\ =29\end{array}$

Q.141

${\int }_2^8|\mathrm{x}-5|\mathrm{dx}$

Ans.

$\begin{array}{l}\text{Let I}={\int }_2^8|\mathrm{x}-5|\mathrm{dx}\\ \text{Since},\text{}(\mathrm{x}-5)\le 0\text{on\hspace{0.17em}}[2,5]\text{and}(\mathrm{x}-5)\ge 0\text{on\hspace{0.17em}}[5,8].\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}I}={\int }_2^5-(\mathrm{x}-5)\mathrm{dx}+{\int }_5^8(\mathrm{x}-5)\mathrm{dx}\\ ={{[-\frac{{\mathrm{x}}^2}{2}+5\mathrm{x}]}_2}^5+{{[\frac{{\mathrm{x}}^2}{2}-5\mathrm{x}]}_5}^8\\ =[-\frac{5^2}{2}+5\left(5\right)-\{-\frac{2^2}{2}+5\left(2\right)\}]+[\frac{8^2}{2}-5\left(8\right)-\{\frac{5^2}{2}-5\left(5\right)\}]\\ =[-\frac{25}{2}+25+\frac{4}{2}-10]+[\frac{64}{2}-40-\frac{25}{2}+25]\\ =[\frac{25}{2}-8]+[-8+\frac{25}{2}]\\ =25-16\\ =9\end{array}$

Q.142

${\int }_0^1\mathrm{x}{(1-\mathrm{x})}^{\mathrm{n}}\mathrm{dx}$

Ans.

$\begin{array}{l}\text{Let I}={\int }_0^1\mathrm{x}{(1-\mathrm{x})}^{\mathrm{n}}\mathrm{dx}\\ ={\int }_0^1(1-\mathrm{x}){\{1-(1-\mathrm{x})\}}^{\mathrm{n}}\mathrm{dx}[{\mathrm{P}}_4:{\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_0^{\mathrm{a}}\mathrm{f}(\mathrm{a}-\mathrm{x})\mathrm{dx}]\\ ={\int }_0^1(1-\mathrm{x}){\mathrm{x}}^{\mathrm{n}}\mathrm{dx}\\ ={\int }_0^1({\mathrm{x}}^{\mathrm{n}}-{\mathrm{x}}^{\mathrm{n}+1})\mathrm{dx}\\ ={{[\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}-\frac{{\mathrm{x}}^{\mathrm{n}+2}}{\mathrm{n}+2}]}_0}^1\\ =[\frac{1^{\mathrm{n}+1}}{\mathrm{n}+1}-\frac{1^{\mathrm{n}+2}}{\mathrm{n}+2}]-[\frac{0^{\mathrm{n}+1}}{\mathrm{n}+1}-\frac{0^{\mathrm{n}+2}}{\mathrm{n}+2}]\\ =\frac{\mathrm{n}+2-\mathrm{n}-1}{(\mathrm{n}+1)(\mathrm{n}+2)}=\frac{1}{(\mathrm{n}+1)(\mathrm{n}+2)}\end{array}$

Q.143

${\int }_0^{\frac{\mathrm{\pi }}{4}}\log (1+\mathrm{tanx})\mathrm{dx}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{4}}\log (1+\mathrm{tanx})\mathrm{dx}\\ ={\int }_0^{\frac{\mathrm{\pi }}{4}}\log \{1+\tan (\frac{\mathrm{\pi }}{4}-\mathrm{x})\}\mathrm{dx}[{\mathrm{P}}_4:{\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_0^{\mathrm{a}}\mathrm{f}(\mathrm{a}-\mathrm{x})\mathrm{dx}]\\ ={\int }_0^{\frac{\mathrm{\pi }}{4}}\log \{1+\frac{\tan \frac{\mathrm{\pi }}{4}-\mathrm{tanx}}{1+\tan \frac{\mathrm{\pi }}{4}\mathrm{tanx}}\}\mathrm{dx}\\ ={\int }_0^{\frac{\mathrm{\pi }}{4}}\log \{1+\frac{1-\mathrm{tanx}}{1+1.\mathrm{tanx}}\}\mathrm{dx}\\ ={\int }_0^{\frac{\mathrm{\pi }}{4}}\log \left\{\frac{1+\mathrm{tanx}+1-\mathrm{tanx}}{1+\mathrm{tanx}}\right\}\mathrm{dx}\\ ={\int }_0^{\frac{\mathrm{\pi }}{4}}\log \left(\frac{2}{1+\mathrm{tanx}}\right)\mathrm{dx}\\ \mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{4}}\log 2\mathrm{dx}-{\int }_0^{\frac{\mathrm{\pi }}{4}}\log (1+\mathrm{tanx})\mathrm{dx}\\ \mathrm{I}=\log 2{{\left[\mathrm{x}\right]}_0}^{\frac{\mathrm{\pi }}{4}}-\mathrm{I}\\ 2\mathrm{I}=[\frac{\mathrm{\pi }}{4}-0]\log 2\\ \mathrm{I}=\frac{\mathrm{\pi }}{8}\log 2\end{array}$

Q.144

${\int }_0^2\mathrm{x}\sqrt{2-\mathrm{x}}\mathrm{dx}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_0^2\mathrm{x}\sqrt{2-\mathrm{x}}\mathrm{dx}\\ \mathrm{I}={\int }_0^2(2-\mathrm{x})\sqrt{2-(2-\mathrm{x})}\mathrm{dx}[{\mathrm{P}}_4:{\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_0^{\mathrm{a}}\mathrm{f}(\mathrm{a}-\mathrm{x})\mathrm{dx}]\\ \mathrm{I}={\int }_0^2(2-\mathrm{x})\sqrt{\mathrm{x}}\mathrm{dx}\\ \mathrm{I}={\int }_0^2(2{\mathrm{x}}^{\frac{1}{2}}-{\mathrm{x}}^{\frac{3}{2}})\mathrm{dx}\\ ={{[2.\frac{{\mathrm{x}}^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{{\mathrm{x}}^{\frac{3}{2}+1}}{\frac{3}{2}+1}]}_0}^2\\ =[2.\frac{2^{\frac{3}{2}}}{\frac{3}{2}}-\frac{2^{\frac{5}{2}}}{\frac{5}{2}}]-[2.\frac{0^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{0^{\frac{3}{2}+1}}{\frac{3}{2}+1}]\\ =[\frac{2^2}{3}.2^{\frac{3}{2}}-\frac{2{.2}^{\frac{5}{2}}}{5}]\\ =\frac{2^{\frac{7}{2}}}{3}-\frac{2^{\frac{7}{2}}}{5}\\ =2^{\frac{7}{2}}(\frac{1}{3}-\frac{1}{5})\\ =2^{\frac{7}{2}}\left(\frac{5-3}{15}\right)\\ =2^{\frac{7}{2}}\left(\frac{2}{15}\right)\\ =\frac{2^{\frac{9}{2}}}{15}\\ \mathrm{I}=\frac{16\sqrt{2}}{15}\end{array}$

Q.145

${\int }_0^{\frac{\mathrm{\pi }}{2}}(2\log \sin \mathrm{x}-\log \sin 2\mathrm{x})\mathrm{dx}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}(2\mathrm{logsinx}–\mathrm{logsin}2\mathrm{x})\mathrm{dx}\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}(2\mathrm{logsinx}-\log 2\mathrm{sinxcosx})\mathrm{dx}\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}(2\mathrm{logsinx}-\log 2-\mathrm{logsinx}-\mathrm{logcosx})\mathrm{dx}\\ \mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}(\mathrm{logsinx}-\log 2-\mathrm{logcosx})\mathrm{dx}...\left(\mathrm{i}\right)\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\{\mathrm{logsin}(\frac{\mathrm{\pi }}{2}-\mathrm{x})-\log 2-\mathrm{logcos}(\frac{\mathrm{\pi }}{2}-\mathrm{x})\}\mathrm{dx}\\ [{\mathrm{P}}_4:{\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_0^{\mathrm{a}}\mathrm{f}(\mathrm{a}-\mathrm{x})\mathrm{dx}]\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\{\mathrm{logcosx}-\log 2-\mathrm{logsinx}\}\mathrm{dx}...\left(\mathrm{ii}\right)\\ \text{Adding equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{ we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2I}={\int }_0^{\frac{\mathrm{\pi }}{2}}(\mathrm{logsinx}-\log 2-\mathrm{logcosx}+\mathrm{logcosx}-\log 2-\mathrm{logsinx})\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2I}={\int }_0^{\frac{\mathrm{\pi }}{2}}\{-2\log 2\}\mathrm{dx}\\ =-2\log 2{{\left[\mathrm{x}\right]}_0}^{\frac{\mathrm{\pi }}{2}}\\ =-2\log 2[\frac{\mathrm{\pi }}{2}-0]\\ \mathrm{I}=-\left(\frac{\mathrm{\pi }}{2}\right)\log 2=\left(\frac{\mathrm{\pi }}{2}\right)\log \frac{1}{2}\end{array}$

Q.146

${\int }_{–\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}{\sin }^2\mathrm{x}\mathrm{dx}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_{–\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}{\sin }^2\mathrm{x}\mathrm{dx}\\ \text{Here},\text{f}\left(\mathrm{x}\right)={\sin }^2\mathrm{x}\\ \text{so},\mathrm{f}(-\mathrm{x})={\sin }^2(-\mathrm{x})\\ ={\sin }^2\mathrm{x}\\ =\mathrm{f}\left(\mathrm{x}\right)\\ \therefore \mathrm{I}=2{\int }_0^{\frac{\mathrm{\pi }}{2}}{\sin }^2\mathrm{x}\mathrm{dx}[{\int }_{-\mathrm{a}}^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=2{\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx},\text{\hspace{0.17em}\hspace{0.17em}if\hspace{0.17em}f}(-\mathrm{x})=\mathrm{f}\left(\mathrm{x}\right)]\\ \mathrm{I}=2{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1-\cos 2\mathrm{x}}{2}\mathrm{dx}\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}(1-\cos 2\mathrm{x})\mathrm{dx}\\ ={{[\mathrm{x}-\frac{\sin 2\mathrm{x}}{2}]}_0}^{\frac{\mathrm{\pi }}{2}}\\ =[\frac{\mathrm{\pi }}{2}-\frac{1}{2}\sin 2\left(\frac{\mathrm{\pi }}{2}\right)]-[0-\frac{1}{2}\sin 2\left(0\right)]\\ =\frac{\mathrm{\pi }}{2}-\frac{1}{2}\mathrm{sin\pi }\\ =\frac{\mathrm{\pi }}{2}\end{array}$

Q.147

${\int }_0^{\mathrm{\pi }}\frac{\mathrm{x}}{1+\mathrm{sinx}}\mathrm{dx}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_0^{\mathrm{\pi }}\frac{\mathrm{x}}{1+\mathrm{sinx}}\mathrm{dx}\\ ={\int }_0^{\mathrm{\pi }}\frac{\mathrm{\pi }-\mathrm{x}}{1+\sin (\mathrm{\pi }-\mathrm{x})}\mathrm{dx}\\ ={\int }_0^{\mathrm{\pi }}\frac{\mathrm{\pi }-\mathrm{x}}{1+\mathrm{sinx}}\mathrm{dx}\\ \mathrm{I}={\int }_0^{\mathrm{\pi }}\frac{\mathrm{\pi }}{1+\mathrm{sinx}}\mathrm{dx}-{\int }_0^{\mathrm{\pi }}\frac{\mathrm{x}}{1+\mathrm{sinx}}\mathrm{dx}\\ \mathrm{I}={\int }_0^{\mathrm{\pi }}\frac{\mathrm{\pi }}{1+\mathrm{sinx}}\mathrm{dx}-\mathrm{I}\\ 2\mathrm{I}={\int }_0^{\mathrm{\pi }}\frac{\mathrm{\pi }(1-\mathrm{sinx})}{(1+\mathrm{sinx})(1-\mathrm{sinx})}\mathrm{dx}\\ ={\int }_0^{\mathrm{\pi }}\frac{\mathrm{\pi }(1-\mathrm{sinx})}{1-{\sin }^2\mathrm{x}}\mathrm{dx}\\ =\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\frac{(1-\mathrm{sinx})}{{\cos }^2\mathrm{x}}\mathrm{dx}\\ =\mathrm{\pi }{\int }_0^{\mathrm{\pi }}({\sec }^2\mathrm{x}-\mathrm{secxtanx})\mathrm{dx}\\ =\mathrm{\pi }{{[\mathrm{tanx}-\mathrm{secx}]}_0}^{\mathrm{\pi }}\\ =\mathrm{\pi }\{(\mathrm{tan\pi }-\mathrm{sec\pi })-(\tan 0-\sec 0)\}\\ =\mathrm{\pi }\{0-(-1)-0+1\}\\ =2\mathrm{\pi }\\ \mathrm{{\rm I}}=\mathrm{\pi }\end{array}$

Q.148

${\int }_{–\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}{\sin }^7\mathrm{x}\mathrm{dx}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_{–\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}{\sin }^7\mathrm{x}\mathrm{dx}\\ \text{Here},\text{f}\left(\mathrm{x}\right)={\sin }^7\mathrm{x}\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}\mathrm{f}(-\mathrm{x})={\sin }^7(-\mathrm{x})\\ =-{\sin }^7\mathrm{x}=-\mathrm{f}\left(\mathrm{x}\right)\\ \text{So},\text{ f}\left(\mathrm{x}\right)\text{\hspace{0.17em}is an odd function, then}{\int }_{-\mathrm{a}}^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=0\\ \therefore \mathrm{I}={\int }_{–\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}{\sin }^7\mathrm{x}\mathrm{dx}=0\end{array}$

Q.149

${\int }_0^{2\mathrm{\pi }}{\cos }^5\mathrm{x}\mathrm{dx}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_0^{2\mathrm{\pi }}{\cos }^5\mathrm{x}\mathrm{dx}\\ \text{since},{\text{cos}}^{\text{5}}(2\mathrm{\pi }-\mathrm{x})={\cos }^5\mathrm{x}\\ \because {\int }_0^{2\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\{\begin{array}{l}2{\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx},\text{\hspace{0.17em}if\hspace{0.17em}}\mathrm{f}(2\mathrm{a}-\mathrm{x})=\mathrm{f}\left(\mathrm{x}\right)\\ 0,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}if\hspace{0.17em}\hspace{0.17em}}\mathrm{f}(2\mathrm{a}-\mathrm{x})=-\mathrm{f}\left(\mathrm{x}\right)\end{array}\\ \therefore {\int }_0^{2\mathrm{\pi }}{\cos }^5\mathrm{x}\mathrm{dx}=2{\int }_0^{\mathrm{\pi }}{\cos }^5\mathrm{x}\mathrm{dx}\\ =0[\because {\cos }^5(\mathrm{\pi }-\mathrm{x})=-{\cos }^5\mathrm{x}]\end{array}$

Q.150

${\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sin \mathrm{x}-\cos \mathrm{x}}{1+\sin \mathrm{x}\cos \mathrm{x}}\mathrm{dx}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\mathrm{sinx}-\mathrm{cosx}}{1+\mathrm{sinxcosx}}\mathrm{dx}...\left(\mathrm{i}\right)\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sin (\frac{\mathrm{\pi }}{2}-\mathrm{x})-\cos (\frac{\mathrm{\pi }}{2}-\mathrm{x})}{1+\sin (\frac{\mathrm{\pi }}{2}-\mathrm{x})\cos (\frac{\mathrm{\pi }}{2}-\mathrm{x})}\mathrm{dx}\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\mathrm{cosx}-\mathrm{sinx}}{1+\mathrm{cosxsinx}}\mathrm{dx}...\left(\mathrm{ii}\right)\\ \text{Adding equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2I}={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\mathrm{sinx}-\mathrm{cosx}+\mathrm{cosx}-\mathrm{sinx}}{1+\mathrm{sinxcosx}}\mathrm{dx}\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{0}{1+\mathrm{sinxcosx}}\mathrm{dx}\\ =0\\ \mathrm{I}=0\\ \therefore {\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\mathrm{sinx}-\mathrm{cosx}}{1+\mathrm{sinxcosx}}\mathrm{dx}=0\end{array}$

Q.151

${\int }_0^{\mathrm{\pi }}\log (1+\cos \mathrm{x})\mathrm{dx}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_0^{\mathrm{\pi }}\log (1+\mathrm{cosx})\mathrm{dx}...\left(\mathrm{i}\right)\\ ={\int }_0^{\mathrm{\pi }}\log \{1+\cos (\mathrm{\pi }-\mathrm{x})\}\mathrm{dx}\\ ={\int }_0^{\mathrm{\pi }}\log (1-\mathrm{cosx})\mathrm{dx}...\left(\mathrm{ii}\right)\\ \text{Adding equation}\left(\mathrm{i}\right)\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}equation}\left(\mathrm{ii}\right),\text{ we get}\\ \text{\hspace{0.17em}\hspace{0.17em}2I}={\int }_0^{\mathrm{\pi }}\log (1+\mathrm{cosx})\mathrm{dx}+{\int }_0^{\mathrm{\pi }}\log (1-\mathrm{cosx})\mathrm{dx}\\ ={\int }_0^{\mathrm{\pi }}\log \left\{(1+\mathrm{cosx})(1-\mathrm{cosx})\right\}\mathrm{dx}\\ ={\int }_0^{\mathrm{\pi }}\log (1-{\cos }^2\mathrm{x})\mathrm{dx}\\ ={\int }_0^{\mathrm{\pi }}{\mathrm{logsin}}^2\mathrm{x}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}2I}=2{\int }_0^{\mathrm{\pi }}\mathrm{logsinx}\mathrm{dx}\\ \mathrm{I}={\int }_0^{\mathrm{\pi }}\mathrm{logsinx}\mathrm{dx}...\left(\mathrm{iii}\right)\\ \mathrm{I}=2{\int }_0^{\frac{\mathrm{\pi }}{2}}\mathrm{logsinx}\mathrm{dx}...\left(\mathrm{iv}\right)[\because \sin (\mathrm{\pi }-\mathrm{x})=\mathrm{sinx}]\\ \mathrm{I}=2{\int }_0^{\frac{\mathrm{\pi }}{2}}\mathrm{logsin}(\frac{\mathrm{\pi }}{2}-\mathrm{x})\mathrm{dx}\\ =2{\int }_0^{\frac{\mathrm{\pi }}{2}}\mathrm{logcosx}\mathrm{dx}...\left(\mathrm{v}\right)\\ \text{Adding equation}\left(\mathrm{iv}\right)\text{\hspace{0.17em}\hspace{0.17em}}\\ \text{2I}=2{\int }_0^{\frac{\mathrm{\pi }}{2}}(\mathrm{logsinx}+\mathrm{logcosx})\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}I}={\int }_0^{\frac{\mathrm{\pi }}{2}}(\mathrm{logsinx}+\mathrm{logcosx})\mathrm{dx}\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\log \left(\frac{2\mathrm{sinxcosx}}{2}\right)\mathrm{dx}\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\mathrm{logsin}2\mathrm{x}\mathrm{dx}-{\int }_0^{\frac{\mathrm{\pi }}{2}}\log 2\mathrm{dx}\\ \text{Let t}=\text{2x}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=2\\ \text{When x}=0,\mathrm{t}=0\text{and when x}=\frac{\mathrm{\pi }}{2},\text{\hspace{0.17em}t}=\mathrm{\pi }.\\ \mathrm{{\rm I}}={\int }_0^{\mathrm{\pi }}\mathrm{logsint}\frac{\mathrm{dt}}{2}-\log 2{{\left[\mathrm{x}\right]}_0}^{\frac{\mathrm{\pi }}{2}}\\ \mathrm{{\rm I}}=\frac{1}{2}{\int }_0^{\mathrm{\pi }}\mathrm{logsinx}\mathrm{dx}-\log 2[\frac{\mathrm{\pi }}{2}-0][\therefore {\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}]\\ \mathrm{I}=\frac{1}{2}\mathrm{I}-\frac{\mathrm{\pi }}{2}\log 2\\ \mathrm{I}-\frac{1}{2}\mathrm{I}=-\frac{\mathrm{\pi }}{2}\log 2\\ \frac{1}{2}\mathrm{I}=-\frac{\mathrm{\pi }}{2}\log 2\\ \mathrm{I}=-\mathrm{\pi log}2\end{array}$

Q.152

${\int }_0^{\mathrm{a}}\frac{\sqrt{\mathrm{x}}}{\sqrt{\mathrm{x}}+\sqrt{\mathrm{x}-\mathrm{a}}}\mathrm{dx}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_0^{\mathrm{a}}\frac{\sqrt{\mathrm{x}}}{\sqrt{\mathrm{x}}+\sqrt{\mathrm{a}-\mathrm{x}}}\mathrm{dx}...\left(\mathrm{i}\right)\\ ={\int }_0^{\mathrm{a}}\frac{\sqrt{\mathrm{a}-\mathrm{x}}}{\sqrt{\mathrm{a}-\mathrm{x}}+\sqrt{\mathrm{a}-(\mathrm{a}-\mathrm{x})}}\mathrm{dx}\\ ={\int }_0^{\mathrm{a}}\frac{\sqrt{\mathrm{a}-\mathrm{x}}}{\sqrt{\mathrm{a}-\mathrm{x}}+\sqrt{\mathrm{x}}}\mathrm{dx}...\left(\mathrm{ii}\right)\\ \text{Adding equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2I}={\int }_0^{\mathrm{a}}\frac{\sqrt{\mathrm{x}}+\sqrt{\mathrm{a}-\mathrm{x}}}{\sqrt{\mathrm{x}}+\sqrt{\mathrm{a}-\mathrm{x}}}\mathrm{dx}\\ ={\int }_0^{\mathrm{a}}1\mathrm{dx}\\ ={{\left[\mathrm{x}\right]}_0}^{\mathrm{a}}\\ \mathrm{I}=\frac{\mathrm{a}}{2}\end{array}\mathrm{a}$

Q.153

${\int }_0^4|\mathrm{x}-1|\mathrm{dx}$

Ans.

$\begin{array}{l}\text{Let I}={\int }_0^4|\mathrm{x}-1|\mathrm{dx}\\ \text{Since},\text{}(\mathrm{x}-1)\le 0\text{on\hspace{0.17em}}[0,2]\text{and}(\mathrm{x}-1)\ge 0\text{on\hspace{0.17em}}[1,4].\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}I}={\int }_0^1-(\mathrm{x}-1)\mathrm{dx}+{\int }_1^4(\mathrm{x}-1)\mathrm{dx}\\ ={{[-\frac{{\mathrm{x}}^2}{2}+\mathrm{x}]}_0}^1+{{[\frac{{\mathrm{x}}^2}{2}-\mathrm{x}]}_1}^4\\ =[-\frac{1^2}{2}+1+0-0]+[\frac{4^2}{2}-4-\frac{1^2}{2}+1]\\ =-\frac{1}{2}+1+8-4-\frac{1}{2}+1\\ =5\\ \therefore {\int }_0^4|\mathrm{x}-1|\mathrm{dx}=5\end{array}$

Q.154

$\begin{array}{l}\text{Show that}{\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}\right)\mathrm{dx}=2{\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx},\text{\hspace{0.17em}if f and g are defined}\\ \text{as}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}(\mathrm{a}-\mathrm{x})\mathrm{}\text{and}\mathrm{g}\left(\mathrm{x}\right)+\mathrm{g}(\mathrm{a}-\mathrm{x})=4\end{array}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}\right)\mathrm{dx}...\left(\mathrm{i}\right)\\ ={\int }_0^{\mathrm{a}}\mathrm{f}(\mathrm{a}-\mathrm{x})\mathrm{g}(\mathrm{a}-\mathrm{x})\mathrm{dx}[\because {\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_0^{\mathrm{a}}\mathrm{f}(\mathrm{a}-\mathrm{x})\mathrm{dx}]\\ ={\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{g}(\mathrm{a}-\mathrm{x})\mathrm{dx}...\left(\mathrm{ii}\right)[\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}(\mathrm{a}-\mathrm{x})]\\ \text{Adding equation}\left(\text{i}\right)\text{and}\left(\text{ii}\right)\text{, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2I}={\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}\right)\mathrm{dx}+{\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{g}(\mathrm{a}-\mathrm{x})\mathrm{dx}\\ ={\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\{\mathrm{g}\left(\mathrm{x}\right)+\mathrm{g}(\mathrm{a}-\mathrm{x})\}\mathrm{dx}\\ ={\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\left(4\right)\mathrm{dx}[\because \mathrm{g}\left(\mathrm{x}\right)+\mathrm{g}(\mathrm{a}-\mathrm{x})=4]\\ \mathrm{I}=2{\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.155

$\begin{array}{l}\mathrm{Choose}\mathrm{the}\mathrm{correct}\mathrm{answer}\\ \text{The value of}{\int }_{-\frac{\text{π}}{2}}^{\frac{\text{π}}{2}}\left({\text{x}}^{\text{3}}\text{+\hspace{0.33em}xcosx}+{\tan }^5\text{x}+1\right)\mathrm{dx}\mathrm{is}\\ \left(\mathrm{A}\right)0\\ \left(\mathrm{B}\right)2\\ \left(\mathrm{C}\right)\mathrm{\pi }\\ \left(\mathrm{D}\right)1\end{array}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}({\mathrm{x}}^3+\mathrm{xcosx}+{\tan }^5\mathrm{x}+1)\mathrm{dx}\\ ={\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}{\mathrm{x}}^3\mathrm{dx}+{\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\mathrm{xcosxdx}+{\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}{\tan }^5\mathrm{xdx}+{\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}1\mathrm{dx}\\ \text{Since},\text{if f}\left(\mathrm{x}\right)\text{is even function then,}{\int }_{-\mathrm{a}}^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=2{\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\\ \text{and if f}\left(\text{x}\right)\text{is odd function then, \hspace{0.17em}\hspace{0.17em}}{\int }_{-\mathrm{a}}^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=0\\ \therefore \mathrm{I}=0+0+0+2{\int }_0^{\frac{\mathrm{\pi }}{2}}1\mathrm{dx}\\ =2{{\left[\mathrm{x}\right]}_0}^{\frac{\mathrm{\pi }}{2}}\\ =2\left(\frac{\mathrm{\pi }}{2}\right)\\ =\mathrm{\pi }\\ \text{Therefore, the correct option is C.}\end{array}$

Q.156

$\begin{array}{l}\mathrm{Choose}\mathrm{the}\mathrm{correct}\mathrm{answer}\\ \text{The value of}{\int }_0^{\frac{\mathrm{\pi }}{2}}\text{log}\left(\frac{\text{4\hspace{0.33em}+\hspace{0.33em}3sinx}}{\text{4\hspace{0.33em}+\hspace{0.33em}3cosx}}\right)\text{dx is}\\ \left(\mathrm{A}\right)2\\ \left(\mathrm{B}\right)\frac{3}{4}\\ \left(\mathrm{C}\right)0\\ \left(\mathrm{D}\right)-2\end{array}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}\log \left(\frac{4+3\mathrm{sinx}}{4+3\mathrm{cosx}}\right)\mathrm{dx}...\left(\mathrm{i}\right)\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\log \left\{\frac{4+3\sin (\frac{\mathrm{\pi }}{2}-\mathrm{x})}{4+3\cos (\frac{\mathrm{\pi }}{2}-\mathrm{x})}\right\}\mathrm{dx}\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\log \left(\frac{4+3\mathrm{cosx}}{4+3\mathrm{sinx}}\right)\mathrm{dx}...\left(\mathrm{ii}\right)\\ \text{Adding equation}\left(\text{i}\right)\text{and equation}\left(\text{ii}\right)\text{, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2I}={\int }_0^{\frac{\mathrm{\pi }}{2}}\{\log \left(\frac{4+3\mathrm{sinx}}{4+3\mathrm{cosx}}\right)+\log \left(\frac{4+3\mathrm{cosx}}{4+3\mathrm{sinx}}\right)\}\mathrm{dx}\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\left\{\log (\frac{4+3\mathrm{sinx}}{4+3\mathrm{cosx}}\times \frac{4+3\mathrm{cosx}}{4+3\mathrm{sinx}})\right\}\mathrm{dx}\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\log \left(1\right)\mathrm{dx}\\ \mathrm{I}=0\\ \text{Therefore, the correction option is C.}\end{array}$

Q.157

$\mathrm{Integrate}\mathrm{the}\mathrm{functions}\frac{1}{\mathrm{x}-{\mathrm{x}}^3}$

Ans.

$\begin{array}{l}\text{We\hspace{0.17em}\hspace{0.17em}have}\\ \frac{1}{\mathrm{x}-{\mathrm{x}}^3}=\frac{1}{\mathrm{x}(1-\mathrm{x})(1+\mathrm{x})}\\ =\frac{\mathrm{A}}{\mathrm{x}}+\frac{\mathrm{B}}{(1-\mathrm{x})}+\frac{\mathrm{C}}{(1+\mathrm{x})}\\ 1=\mathrm{A}(1-\mathrm{x})(1+\mathrm{x})+\mathrm{Bx}(1+\mathrm{x})+\mathrm{Cx}(1-\mathrm{x})...\left(\mathrm{i}\right)\\ \text{Substituting x = 0, 1 and\hspace{0.17em}–1 respectively in equation}\left(\text{i}\right)\text{, we get}\\ \text{A}=1\text{,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}B}=\frac{1}{2}\text{and C}=-\frac{1}{2}\\ \Rightarrow \frac{1}{\mathrm{x}-{\mathrm{x}}^3}=\frac{1}{\mathrm{x}}+\frac{1}{2(1-\mathrm{x})}-\frac{1}{2(1+\mathrm{x})}\\ \therefore \int \frac{1}{\mathrm{x}-{\mathrm{x}}^3}\mathrm{dx}=\int \frac{1}{\mathrm{x}}\mathrm{dx}+\frac{1}{2}\int \frac{1}{1-\mathrm{x}}\mathrm{dx}-\frac{1}{2}\int \frac{1}{1+\mathrm{x}}\mathrm{dx}\\ =\log \left|\mathrm{x}\right|-\frac{1}{2}\log |1-\mathrm{x}|-\frac{1}{2}\log |1+\mathrm{x}|+\mathrm{D}\\ =\frac{1}{2}(2\log \left|\mathrm{x}\right|-\log \left|(1-\mathrm{x})(1+\mathrm{x})\right|)+\mathrm{D}\\ =\frac{1}{2}\log \left|\frac{{\mathrm{x}}^2}{1-{\mathrm{x}}^2}\right|+\mathrm{D}\end{array}$

Q.158

$\mathrm{Integrate}\mathrm{the}\mathrm{functions}\frac{1}{\sqrt{\mathrm{x}+\mathrm{a}}+\sqrt{\mathrm{x}+\mathrm{b}}}$

Ans.

$\begin{array}{l}\text{We have}\frac{1}{\sqrt{\mathrm{x}+\mathrm{a}}+\sqrt{\mathrm{x}+\mathrm{b}}}=\frac{1}{\sqrt{\mathrm{x}+\mathrm{a}}+\sqrt{\mathrm{x}+\mathrm{b}}}\times \frac{\sqrt{\mathrm{x}+\mathrm{a}}-\sqrt{\mathrm{x}+\mathrm{b}}}{\sqrt{\mathrm{x}+\mathrm{a}}-\sqrt{\mathrm{x}+\mathrm{b}}}\\ =\frac{\sqrt{\mathrm{x}+\mathrm{a}}-\sqrt{\mathrm{x}+\mathrm{b}}}{(\mathrm{x}+\mathrm{a})-(\mathrm{x}+\mathrm{b})}\\ =\frac{\sqrt{\mathrm{x}+\mathrm{a}}-\sqrt{\mathrm{x}+\mathrm{b}}}{(\mathrm{a}-\mathrm{b})}\\ \int \frac{1}{\sqrt{\mathrm{x}+\mathrm{a}}+\sqrt{\mathrm{x}+\mathrm{b}}}\mathrm{dx}=\frac{1}{(\mathrm{a}-\mathrm{b})}\int (\sqrt{\mathrm{x}+\mathrm{a}}-\sqrt{\mathrm{x}+\mathrm{b}})\mathrm{dx}\\ =\frac{1}{(\mathrm{a}-\mathrm{b})}\int \{{(\mathrm{x}+\mathrm{a})}^{\frac{1}{2}}-{(\mathrm{x}+\mathrm{b})}^{\frac{1}{2}}\}\mathrm{dx}\\ =\frac{1}{(\mathrm{a}-\mathrm{b})}\{\frac{{(\mathrm{x}+\mathrm{a})}^{\frac{3}{2}}}{\frac{3}{2}}-\frac{{(\mathrm{x}+\mathrm{b})}^{\frac{3}{2}}}{\frac{3}{2}}\}+\mathrm{C}\\ \therefore \int \frac{1}{\sqrt{\mathrm{x}+\mathrm{a}}+\sqrt{\mathrm{x}+\mathrm{b}}}\mathrm{dx}=\frac{2}{3(\mathrm{a}-\mathrm{b})}\{{(\mathrm{x}+\mathrm{a})}^{\frac{3}{2}}-{(\mathrm{x}+\mathrm{b})}^{\frac{3}{2}}\}+\mathrm{C}\end{array}$

Q.159

$\mathrm{Integrate}\mathrm{the}\mathrm{functions}\frac{1}{\mathrm{x}\sqrt{\mathrm{ax}-{\mathrm{x}}^2}}$

Ans.

$\begin{array}{l}\text{We\hspace{0.17em}have,}\\ \int \frac{1}{\mathrm{x}\sqrt{\mathrm{ax}-{\mathrm{x}}^2}}\mathrm{dx}=\int \frac{1}{\mathrm{x}\sqrt{{\mathrm{x}}^2(\frac{\mathrm{a}}{\mathrm{x}}-1)}}\mathrm{dx}\\ =\int \frac{1}{{\mathrm{x}}^2\sqrt{(\frac{\mathrm{a}}{\mathrm{x}}-1)}}\mathrm{dx}\\ =\int \frac{1}{{\mathrm{x}}^2\sqrt{\mathrm{t}}}\times -\frac{{\mathrm{x}}^2}{\mathrm{a}}\mathrm{dt}\left[\begin{array}{l}\text{Let t}=\frac{\mathrm{a}}{\mathrm{x}}-1\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=-\frac{\mathrm{a}}{{\mathrm{x}}^2}\\ \Rightarrow \mathrm{dx}=-\frac{{\mathrm{x}}^2}{\mathrm{a}}\mathrm{dt}\end{array}\right]\\ =-\frac{1}{\mathrm{a}}\int {\mathrm{t}}^{-\frac{1}{2}}\mathrm{dt}=-\frac{1}{\mathrm{a}}\left[\frac{{\mathrm{t}}^{\frac{1}{2}}}{\frac{1}{2}}\right]+\mathrm{C}\\ =-\frac{2}{\mathrm{a}}\sqrt{(\frac{\mathrm{a}}{\mathrm{x}}-1)}+\mathrm{C}\\ =-\frac{2}{\mathrm{a}}\sqrt{\left(\frac{\mathrm{a}-\mathrm{x}}{\mathrm{x}}\right)}+\mathrm{C}\end{array}$

Q.160

$\mathrm{Integrate}\mathrm{the}\mathrm{functions}\frac{1}{{\mathrm{x}}^2{({\mathrm{x}}^4+1)}^{\frac{3}{4}}}$

Ans.

$\begin{array}{l}\text{We have,}\\ \int \frac{1}{{\mathrm{x}}^2{({\mathrm{x}}^4+1)}^{\frac{3}{4}}}\mathrm{dx}=\int \frac{1}{{\mathrm{x}}^2{\left\{{\mathrm{x}}^4(1+\frac{1}{{\mathrm{x}}^4})\right\}}^{\frac{3}{4}}}\mathrm{dx}\\ =\int \frac{1}{{\mathrm{x}}^2.{\mathrm{x}}^3{(1+\frac{1}{{\mathrm{x}}^4})}^{\frac{3}{4}}}\mathrm{dx}\\ =\int \frac{1}{{\mathrm{x}}^5{\mathrm{t}}^{\frac{3}{4}}}\frac{{\mathrm{x}}^5\mathrm{dt}}{-4}\left[\begin{array}{l}\text{Let\hspace{0.33em}\hspace{0.33em}t}=1+\frac{1}{{\text{x}}^4}\\ \Rightarrow \frac{dt}{dx}=-\frac{4}{{\text{x}}^5}\Rightarrow dx=\frac{{\text{x}}^{5}dt}{-4}\end{array}\right]\\ =-\frac{1}{4}\int {\mathrm{t}}^{-\frac{3}{4}}\mathrm{dt}\\ =-\frac{1}{4}\left[\frac{{\mathrm{t}}^{\frac{1}{4}}}{\frac{1}{4}}\right]+\mathrm{C}\\ =-{(1+\frac{1}{{\mathrm{x}}^4})}^{\frac{1}{4}}+\mathrm{C}\end{array}$

Q.161

$\mathrm{Integrate}\mathrm{the}\mathrm{functions}\frac{1}{{\mathrm{x}}^{\frac{1}{2}}+{\mathrm{x}}^{\frac{1}{3}}}$

Ans.

$\begin{array}{l}\text{We have,}\\ \int \frac{1}{{\mathrm{x}}^{\frac{1}{2}}+{\mathrm{x}}^{\frac{1}{3}}}\mathrm{dx}=\int \frac{1}{{\mathrm{x}}^{\frac{1}{3}}({\mathrm{x}}^{\frac{1}{2}-\frac{1}{3}}+1)}\mathrm{dx}\\ =\int \frac{1}{{\mathrm{x}}^{\frac{1}{3}}(1+{\mathrm{x}}^{\frac{1}{6}})}\mathrm{dx}\\ \text{Putting x}={\mathrm{t}}^6\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=6{\mathrm{t}}^5\Rightarrow \mathrm{dx}=6{\mathrm{t}}^5\mathrm{dt}\\ =\int \frac{1}{{\mathrm{t}}^2(1+\mathrm{t})}6{\mathrm{t}}^5\mathrm{dt}\\ =6\int \frac{{\mathrm{t}}^3}{(1+\mathrm{t})}\mathrm{dt}\\ =6\int \{({\mathrm{t}}^2-\mathrm{t}+1)-\frac{1}{\mathrm{t}+1}\}\mathrm{dx}\\ =6[\frac{{\mathrm{t}}^3}{3}-\frac{{\mathrm{t}}^2}{2}+\mathrm{t}-\log (\mathrm{t}+1)]+\mathrm{C}\\ =6[\frac{{\left({\mathrm{x}}^{\frac{1}{6}}\right)}^3}{3}-\frac{{\left({\mathrm{x}}^{\frac{1}{6}}\right)}^2}{2}+\left({\mathrm{x}}^{\frac{1}{6}}\right)-\log \{\left({\mathrm{x}}^{\frac{1}{6}}\right)+1\}]+\mathrm{C}\\ =6[\frac{\sqrt{\mathrm{x}}}{3}-\frac{\sqrt[3]{\mathrm{x}}}{2}+\left({\mathrm{x}}^{\frac{1}{6}}\right)-\log \{\left({\mathrm{x}}^{\frac{1}{6}}\right)+1\}]+\mathrm{C}\\ =6\sqrt{\mathrm{x}}-3\sqrt[3]{\mathrm{x}}+6{\mathrm{x}}^{\frac{1}{6}}-6\log (1+{\mathrm{x}}^{\frac{1}{6}})+\mathrm{C}\end{array}$

Q.162

$\mathrm{Integrate}\mathrm{the}\mathrm{functions}\frac{5\mathrm{x}}{(\mathrm{x}+1)({\mathrm{x}}^2+9)}$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{5\mathrm{x}}{(\mathrm{x}+1)({\mathrm{x}}^2+9)}=\frac{\mathrm{A}}{\mathrm{x}+1}+\frac{\mathrm{Bx}+\mathrm{C}}{{\mathrm{x}}^2+9}...\left(\mathrm{i}\right)\\ \frac{5\mathrm{x}}{(\mathrm{x}+1)({\mathrm{x}}^2+9)}=\frac{\mathrm{A}({\mathrm{x}}^2+9)+(\mathrm{Bx}+\mathrm{C})(\mathrm{x}+1)}{(\mathrm{x}+1)({\mathrm{x}}^2+9)}\\ \Rightarrow 5\mathrm{x}=\mathrm{A}({\mathrm{x}}^2+9)+(\mathrm{Bx}+\mathrm{C})(\mathrm{x}+1)\\ \Rightarrow 5\mathrm{x}={\mathrm{x}}^2(\mathrm{A}+\mathrm{B})+\mathrm{x}(\mathrm{B}+\mathrm{C})+(\mathrm{C}+9\mathrm{A})\\ {\text{Comparing coefficients of x}}^{\text{2}}\text{, x and constant terms, we get}\\ \text{A}+\mathrm{B}=0,\mathrm{B}+\mathrm{C}=5\mathrm{and}\mathrm{C}+9\mathrm{A}=0\\ \text{Solving the above equations, we get}\\ \text{A}=-\frac{1}{2},\mathrm{B}=\frac{1}{2},\mathrm{C}=\frac{9}{2}\\ \text{From equation}\left(\mathrm{i}\right),\text{we have}\\ \frac{5\mathrm{x}}{(\mathrm{x}+1)({\mathrm{x}}^2+9)}=-\frac{1}{2(\mathrm{x}+1)}+\frac{\mathrm{x}+9}{2({\mathrm{x}}^2+9)}\\ \int \frac{5\mathrm{x}}{(\mathrm{x}+1)({\mathrm{x}}^2+9)}\mathrm{dx}=-\frac{1}{2}\int \frac{1}{\mathrm{x}+1}\mathrm{dx}+\frac{1}{2}\int \frac{\mathrm{x}}{{\mathrm{x}}^2+9}\mathrm{dx}+\frac{9}{2}\int \frac{1}{{\mathrm{x}}^2+3^3}\mathrm{dx}\\ =-\frac{1}{2}\log |\mathrm{x}+1|+\frac{1}{2}\times \frac{1}{2}\log |{\mathrm{x}}^2+9|+\frac{9}{2}\times \frac{1}{3}{\tan }^{-1}\left(\frac{\mathrm{x}}{3}\right)+\mathrm{C}\\ =-\frac{1}{2}\log |\mathrm{x}+1|+\frac{1}{4}\log ({\mathrm{x}}^2+9)+\frac{3}{2}{\tan }^{-1}\left(\frac{\mathrm{x}}{3}\right)+\mathrm{C}\end{array}$

Q.163

$\mathrm{Integrate}\mathrm{the}\mathrm{functions}\frac{\mathrm{sinx}}{\sin (\mathrm{x}–\mathrm{a})}$

Ans.

$\begin{array}{l}\text{We have,\hspace{0.17em}\hspace{0.17em}}\int \frac{\mathrm{sinx}}{\sin (\mathrm{x}-\mathrm{a})}\mathrm{dx}\\ \text{Let\hspace{0.17em}\hspace{0.17em}}\mathrm{t}=\mathrm{x}-\mathrm{a}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=1\\ \therefore \int \frac{\mathrm{sinx}}{\sin (\mathrm{x}-\mathrm{a})}\mathrm{dx}=\int \frac{\sin (\mathrm{t}+\mathrm{a})}{\mathrm{sint}}\mathrm{dt}\\ =\int \frac{\mathrm{sintcosa}+\mathrm{sinacost}}{\mathrm{sint}}\mathrm{dt}\\ =\int (\mathrm{cosa}+\mathrm{sinacott})\mathrm{dt}\\ =\mathrm{cosa}\left(\mathrm{t}\right)+\mathrm{sina}\left(\mathrm{logsint}\right)+\mathrm{C}\\ =\mathrm{cosa}(\mathrm{x}-\mathrm{a})+\mathrm{sina}\left\{\mathrm{logsin}(\mathrm{x}-\mathrm{a})\right\}+\mathrm{C}\\ =\mathrm{xcosa}-\mathrm{acosa}+\mathrm{sinalogsin}(\mathrm{x}-\mathrm{a})+\mathrm{C}\\ =\mathrm{xcosa}+\mathrm{sinalogsin}(\mathrm{x}-\mathrm{a})+\mathrm{C}-\mathrm{acosa}\\ =\mathrm{xcosa}+\mathrm{sinalogsin}(\mathrm{x}-\mathrm{a})+{\mathrm{C}}_1\\ [\mathrm{Let}{\mathrm{C}}_1=\mathrm{C}-\mathrm{acosa}]\end{array}$

Q.164

$\mathrm{Integrate}\mathrm{the}\mathrm{functions}\frac{{\mathrm{e}}^{5\mathrm{logx}}-{\mathrm{e}}^{4\mathrm{logx}}}{{\mathrm{e}}^{3\mathrm{logx}}-{\mathrm{e}}^{2\mathrm{logx}}}$

Ans.

$\begin{array}{l}\text{Here},\frac{{\mathrm{e}}^{5\mathrm{logx}}-{\mathrm{e}}^{4\mathrm{logx}}}{{\mathrm{e}}^{3\mathrm{logx}}-{\mathrm{e}}^{2\mathrm{logx}}}=\frac{{\mathrm{e}}^{4\mathrm{logx}}({\mathrm{e}}^{\mathrm{logx}}-1)}{{\mathrm{e}}^{2\mathrm{logx}}({\mathrm{e}}^{\mathrm{logx}}-1)}\\ ={\mathrm{e}}^{2\mathrm{logx}}\\ ={\mathrm{e}}^{{\mathrm{logx}}^2}\\ \therefore \int \frac{{\mathrm{e}}^{5\mathrm{logx}}-{\mathrm{e}}^{4\mathrm{logx}}}{{\mathrm{e}}^{3\mathrm{logx}}-{\mathrm{e}}^{2\mathrm{logx}}}\mathrm{dx}=\int {\mathrm{e}}^{{\mathrm{logx}}^2}\mathrm{dx}\\ =\int {\mathrm{x}}^2\mathrm{dx}\\ =\frac{{\mathrm{x}}^3}{3}+\mathrm{C}\end{array}$

Q.165

$\mathrm{Integrate}\mathrm{the}\mathrm{functions}\frac{\mathrm{cosx}}{\sqrt{4-{\sin }^2\mathrm{x}}}$

Ans.

$\begin{array}{l}\text{Given},\int \frac{\mathrm{cosx}}{\sqrt{4-{\sin }^2\mathrm{x}}}\mathrm{dx}\\ \text{Let \hspace{0.17em}\hspace{0.17em}}\mathrm{t}=\mathrm{sinx}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=\mathrm{cosx}\\ \therefore \int \frac{\mathrm{cosx}}{\sqrt{4-{\sin }^2\mathrm{x}}}\mathrm{dx}=\int \frac{\mathrm{cosx}}{\sqrt{2^2-{\mathrm{t}}^2}}\frac{\mathrm{dt}}{\mathrm{cosx}}\\ ={\sin }^{-1}\left(\frac{\mathrm{t}}{2}\right)+\mathrm{C}\\ ={\sin }^{-1}\left(\frac{\mathrm{sinx}}{2}\right)+\mathrm{C}\end{array}$

Q.166

$\mathrm{Integrate}\mathrm{the}\mathrm{functions}\frac{{\sin }^8\mathrm{x}-{\cos }^8\mathrm{x}}{1-2{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}$

Ans.

$\begin{array}{l}\text{Here},\\ \frac{{\sin }^8\mathrm{x}-{\cos }^8\mathrm{x}}{1-2{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}=\frac{{\left({\sin }^4\mathrm{x}\right)}^2-{\left({\cos }^4\mathrm{x}\right)}^2}{1-2{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}\\ =\frac{({\sin }^4\mathrm{x}+{\cos }^4\mathrm{x})({\sin }^4\mathrm{x}-{\cos }^4\mathrm{x})}{{({\sin }^2\mathrm{x}+{\cos }^2\mathrm{x})}^2-2{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}\\ =\frac{({\sin }^4\mathrm{x}+{\cos }^4\mathrm{x})({\sin }^2\mathrm{x}-{\cos }^2\mathrm{x})({\sin }^2\mathrm{x}+{\cos }^2\mathrm{x})}{{\sin }^4\mathrm{x}+{\cos }^4\mathrm{x}+2{\sin }^2{\mathrm{xcos}}^2\mathrm{x}-2{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}\\ =\frac{({\sin }^4\mathrm{x}+{\cos }^4\mathrm{x})({\sin }^2\mathrm{x}-{\cos }^2\mathrm{x})({\sin }^2\mathrm{x}+{\cos }^2\mathrm{x})}{({\sin }^4\mathrm{x}+{\cos }^4\mathrm{x})}\\ =({\sin }^2\mathrm{x}-{\cos }^2\mathrm{x})\\ =-\cos 2\mathrm{x}\\ \therefore \int \frac{{\sin }^8\mathrm{x}-{\cos }^8\mathrm{x}}{1-2{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}\mathrm{dx}\\ =-\int \cos 2\mathrm{x}\mathrm{dx}\\ =-\left(\frac{\sin 2\mathrm{x}}{2}\right)+\mathrm{C}\end{array}$

Q.167

$\mathrm{Integrate}\mathrm{the}\mathrm{functions}\frac{1}{\cos (\mathrm{x}+\mathrm{a})\cos (\mathrm{x}+\mathrm{b})}$

Ans.

$\begin{array}{l}\mathrm{Since},\frac{1}{\cos (\mathrm{x}+\mathrm{a})\cos (\mathrm{x}+\mathrm{b})}\\ =\frac{1}{\sin (\mathrm{b}-\mathrm{a})}\left(\frac{\sin (\mathrm{b}-\mathrm{a})}{\cos (\mathrm{x}+\mathrm{a})\cos (\mathrm{x}+\mathrm{b})}\right)\\ =\frac{1}{\cos (\mathrm{b}-\mathrm{a})}\left\{\frac{\sin (\mathrm{x}+\mathrm{b}-\mathrm{a}-\mathrm{x})}{\cos (\mathrm{x}+\mathrm{a})\cos (\mathrm{x}+\mathrm{b})}\right\}\\ =\frac{1}{\sin (\mathrm{b}-\mathrm{a})}\left[\frac{\sin \{(\mathrm{x}+\mathrm{b})-(\mathrm{x}+\mathrm{a})\}}{\cos (\mathrm{x}+\mathrm{a})\cos (\mathrm{x}+\mathrm{b})}\right]\\ =\frac{1}{\sin (\mathrm{b}-\mathrm{a})}\left\{\frac{\sin (\mathrm{x}+\mathrm{b})\cos (\mathrm{x}+\mathrm{a})-\cos (\mathrm{x}+\mathrm{b})\sin (\mathrm{x}+\mathrm{a})}{\cos (\mathrm{x}+\mathrm{a})\cos (\mathrm{x}+\mathrm{b})}\right\}\\ =\frac{1}{\sin (\mathrm{b}-\mathrm{a})}\{\tan (\mathrm{x}+\mathrm{b})-\tan (\mathrm{x}+\mathrm{a})\}\\ \therefore \int \frac{1}{\cos (\mathrm{x}+\mathrm{a})\cos (\mathrm{x}+\mathrm{b})}\mathrm{dx}\\ =\frac{1}{\sin (\mathrm{b}-\mathrm{a})}\int \{\tan (\mathrm{x}+\mathrm{b})-\tan (\mathrm{x}+\mathrm{a})\}\mathrm{dx}\\ =\frac{1}{\sin (\mathrm{b}-\mathrm{a})}\{-\mathrm{logcos}(\mathrm{x}+\mathrm{b})+\mathrm{logcos}(\mathrm{x}+\mathrm{a})\}+\mathrm{C}\\ =\frac{1}{\sin (\mathrm{b}-\mathrm{a})}\log \left|\frac{\cos (\mathrm{x}+\mathrm{a})}{\cos (\mathrm{x}+\mathrm{b})}\right|+\mathrm{C}\end{array}$

Q.168

$\mathrm{Integrate}\mathrm{the}\mathrm{functions}\frac{{\mathrm{x}}^3}{\sqrt{1-{\mathrm{x}}^8}}$

Ans.

$\begin{array}{l}\text{Since},\frac{{\mathrm{x}}^3}{\sqrt{1-{\mathrm{x}}^8}}=\frac{{\mathrm{x}}^3}{\sqrt{1-{\left({\mathrm{x}}^4\right)}^2}}\\ \text{Let\hspace{0.17em}\hspace{0.17em}}\mathrm{t}={\mathrm{x}}^4\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=4{\mathrm{x}}^3\Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{4{\mathrm{x}}^3}\\ \therefore \int \frac{{\mathrm{x}}^3}{\sqrt{1-{\mathrm{x}}^8}}\mathrm{dx}=\int \frac{{\mathrm{x}}^3}{\sqrt{1-{\mathrm{t}}^2}}\frac{\mathrm{dt}}{4{\mathrm{x}}^3}\\ =\frac{1}{4}\int \frac{1}{\sqrt{1-{\mathrm{t}}^2}}\mathrm{dt}\\ =\frac{1}{4}{\sin }^{-1}\mathrm{t}+\mathrm{C}\\ =\frac{1}{4}{\sin }^{-1}{\mathrm{x}}^4+\mathrm{C}\end{array}$

Q.169

$\mathrm{Integrate}\mathrm{the}\mathrm{functions}\frac{{\mathrm{e}}^{\mathrm{x}}}{(1+{\mathrm{e}}^{\mathrm{x}})(2+{\mathrm{e}}^{\mathrm{x}})}$

Ans.

$\begin{array}{l}\text{Given},\int \frac{{\mathrm{e}}^{\mathrm{x}}}{(1+{\mathrm{e}}^{\mathrm{x}})(2+{\mathrm{e}}^{\mathrm{x}})}\mathrm{dx}\\ \text{Let\hspace{0.17em}\hspace{0.17em}}\mathrm{t}={\mathrm{e}}^{\mathrm{x}}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}={\mathrm{e}}^{\mathrm{x}}\Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{{\mathrm{e}}^{\mathrm{x}}}\\ \therefore \int \frac{{\mathrm{e}}^{\mathrm{x}}}{(1+{\mathrm{e}}^{\mathrm{x}})(2+{\mathrm{e}}^{\mathrm{x}})}\mathrm{dx}=\int \frac{{\mathrm{e}}^{\mathrm{x}}}{(1+\mathrm{t})(2+\mathrm{t})}\frac{\mathrm{dt}}{{\mathrm{e}}^{\mathrm{x}}}\\ =\int (\frac{1}{2+\mathrm{t}}-\frac{1}{1+\mathrm{t}})\mathrm{dt}\\ =\int \frac{1}{2+\mathrm{t}}\mathrm{dt}-\int \frac{1}{1+\mathrm{t}}\mathrm{dt}\\ =\log |2+\mathrm{t}|-\log |1+\mathrm{t}|+\mathrm{C}\\ =\log \left|\frac{2+\mathrm{t}}{1+\mathrm{t}}\right|+\mathrm{C}\\ =\log \left|\frac{2+{\mathrm{e}}^{\mathrm{x}}}{1+{\mathrm{e}}^{\mathrm{x}}}\right|+\mathrm{C}\end{array}$

Q.170

$\mathrm{Integrate}\mathrm{the}\mathrm{functions}\frac{1}{({\mathrm{x}}^2+1)({\mathrm{x}}^2+4)}$

Ans.

$\begin{array}{l}\mathrm{Let}\frac{1}{({\mathrm{x}}^2+1)({\mathrm{x}}^2+4)}=\frac{\mathrm{Ax}+\mathrm{B}}{({\mathrm{x}}^2+1)}+\frac{\mathrm{Cx}+\mathrm{D}}{({\mathrm{x}}^2+4)}\\ 1=(\mathrm{Ax}+\mathrm{B})({\mathrm{x}}^2+4)+(\mathrm{Cx}+\mathrm{D})({\mathrm{x}}^2+1)\\ 1={\mathrm{x}}^3(\mathrm{A}+\mathrm{C})+{\mathrm{x}}^2(\mathrm{B}+\mathrm{D})+\mathrm{x}(4\mathrm{A}+\mathrm{C})+(4\mathrm{B}+\mathrm{D})\\ {\text{Comparing coefficients of x}}^{\text{2}}\text{, x and constant both sides, we get}\\ \text{A}+\text{C}=\text{0,\hspace{0.17em}\hspace{0.17em}B}+\mathrm{D}=0,4\mathrm{A}+\mathrm{C}=0\mathrm{and}4\mathrm{B}+\mathrm{D}=1\\ \text{Solving the above equations, we get}\\ \text{A}=0,\mathrm{B}=\frac{1}{3},\mathrm{C}=0\text{and D}=-\frac{1}{3}\\ \therefore \frac{1}{({\mathrm{x}}^2+1)({\mathrm{x}}^2+4)}=\frac{1}{3({\mathrm{x}}^2+1)}-\frac{1}{3({\mathrm{x}}^2+4)}\\ \Rightarrow \int \frac{1}{({\mathrm{x}}^2+1)({\mathrm{x}}^2+4)}\mathrm{dx}=\frac{1}{3}\int \frac{1}{({\mathrm{x}}^2+1)}\mathrm{dx}-\frac{1}{3}\int \frac{1}{({\mathrm{x}}^2+4)}\mathrm{dx}\\ =\frac{1}{3}{\tan }^{-1}\left(\mathrm{x}\right)-\frac{1}{3\times 2}{\tan }^{-1}\left(\frac{\mathrm{x}}{2}\right)+\mathrm{C}\\ =\frac{1}{3}{\tan }^{-1}\left(\mathrm{x}\right)-\frac{1}{6}{\tan }^{-1}\left(\frac{\mathrm{x}}{2}\right)+\mathrm{C}\end{array}$

Q.171

$\mathrm{Integrate}\mathrm{the}\mathrm{functions}{\cos }^3\mathrm{x}{}_{\mathrm{e}}\mathrm{logsinx}$

Ans.

$\begin{array}{l}{\cos }^3\mathrm{x}{\mathrm{e}}^{\mathrm{logsinx}}={\cos }^3\mathrm{xsinx}[\because {\mathrm{e}}^{\mathrm{logx}}=\mathrm{x}]\\ \therefore \int {\cos }^3\mathrm{x}{\mathrm{e}}^{\mathrm{logsinx}}\mathrm{dx}=\int {\cos }^3\mathrm{xsinx}\mathrm{dx}\\ =-\int {\mathrm{t}}^3\mathrm{sinx}\frac{\mathrm{dt}}{\mathrm{sinx}}\left[\mathrm{Let}\mathrm{t}=\mathrm{cosx}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=-\mathrm{sinx}\right]\\ =-\int {\mathrm{t}}^3\mathrm{dt}\\ =-\frac{{\mathrm{t}}^4}{4}+\mathrm{C}\\ =-\int {\mathrm{t}}^3\mathrm{dt}\\ =-\frac{{\cos }^4\mathrm{x}}{4}+\mathrm{C}\end{array}$

Q.172

$\mathrm{Integrate}\mathrm{the}\mathrm{functions}{\mathrm{e}}^3\mathrm{logx}{\left({\mathrm{x}}^4+1\right)}^{-1}$

Ans.

$\begin{array}{l}{\mathrm{e}}^{3\mathrm{logx}}{({\mathrm{x}}^4+1)}^{-1}={\mathrm{e}}^{{\mathrm{logx}}^3}{({\mathrm{x}}^4+1)}^{-1}\\ =\frac{{\mathrm{x}}^3}{{\mathrm{x}}^4+1}[\because {\mathrm{e}}^{\mathrm{logx}}=\mathrm{x}]\\ \therefore \int {\mathrm{e}}^{3\mathrm{logx}}{({\mathrm{x}}^4+1)}^{-1}\mathrm{dx}=\int \frac{{\mathrm{x}}^3}{{\mathrm{x}}^4+1}\mathrm{dx}\\ =\int \frac{{\mathrm{x}}^3}{\mathrm{t}}\frac{\mathrm{dt}}{4{\mathrm{x}}^3}\left[\because \text{Let\hspace{0.17em}\hspace{0.17em}}\mathrm{t}={\mathrm{x}}^4+1\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=4{\mathrm{x}}^3\right]\\ =\int \frac{1}{\mathrm{t}}\mathrm{dt}\\ =\mathrm{logt}+\mathrm{C}\\ =\log ({\mathrm{x}}^4+1)+\mathrm{C}\end{array}$

Q.173

$\mathrm{Integrate}\mathrm{the}\mathrm{functions}\mathrm{f}‘\left(\mathrm{ax}+\mathrm{b}\right){\left[\mathrm{f}\left(\mathrm{ax}+\mathrm{b}\right)\right]}^{\mathrm{n}}$

Ans.

$\begin{array}{l}\int \mathrm{f}’(\mathrm{ax}+\mathrm{b}){\left[\mathrm{f}(\mathrm{ax}+\mathrm{b})\right]}^{\mathrm{n}}\mathrm{dx}\\ \mathrm{Let}\mathrm{t}=\mathrm{f}(\mathrm{ax}+\mathrm{b})\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=\mathrm{af}‘(\mathrm{ax}+\mathrm{b})\\ \therefore \int \mathrm{f}’(\mathrm{ax}+\mathrm{b}){\left[\mathrm{f}(\mathrm{ax}+\mathrm{b})\right]}^{\mathrm{n}}\mathrm{dx}=\int \mathrm{f}’(\mathrm{ax}+\mathrm{b}){\mathrm{t}}^{\mathrm{n}}\frac{\mathrm{dt}}{\mathrm{af}‘(\mathrm{ax}+\mathrm{b})}\\ =\frac{1}{\mathrm{a}}\int {\mathrm{t}}^{\mathrm{n}}\mathrm{dx}\\ =\frac{1}{\mathrm{a}}\frac{{\mathrm{t}}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{C}\\ =\frac{1}{\mathrm{a}}\frac{{\left[\mathrm{f}(\mathrm{ax}+\mathrm{b})\right]}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{C}\end{array}$

Q.174

$\mathrm{Integrate}\mathrm{the}\mathrm{functions}\frac{1}{\sqrt{{\sin }^3\mathrm{xsin}(\mathrm{x}+\mathrm{\alpha })}}$

Ans.

$\begin{array}{l}\int \frac{1}{\sqrt{{\sin }^3\mathrm{xsin}(\mathrm{x}+\mathrm{\alpha })}}\mathrm{dx}=\int \frac{1}{\sqrt{{\sin }^3\mathrm{x}(\mathrm{sinxcos\alpha }+\mathrm{cosxsin\alpha })}}\mathrm{dx}\\ =\int \frac{1}{\sqrt{{\sin }^4\mathrm{x}(\mathrm{cos\alpha }+\mathrm{cotxsin\alpha })}}\mathrm{dx}\\ =\int \frac{{\mathrm{cosec}}^2\mathrm{x}}{\sqrt{(\mathrm{cos\alpha }+\mathrm{cotxsin\alpha })}}\mathrm{dx}\\ \left[\mathrm{Let}\mathrm{t}=\mathrm{cos\alpha }+\mathrm{cotxsin\alpha }\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=-{\mathrm{cosec}}^2\mathrm{x}.\mathrm{sin\alpha }\right]\\ =\int \frac{{\mathrm{cosec}}^2\mathrm{x}}{\sqrt{\mathrm{t}}}\frac{\mathrm{dt}}{-{\mathrm{cosec}}^2\mathrm{x}.\mathrm{sin\alpha }}\\ =-\frac{1}{\mathrm{sin\alpha }}\int {\mathrm{t}}^{-\frac{1}{2}}\mathrm{dt}\\ =-\frac{1}{\mathrm{sin\alpha }}\left[\frac{{\mathrm{t}}^{\frac{1}{2}}}{\frac{1}{2}}\right]+\mathrm{C}\\ =-\frac{2}{\mathrm{sin\alpha }}\left[\sqrt{\mathrm{cos\alpha }+\mathrm{cotxsin\alpha }}\right]+\mathrm{C}\\ =-\frac{2}{\mathrm{sin\alpha }}\left[\sqrt{\mathrm{cos\alpha }+\frac{\mathrm{cosx}}{\mathrm{sinx}}\mathrm{sin\alpha }}\right]+\mathrm{C}\\ =-\frac{2}{\mathrm{sin\alpha }}\left[\sqrt{\frac{\mathrm{sinxcos\alpha }+\mathrm{cosxsin\alpha }}{\mathrm{sinx}}}\right]+\mathrm{C}\\ =-\frac{2}{\mathrm{sin\alpha }}\left[\sqrt{\frac{\sin (\mathrm{x}+\mathrm{\alpha })}{\mathrm{sinx}}}\right]+\mathrm{C}\end{array}$

Q.175

$$\begin{gathered} \mathrm{Integrate}\mathrm{the}\mathrm{functions} \\ \frac{{\text{sin}}^{\text{-1}}\sqrt{\text{x}}-{\text{\hspace{0.17em}cos}}^{–\text{1}}\sqrt{\text{x}}}{{\text{sin}}^{\text{-1}}\sqrt{\text{x}}{\text{\hspace{0.17em}+\hspace{0.17em}cos}}^{–\text{1}}\sqrt{\text{x}}}\text{,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}\in \left[\text{0,\hspace{0.17em}1}\right] \end{gathered}$$

Ans.

$\begin{array}{l}\frac{{\sin }^{–1}\sqrt{\mathrm{x}}-{\cos }^{–1}\sqrt{\mathrm{x}}}{{\sin }^{–1}\sqrt{\mathrm{x}}+{\cos }^{–1}\sqrt{\mathrm{x}}},\mathrm{x}\in [0,1]\\ =\frac{{\sin }^{–1}\sqrt{\mathrm{x}}-{\cos }^{–1}\sqrt{\mathrm{x}}}{\frac{\mathrm{\pi }}{2}}[\because {\sin }^{–1}\sqrt{\mathrm{x}}+{\cos }^{–1}\sqrt{\mathrm{x}}=\frac{\mathrm{\pi }}{2}]\\ =\frac{2}{\mathrm{\pi }}(\frac{\mathrm{\pi }}{2}-{\cos }^{–1}\sqrt{\mathrm{x}}-{\cos }^{–1}\sqrt{\mathrm{x}})\\ =\frac{2}{\mathrm{\pi }}(\frac{\mathrm{\pi }}{2}-2{\cos }^{–1}\sqrt{\mathrm{x}})\\ \therefore \int \frac{{\sin }^{–1}\sqrt{\mathrm{x}}-{\cos }^{–1}\sqrt{\mathrm{x}}}{{\sin }^{–1}\sqrt{\mathrm{x}}+{\cos }^{–1}\sqrt{\mathrm{x}}}\mathrm{dx}\\ =\frac{2}{\mathrm{\pi }}\int (\frac{\mathrm{\pi }}{2}-2{\cos }^{–1}\sqrt{\mathrm{x}})\mathrm{dx}\\ =\int 1\mathrm{dx}-\frac{4}{\mathrm{\pi }}\int {\cos }^{–1}\sqrt{\mathrm{x}}\mathrm{dx}\\ =\mathrm{x}-\frac{4}{\mathrm{\pi }}\int {\cos }^{–1}\mathrm{t}.2\mathrm{tdt}\left[\mathrm{Let}\mathrm{t}=\sqrt{\mathrm{x}}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=\frac{1}{2\sqrt{\mathrm{x}}}=\frac{1}{2\mathrm{t}}\right]\\ =\mathrm{x}-\frac{8}{\mathrm{\pi }}\int {\mathrm{tcos}}^{–1}\mathrm{t}\mathrm{dt}\\ =\mathrm{x}-\frac{8}{\mathrm{\pi }}[{\cos }^{–1}\mathrm{t}\int \mathrm{t}\mathrm{dt}-\int \{\frac{\mathrm{d}}{\mathrm{dt}}{\cos }^{–1}\mathrm{t}\int \mathrm{t}\mathrm{dt}\}\mathrm{dt}]\\ =\mathrm{x}-\frac{8}{\mathrm{\pi }}[{\cos }^{–1}\mathrm{t}.\frac{{\mathrm{t}}^2}{2}-\int \{-\frac{1}{\sqrt{1-{\mathrm{t}}^2}}.\frac{{\mathrm{t}}^2}{2}\}\mathrm{dt}]\\ =\mathrm{x}-\frac{8}{\mathrm{\pi }}[{\cos }^{–1}\mathrm{t}.\frac{{\mathrm{t}}^2}{2}+\frac{1}{2}\int \left\{\frac{{\mathrm{t}}^2}{\sqrt{1-{\mathrm{t}}^2}}\right\}\mathrm{dt}]\\ =\mathrm{x}-\frac{8}{\mathrm{\pi }}[{\cos }^{–1}\mathrm{t}.\frac{{\mathrm{t}}^2}{2}-\frac{1}{2}\int \left\{\frac{1-{\mathrm{t}}^2-1}{\sqrt{1-{\mathrm{t}}^2}}\right\}\mathrm{dt}]\\ =\mathrm{x}-\frac{8}{\mathrm{\pi }}[\frac{1}{2}{\mathrm{t}}^2{\cos }^{–1}\mathrm{t}-\frac{1}{2}\int \{\frac{1-{\mathrm{t}}^2}{\sqrt{1-{\mathrm{t}}^2}}-\frac{1}{\sqrt{1-{\mathrm{t}}^2}}\}\mathrm{dt}]\\ =\mathrm{x}-\frac{8}{\mathrm{\pi }}[\frac{1}{2}{\mathrm{t}}^2{\cos }^{–1}\mathrm{t}-\frac{1}{2}\int \{\sqrt{1-{\mathrm{t}}^2}-\frac{1}{\sqrt{1-{\mathrm{t}}^2}}\}\mathrm{dt}]\\ =\mathrm{x}-\frac{8}{\mathrm{\pi }}.\frac{1}{2}{\mathrm{t}}^2{\cos }^{–1}\mathrm{t}+\frac{8}{\mathrm{\pi }}.\frac{1}{2}\left(\begin{array}{l}\frac{\mathrm{t}}{2}\sqrt{1-{\mathrm{t}}^2}+\frac{1}{2}{\sin }^{-1}\mathrm{t}\\ -{\sin }^{-1}\mathrm{t}\end{array}\right)+\mathrm{C}\\ \\ =\mathrm{x}-\frac{4}{\mathrm{\pi }}{\mathrm{xcos}}^{–1}\sqrt{\mathrm{x}}+\frac{4}{\mathrm{\pi }}(\frac{\sqrt{\mathrm{x}}}{2}\sqrt{1-\mathrm{x}}-\frac{1}{2}{\sin }^{-1}\sqrt{\mathrm{x}})+\mathrm{C}\\ =\mathrm{x}-\frac{4}{\mathrm{\pi }}\mathrm{x}(\frac{\mathrm{\pi }}{2}-{\sin }^{-1}\sqrt{\mathrm{x}})+\frac{2}{\mathrm{\pi }}\sqrt{\mathrm{x}-{\mathrm{x}}^2}-\frac{2}{\mathrm{\pi }}{\sin }^{-1}\sqrt{\mathrm{x}}+\mathrm{C}\\ =\mathrm{x}-2\mathrm{x}+\frac{2}{\mathrm{\pi }}\sqrt{\mathrm{x}-{\mathrm{x}}^2}+\frac{4}{\mathrm{\pi }}{\mathrm{xsin}}^{-1}\sqrt{\mathrm{x}}-\frac{2}{\mathrm{\pi }}{\sin }^{-1}\sqrt{\mathrm{x}}+\mathrm{C}\\ =-\mathrm{x}+\frac{2}{\mathrm{\pi }}\sqrt{\mathrm{x}-{\mathrm{x}}^2}+\frac{2}{\mathrm{\pi }}(2\mathrm{x}-1){\sin }^{-1}\sqrt{\mathrm{x}}+\mathrm{C}\end{array}$

Q.176

$\mathrm{Integrate}\mathrm{the}\mathrm{functions}\sqrt{\frac{1-\sqrt{\mathrm{x}}}{1+\sqrt{\mathrm{x}}}}$

Ans.

$\begin{array}{l}\sqrt{\frac{1-\sqrt{\mathrm{x}}}{1+\sqrt{\mathrm{x}}}}\\ \mathrm{Let}\mathrm{x}={\cos }^2\mathrm{t}\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=-2\mathrm{costsint}\\ \int \sqrt{\frac{1-\sqrt{\mathrm{x}}}{1+\sqrt{\mathrm{x}}}}\mathrm{dx}=\int \sqrt{\frac{1-\sqrt{{\cos }^2\mathrm{t}}}{1+\sqrt{{\cos }^2\mathrm{t}}}}\times -2\mathrm{costsint}\mathrm{dt}\\ =\int \sqrt{\frac{1-\mathrm{cost}}{1+\mathrm{cost}}}\times -2\mathrm{costsint}\mathrm{dt}\\ =\int \sqrt{\frac{2{\sin }^2\left(\frac{\mathrm{t}}{2}\right)}{2{\cos }^2\left(\frac{\mathrm{t}}{2}\right)}}\times -2\mathrm{costsint}\mathrm{dt}\\ =\int \frac{\sin \left(\frac{\mathrm{t}}{2}\right)}{\cos \left(\frac{\mathrm{t}}{2}\right)}\times -2\mathrm{cost}\left\{2\sin \left(\frac{\mathrm{t}}{2}\right)\cos \left(\frac{\mathrm{t}}{2}\right)\right\}\mathrm{dt}\\ =-4\int \frac{\sin \left(\frac{\mathrm{t}}{2}\right)}{\cos \left(\frac{\mathrm{t}}{2}\right)}\times \mathrm{cost}\left\{\sin \left(\frac{\mathrm{t}}{2}\right)\cos \left(\frac{\mathrm{t}}{2}\right)\right\}\mathrm{dt}\\ =-4\int {\sin }^2\left(\frac{\mathrm{t}}{2}\right)\mathrm{cost}\mathrm{dt}\\ =-4\int {\sin }^2\left(\frac{\mathrm{t}}{2}\right)(2{\cos }^2\left(\frac{\mathrm{t}}{2}\right)-1)\mathrm{dt}\\ =-4\int \{2{\sin }^2\left(\frac{\mathrm{t}}{2}\right){\cos }^2\left(\frac{\mathrm{t}}{2}\right)-{\sin }^2\left(\frac{\mathrm{t}}{2}\right)\}\mathrm{dt}\\ =-2\int 4{\sin }^2\left(\frac{\mathrm{t}}{2}\right){\cos }^2\left(\frac{\mathrm{t}}{2}\right)\mathrm{dt}+4\int {\sin }^2\left(\frac{\mathrm{t}}{2}\right)\mathrm{dt}\\ =-2\int {\sin }^2\mathrm{t}\mathrm{dt}+2\int (1-\mathrm{cost})\mathrm{dt}\\ =-\int (1-\cos 2\mathrm{t})\mathrm{dt}+2\int (1-\mathrm{cost})\mathrm{dt}\\ =-(\mathrm{t}-\frac{\sin 2\mathrm{t}}{2})+2(\mathrm{t}-\mathrm{sint})+\mathrm{C}\\ =\frac{\sin 2\mathrm{t}}{2}-\mathrm{t}+2\mathrm{t}-2\mathrm{sint}+\mathrm{C}\\ =\mathrm{t}+\frac{2\mathrm{sintcost}}{2}-2\mathrm{sint}+\mathrm{C}\\ =\mathrm{t}+\mathrm{sintcost}-2\mathrm{sint}+\mathrm{C}\\ =\mathrm{t}+\sqrt{1-{\cos }^2\mathrm{t}}\mathrm{cost}-2\sqrt{1-{\cos }^2\mathrm{t}}+\mathrm{C}\\ ={\cos }^{-1}\left(\sqrt{\mathrm{x}}\right)+\sqrt{1-\mathrm{x}}\sqrt{\mathrm{x}}-2\sqrt{1-\mathrm{x}}+\mathrm{C}\\ ={\cos }^{-1}\left(\sqrt{\mathrm{x}}\right)+\sqrt{\mathrm{x}-{\mathrm{x}}^2}-2\sqrt{1-\mathrm{x}}+\mathrm{C}\\ =-2\sqrt{1-\mathrm{x}}+\sqrt{\mathrm{x}-{\mathrm{x}}^2}+{\cos }^{-1}\left(\sqrt{\mathrm{x}}\right)+\mathrm{C}\end{array}$

Q.177

$$\begin{gathered} \mathrm{Integrate}\mathrm{the}\mathrm{functions} \\ \frac{2+\sin 2\mathrm{x}}{(1+\cos 2\mathrm{x})}{\mathrm{e}}^{\mathrm{x}} \end{gathered}$$

Ans.

$\begin{array}{l}\mathrm{Let}\\ \mathrm{I}=\int \frac{2+\sin 2\mathrm{x}}{(1+\cos 2\mathrm{x})}{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}\\ =\int \frac{2+2\mathrm{sinxcosx}}{2{\cos }^2\mathrm{x}}{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}\\ =\int \frac{1+\mathrm{sinxcosx}}{{\cos }^2\mathrm{x}}{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}\\ =\int ({\sec }^2\mathrm{x}+\mathrm{tanx})\mathrm{dx}\\ \mathrm{Let}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{tanx}\Rightarrow \mathrm{f}‘\left(\mathrm{x}\right)={\sec }^2\mathrm{x}\\ \therefore \mathrm{I}=\int [\mathrm{f}\left(\mathrm{x}\right)+\mathrm{t}‘\left(\mathrm{x}\right)]{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}\\ ={\mathrm{e}}^{\mathrm{x}}\mathrm{f}\left(\mathrm{x}\right)+\mathrm{C}\\ ={\mathrm{e}}^{\mathrm{x}}\mathrm{tanx}+\mathrm{C}\end{array}$

Q.178

$$\begin{gathered} \mathrm{Integrate}\mathrm{the}\mathrm{functions} \\ \frac{{\mathrm{x}}^2+\mathrm{x}+1}{{(\mathrm{x}+1)}^2(\mathrm{x}+2)} \end{gathered}$$

Ans.

$\begin{array}{l}\mathrm{Let}\frac{{\mathrm{x}}^2+\mathrm{x}+1}{{(\mathrm{x}+1)}^2(\mathrm{x}+2)}=\frac{\mathrm{A}}{(\mathrm{x}+1)}+\frac{\mathrm{B}}{{(\mathrm{x}+1)}^2}+\frac{\mathrm{C}}{(\mathrm{x}+2)}..\left(\mathrm{i}\right)\\ \Rightarrow {\mathrm{x}}^2+\mathrm{x}+1=\mathrm{A}(\mathrm{x}+1)(\mathrm{x}+2)+\mathrm{B}(\mathrm{x}+2)+\mathrm{C}{(\mathrm{x}+1)}^2\\ \Rightarrow {\mathrm{x}}^2+\mathrm{x}+1=\mathrm{A}({\mathrm{x}}^2+3\mathrm{x}+2)+\mathrm{B}(\mathrm{x}+2)+\mathrm{C}({\mathrm{x}}^2+2\mathrm{x}+1)\\ \Rightarrow {\mathrm{x}}^2+\mathrm{x}+1={\mathrm{x}}^2(\mathrm{A}+\mathrm{C})+\mathrm{x}(3\mathrm{A}+\mathrm{B}+2\mathrm{C})+(2\mathrm{A}+2\mathrm{B}+\mathrm{C})\\ {\text{Comparing coefficients of x}}^{\text{2}}\text{, x and constant from both sides,}\\ \text{we get}\\ \text{A}+\text{C}=\text{1,}3\mathrm{A}+\mathrm{B}+2\mathrm{C}=\text{1\hspace{0.17em} and 2A}+2\mathrm{B}+\mathrm{C}=1\\ \text{Solving the above equations, we get}\\ \text{A}=-2,\mathrm{B}=1,\mathrm{C}=3\\ \text{So, from equation}\left(\text{i}\right)\text{, we have}\\ \frac{{\mathrm{x}}^2+\mathrm{x}+1}{{(\mathrm{x}+1)}^2(\mathrm{x}+2)}=\frac{-2}{(\mathrm{x}+1)}+\frac{1}{{(\mathrm{x}+1)}^2}+\frac{3}{(\mathrm{x}+2)}\\ \therefore \int \frac{{\mathrm{x}}^2+\mathrm{x}+1}{{(\mathrm{x}+1)}^2(\mathrm{x}+2)}\mathrm{dx}=\int \frac{-2}{(\mathrm{x}+1)}\mathrm{dx}+\int \frac{1}{{(\mathrm{x}+1)}^2}\mathrm{dx}+\int \frac{3}{(\mathrm{x}+2)}\mathrm{dx}\\ =-2\log |\mathrm{x}+1|-\frac{1}{(\mathrm{x}+1)}+3\log |\mathrm{x}+2|+\mathrm{C}\\ =-2\log |\mathrm{x}+1|+3\log |\mathrm{x}+2|-\frac{1}{(\mathrm{x}+1)}+\mathrm{C}\end{array}$

Q.179

$$\begin{gathered} \mathrm{Integrate}\mathrm{the}\mathrm{functions} \\ {\tan }^{–1}\sqrt{\frac{1-\mathrm{x}}{1+\mathrm{x}}} \end{gathered}$$

Ans.

$\begin{array}{l}\int {\tan }^{–1}\sqrt{\frac{1-\mathrm{x}}{1+\mathrm{x}}}\mathrm{dx}\\ \mathrm{Putting}\text{ x}=\mathrm{cos\theta }\Rightarrow \frac{\mathrm{dx}}{\mathrm{d\theta }}=-\mathrm{sin\theta }\\ \therefore \int {\tan }^{–1}\sqrt{\frac{1-\mathrm{x}}{1+\mathrm{x}}}\mathrm{dx}=\int {\tan }^{–1}\sqrt{\frac{1-\mathrm{cos\theta }}{1+\mathrm{cos\theta }}}\times -\mathrm{sin\theta }\mathrm{d\theta }\\ =\int {\tan }^{–1}\sqrt{\frac{2{\sin }^2\frac{\mathrm{\theta }}{2}}{2{\cos }^2\frac{\mathrm{\theta }}{2}}}\times -\mathrm{sin\theta }\mathrm{d\theta }\\ =\int {\tan }^{–1}\tan \frac{\mathrm{\theta }}{2}\times -\mathrm{sin\theta }\mathrm{d\theta }\\ =-\int \frac{\mathrm{\theta }}{2}\mathrm{sin\theta }\mathrm{d\theta }\\ =-\frac{1}{2}\{\mathrm{\theta }\int \mathrm{sin\theta }\mathrm{d\theta }-\int (\frac{\mathrm{d}}{\mathrm{d\theta }}\mathrm{\theta }\int \mathrm{sin\theta }\mathrm{d\theta })\mathrm{d\theta }\}\\ =-\frac{1}{2}\{\mathrm{\theta }(-\mathrm{cos\theta })-\int 1(-\mathrm{cos\theta })\mathrm{d\theta }\}\\ =-\frac{1}{2}\{\mathrm{\theta }(-\mathrm{cos\theta })+\mathrm{sin\theta }\}+\mathrm{C}\\ =\frac{1}{2}\mathrm{\theta cos\theta }-\frac{1}{2}\sqrt{1-{\cos }^2\mathrm{\theta }}+\mathrm{C}\\ =\frac{1}{2}({\mathrm{xcos}}^{-1}\mathrm{x}-\sqrt{1-{\mathrm{x}}^2})+\mathrm{C}\end{array}$

Q.180

$$\begin{gathered} \mathrm{Integrate}\mathrm{the}\mathrm{functions} \\ \frac{\sqrt{{\mathrm{x}}^2+1}[\log ({\mathrm{x}}^2+1)-2\mathrm{logx}]}{{\mathrm{x}}^4} \end{gathered}$$

Ans.

$\begin{array}{l}\frac{\sqrt{{\mathrm{x}}^2+1}[\log ({\mathrm{x}}^2+1)-2\mathrm{logx}]}{{\mathrm{x}}^4}=\frac{1}{{\mathrm{x}}^4}\sqrt{{\mathrm{x}}^2+1}\{\log ({\mathrm{x}}^2+1)-2\mathrm{logx}\}\\ =\frac{1}{{\mathrm{x}}^3}\sqrt{\frac{{\mathrm{x}}^2+1}{{\mathrm{x}}^2}}\{\log ({\mathrm{x}}^2+1)-{\mathrm{logx}}^2\}\\ =\frac{1}{{\mathrm{x}}^3}\sqrt{1+\frac{1}{{\mathrm{x}}^2}}\log \left(\frac{{\mathrm{x}}^2+1}{{\mathrm{x}}^2}\right)\\ =\frac{1}{{\mathrm{x}}^3}\sqrt{1+\frac{1}{{\mathrm{x}}^2}}\log (1+\frac{1}{{\mathrm{x}}^2})\\ \text{Let\hspace{0.17em}\hspace{0.17em}}\mathrm{t}=1+\frac{1}{{\mathrm{x}}^2}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=0-\frac{2}{{\mathrm{x}}^3}\Rightarrow \mathrm{dx}=-\frac{{\mathrm{x}}^3}{2}\mathrm{dt}\\ \therefore \int \frac{\sqrt{{\mathrm{x}}^2+1}[\log ({\mathrm{x}}^2+1)-2\mathrm{logx}]}{{\mathrm{x}}^4}\mathrm{dx}=\int \frac{1}{{\mathrm{x}}^3}\sqrt{1+\frac{1}{{\mathrm{x}}^2}}\mathrm{logt}\times -\frac{{\mathrm{x}}^3}{2}\mathrm{dt}\\ =-\frac{1}{2}\int \sqrt{\mathrm{t}}\mathrm{logt}\mathrm{dt}\\ =-\frac{1}{2}\{\mathrm{logt}\int {\mathrm{t}}^{\frac{1}{2}}\mathrm{dt}-\int (\frac{\mathrm{d}}{\mathrm{dt}}\mathrm{logt}\int {\mathrm{t}}^{\frac{1}{2}}\mathrm{dt})\mathrm{dt}\}\\ =-\frac{1}{2}\{\mathrm{logt}.\frac{{\mathrm{t}}^{\frac{3}{2}}}{\frac{3}{2}}-\int \frac{1}{\mathrm{t}}.\frac{{\mathrm{t}}^{\frac{3}{2}}}{\frac{3}{2}}\mathrm{dt}\}\\ =-\frac{1}{2}\{\frac{2}{3}\mathrm{logt}.{\mathrm{t}}^{\frac{3}{2}}-\frac{2}{3}\int {\mathrm{t}}^{\frac{1}{2}}\mathrm{dt}\}\\ =-\frac{1}{2}\{\frac{2}{3}{\mathrm{t}}^{\frac{3}{2}}\mathrm{logt}-\frac{2}{3}.\frac{{\mathrm{t}}^{\frac{3}{2}}}{\frac{3}{2}}\}+\mathrm{C}\\ =-\frac{1}{2}\{\frac{2}{3}{\mathrm{t}}^{\frac{3}{2}}\mathrm{logt}-\frac{4}{9}{\mathrm{t}}^{\frac{3}{2}}\}+\mathrm{C}\\ =-\frac{1}{3}{\mathrm{t}}^{\frac{3}{2}}(\mathrm{logt}-\frac{2}{3})+\mathrm{C}\\ =-\frac{1}{3}{(1+\frac{1}{{\mathrm{x}}^2})}^{\frac{3}{2}}\{\log (1+\frac{1}{{\mathrm{x}}^2})-\frac{2}{3}\}+\mathrm{C}\end{array}$

Q.181

$$\begin{gathered} \mathrm{Evaluate}\mathrm{the}\mathrm{definite}\mathrm{integrals} \\ {\int }_{\frac{\mathrm{\pi }}{2}}^{\mathrm{\pi }}{\mathrm{e}}^{\mathrm{x}}\left(\frac{1–\mathrm{sinx}}{1–\mathrm{cosx}}\right)\mathrm{dx} \end{gathered}$$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_{\frac{\mathrm{\pi }}{2}}^{\mathrm{\pi }}{\mathrm{e}}^{\mathrm{x}}\left(\frac{1-\mathrm{sinx}}{1-\mathrm{cosx}}\right)\mathrm{dx}\\ ={\int }_{\frac{\mathrm{\pi }}{2}}^{\mathrm{\pi }}{\mathrm{e}}^{\mathrm{x}}\left(\frac{1-2\sin \frac{\mathrm{x}}{2}\cos \frac{\mathrm{x}}{2}}{2{\sin }^2\frac{\mathrm{x}}{2}}\right)\mathrm{dx}\\ ={\int }_{\frac{\mathrm{\pi }}{2}}^{\mathrm{\pi }}{\mathrm{e}}^{\mathrm{x}}(\frac{1}{2{\sin }^2\frac{\mathrm{x}}{2}}-\frac{\sin \frac{\mathrm{x}}{2}\cos \frac{\mathrm{x}}{2}}{{\sin }^2\frac{\mathrm{x}}{2}})\mathrm{dx}\\ ={\int }_{\frac{\mathrm{\pi }}{2}}^{\mathrm{\pi }}{\mathrm{e}}^{\mathrm{x}}(\frac{1}{2}{\mathrm{cosec}}^2\frac{\mathrm{x}}{2}-\cot \frac{\mathrm{x}}{2})\mathrm{dx}\\ \text{Let\hspace{0.17em}\hspace{0.17em}}\mathrm{f}\left(\mathrm{x}\right)=-\cot \frac{\mathrm{x}}{2}\Rightarrow \mathrm{f}‘\left(\mathrm{x}\right)=-(-\frac{1}{2}{\mathrm{cosec}}^2\frac{\mathrm{x}}{2})=\frac{1}{2}{\mathrm{cosec}}^2\frac{\mathrm{x}}{2}\\ \therefore \mathrm{I}={\int }_{\frac{\mathrm{\pi }}{2}}^{\mathrm{\pi }}{\mathrm{e}}^{\mathrm{x}}\{\mathrm{f}‘\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{x}\right)\}\mathrm{dx}\\ ={{\left[{\mathrm{e}}^{\mathrm{x}}\mathrm{f}\left(\mathrm{x}\right)\right]}_{\frac{\mathrm{\pi }}{2}}}^{\mathrm{\pi }}=-{{\left[{\mathrm{e}}^{\mathrm{x}}\cot \frac{\mathrm{x}}{2}\right]}_{\frac{\mathrm{\pi }}{2}}}^{\mathrm{\pi }}\\ =-[{\mathrm{e}}^{\mathrm{\pi }}\cot \frac{\mathrm{\pi }}{2}-{\mathrm{e}}^{\frac{\mathrm{\pi }}{2}}\cot \left(\frac{\mathrm{\pi }}{4}\right)]\\ =-[{\mathrm{e}}^{\mathrm{\pi }}.0-{\mathrm{e}}^{\frac{\mathrm{\pi }}{2}}.1]\\ ={\mathrm{e}}^{\frac{\mathrm{\pi }}{2}}\end{array}$

Q.182

$$\begin{gathered} \mathrm{Evaluate}\mathrm{the}\mathrm{definite}\mathrm{integrals} \\ {\int }_0^{\frac{\mathrm{\pi }}{4}}\frac{\mathrm{sinxcosx}}{{\cos }^4\mathrm{x}+{\sin }^4\mathrm{x}}\mathrm{dx} \end{gathered}$$

Ans.

$\begin{array}{l}{\int }_0^{\frac{\mathrm{\pi }}{4}}\frac{\mathrm{sinxcosx}}{{\cos }^4\mathrm{x}+{\sin }^4\mathrm{x}}\mathrm{dx}={\int }_0^{\frac{\mathrm{\pi }}{4}}\frac{\mathrm{sinxcosx}}{{\cos }^4\mathrm{x}(1+\frac{{\sin }^4\mathrm{x}}{{\cos }^4\mathrm{x}})}\mathrm{dx}\\ ={\int }_0^{\frac{\mathrm{\pi }}{4}}\frac{{\mathrm{tanxsec}}^2\mathrm{x}}{(1+{\tan }^4\mathrm{x})}\mathrm{dx}\\ ={\int }_0^{\frac{\mathrm{\pi }}{4}}\frac{{\mathrm{tanxsec}}^2\mathrm{x}}{1+{\left({\tan }^2\mathrm{x}\right)}^2}\mathrm{dx}\\ \text{Let\hspace{0.17em}\hspace{0.17em}}\mathrm{t}={\tan }^2\mathrm{x}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=2{\mathrm{tanxsec}}^2\mathrm{x}\\ \text{When\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=0,\mathrm{t}=0\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}when\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\frac{\mathrm{\pi }}{4},\mathrm{t}=1\\ \therefore {\int }_0^{\frac{\mathrm{\pi }}{4}}\frac{\mathrm{sinxcosx}}{{\cos }^4\mathrm{x}+{\sin }^4\mathrm{x}}\mathrm{dx}={\int }_0^1\frac{{\mathrm{tanxsec}}^2\mathrm{x}}{1+{\mathrm{t}}^2}.\frac{\mathrm{dt}}{2{\mathrm{tanxsec}}^2\mathrm{x}}\\ =\frac{1}{2}{\int }_0^1\frac{1}{1+{\mathrm{t}}^2}.\mathrm{dt}\\ =\frac{1}{2}{{\left[{\tan }^{-1}\mathrm{t}\right]}_0}^1\\ =\frac{1}{2}[{\tan }^{-1}1-{\tan }^{-1}0]\\ =\frac{1}{2}\times \frac{\mathrm{\pi }}{4}\\ =\frac{\mathrm{\pi }}{8}\end{array}$

Q.183

$$\begin{gathered} \mathrm{Evaluate}\mathrm{the}\mathrm{definite}\mathrm{integrals} \\ {\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\cos }^2\mathrm{x}.\mathrm{dx}}{{\cos }^2\mathrm{x}+4{\sin }^2\mathrm{x}} \end{gathered}$$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\cos }^2\mathrm{x}.\mathrm{dx}}{{\cos }^2\mathrm{x}+4{\sin }^2\mathrm{x}}\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\cos }^2\mathrm{x}.\mathrm{dx}}{{\cos }^2\mathrm{x}+4(1-{\cos }^2\mathrm{x})}\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\cos }^2\mathrm{x}.\mathrm{dx}}{4-3{\cos }^2\mathrm{x}}\\ =-\frac{1}{3}{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{-3{\cos }^2\mathrm{x}.\mathrm{dx}}{4-3{\cos }^2\mathrm{x}}\\ =-\frac{1}{3}{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{(4-3{\cos }^2\mathrm{x}-4).\mathrm{dx}}{4-3{\cos }^2\mathrm{x}}\\ =-\frac{1}{3}{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{(4-3{\cos }^2\mathrm{x}).\mathrm{dx}}{4-3{\cos }^2\mathrm{x}}+\frac{4}{3}{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\mathrm{dx}}{4-3{\cos }^2\mathrm{x}}\\ =-\frac{1}{3}{\int }_0^{\frac{\mathrm{\pi }}{2}}\mathrm{dx}+\frac{4}{3}{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\sec }^2\mathrm{xdx}}{4{\sec }^2\mathrm{x}-3}\\ =-\frac{1}{3}{{\left[\mathrm{x}\right]}_0}^{\frac{\mathrm{\pi }}{2}}+\frac{4}{3}{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\sec }^2\mathrm{xdx}}{4({\tan }^2\mathrm{x}+1)-3}\\ =-\frac{1}{3}[\frac{\mathrm{\pi }}{2}-0]+\frac{4}{3}{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\sec }^2\mathrm{xdx}}{4{\tan }^2\mathrm{x}+4-3}\\ =-\frac{\mathrm{\pi }}{6}+\frac{4}{3}{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\sec }^2\mathrm{xdx}}{1+{\left(2\mathrm{tanx}\right)}^2}\\ \text{Let\hspace{0.17em}\hspace{0.17em}}\mathrm{t}=2\mathrm{tanx}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=2{\sec }^2\mathrm{x}\\ \text{When}\mathrm{x}=0,\mathrm{t}=0\text{and when x}=\frac{\mathrm{\pi }}{2},\text{ t}=\mathrm{\infty }\\ \therefore \mathrm{I}=-\frac{\mathrm{\pi }}{6}+\frac{4}{3}{\int }_0^{\mathrm{\infty }}\frac{1}{1+{\mathrm{t}}^2}.\frac{\mathrm{dt}}{2}\\ =-\frac{\mathrm{\pi }}{6}+\frac{4}{3}\times \frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{1+{\mathrm{t}}^2}\\ =-\frac{\mathrm{\pi }}{6}+\frac{2}{3}{{\left[{\tan }^{-1}\mathrm{t}\right]}_0}^{\mathrm{\infty }}\\ =-\frac{\mathrm{\pi }}{6}+\frac{2}{3}[{\tan }^{-1}\mathrm{\infty }-{\tan }^{-1}0]\\ =-\frac{\mathrm{\pi }}{6}+\frac{2}{3}(\frac{\mathrm{\pi }}{2}-0)\\ =-\frac{\mathrm{\pi }}{6}+\frac{\mathrm{\pi }}{3}\\ =\frac{\mathrm{\pi }}{6}\end{array}$

Q.184

$$\begin{gathered} \mathrm{Evaluate}\mathrm{the}\mathrm{definite}\mathrm{integrals} \\ {\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\frac{\mathrm{sinx}+\mathrm{cosx}}{\sqrt{\sin 2\mathrm{x}}}\mathrm{dx} \end{gathered}$$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\frac{\mathrm{sinx}+\mathrm{cosx}}{\sqrt{\sin 2\mathrm{x}}}\mathrm{dx}\\ ={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\frac{\mathrm{sinx}+\mathrm{cosx}}{\sqrt{1-(1-\sin 2\mathrm{x})}}\mathrm{dx}\\ ={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\frac{\mathrm{sinx}+\mathrm{cosx}}{\sqrt{1-({\sin }^2\mathrm{x}+{\cos }^2\mathrm{x}-2\mathrm{sinxcosx})}}\mathrm{dx}\\ ={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\frac{\mathrm{sinx}+\mathrm{cosx}}{\sqrt{1-{(\mathrm{sinx}-\mathrm{cosx})}^2}}\mathrm{dx}\\ \mathrm{Let}\mathrm{t}=\mathrm{sinx}-\mathrm{cosx}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=\mathrm{cosx}+\mathrm{sinx}\\ \mathrm{When}\text{x}=\frac{\mathrm{\pi }}{6},\mathrm{t}=\sin \frac{\mathrm{\pi }}{6}-\cos \frac{\mathrm{\pi }}{6}=\frac{1}{2}-\frac{\sqrt{3}}{2}\\ \mathrm{and}\mathrm{when}\mathrm{x}=\frac{\mathrm{\pi }}{3},\mathrm{t}=\sin \frac{\mathrm{\pi }}{3}-\cos \frac{\mathrm{\pi }}{3}=\frac{\sqrt{3}}{2}-\frac{1}{2}=-(\frac{1}{2}-\frac{\sqrt{3}}{2})\\ \therefore \mathrm{I}={\int }_{-(\frac{1}{2}-\frac{\sqrt{3}}{2})}^{(\frac{\sqrt{3}}{2}-\frac{1}{2})}\frac{1}{\sqrt{1-{\mathrm{t}}^2}}\mathrm{dt}\\ \mathrm{Since},\frac{1}{\sqrt{1-{(-\mathrm{t})}^2}}=\frac{1}{\sqrt{1-{\mathrm{t}}^2}},\text{therefore,}\frac{1}{\sqrt{1-{\mathrm{t}}^2}}\text{is an even function.}\\ \mathrm{Since},{\int }_{-\mathrm{a}}^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=2{\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\\ \therefore \mathrm{I}=2{\int }_0^{(\frac{\sqrt{3}}{2}-\frac{1}{2})}\frac{1}{\sqrt{1-{\mathrm{t}}^2}}\mathrm{dt}\\ =2{\left[{\sin }^{-1}\mathrm{t}\right]}_0^{(\frac{\sqrt{3}}{2}-\frac{1}{2})}\\ =2{\sin }^{-1}\left(\frac{\sqrt{3}-1}{2}\right)\end{array}$

Q.185

$$\begin{gathered} \mathrm{Evaluate}\mathrm{the}\mathrm{definite}\mathrm{integrals} \\ {\int }_0^1\frac{\mathrm{dx}}{\sqrt{1+\mathrm{x}}+\sqrt{\mathrm{x}}} \end{gathered}$$

Ans.

$\begin{array}{l}{\int }_0^1\frac{\mathrm{dx}}{\sqrt{1+\mathrm{x}}+\sqrt{\mathrm{x}}}={\int }_0^1\frac{\sqrt{1+\mathrm{x}}-\sqrt{\mathrm{x}}}{{\left(\sqrt{1+\mathrm{x}}\right)}^2-{\left(\sqrt{\mathrm{x}}\right)}^2}\mathrm{dx}\\ ={\int }_0^1\frac{\sqrt{1+\mathrm{x}}-\sqrt{\mathrm{x}}}{1+\mathrm{x}-\mathrm{x}}\mathrm{dx}\\ ={\int }_0^1\sqrt{1+\mathrm{x}}\mathrm{dx}-{\int }_0^1\sqrt{\mathrm{x}}\mathrm{dx}\\ ={\int }_0^1{(1+\mathrm{x})}^{\frac{1}{2}}\mathrm{dx}-{\int }_0^1{\mathrm{x}}^{\frac{1}{2}}\mathrm{dx}\\ ={[\frac{{(1+\mathrm{x})}^{\frac{3}{2}}}{\frac{3}{2}}-\frac{{\mathrm{x}}^{\frac{3}{2}}}{\frac{3}{2}}]}_0^1\\ =\frac{2}{3}{[{(1+\mathrm{x})}^{\frac{3}{2}}-{\mathrm{x}}^{\frac{3}{2}}]}_0^1\\ =\frac{2}{3}\{{(1+1)}^{\frac{3}{2}}-1^{\frac{3}{2}}-{(1+0)}^{\frac{3}{2}}+0^{\frac{3}{2}}\}\\ =\frac{2}{3}\{{\left(2\right)}^{\frac{3}{2}}-1-1\}\\ =\frac{2}{3}\times {\left(2\right)}^{\frac{3}{2}}=\frac{2^{\frac{5}{2}}}{3}\end{array}$

Q.186

$$\begin{gathered} \text{Evaluate the definite integrals} \\ {\text{∫}}_0^{\frac{\mathrm{\pi }}{4}}\frac{\mathrm{sinx}+\mathrm{cosx}}{9+16\sin 2\mathrm{x}}\mathrm{dx} \end{gathered}$$

Ans.

$\begin{array}{l}{\int }_0^{\frac{\mathrm{\pi }}{4}}\frac{\mathrm{sinx}+\mathrm{cosx}}{9+16\sin 2\mathrm{x}}\mathrm{dx}={\int }_0^{\frac{\mathrm{\pi }}{4}}\frac{\mathrm{sinx}+\mathrm{cosx}}{9+16(1-1+\sin 2\mathrm{x})}\mathrm{dx}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\; \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\int }_0^{\frac{\mathrm{\pi }}{4}}\frac{\mathrm{sinx}+\mathrm{cosx}}{9+16-16({\sin }^2\mathrm{x}+{\cos }^2\mathrm{x}-2\mathrm{sinxcosx})}\mathrm{dx}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\; \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\int }_0^{\frac{\mathrm{\pi }}{4}}\frac{\mathrm{sinx}+\mathrm{cosx}}{25-16{(\mathrm{sinx}-\mathrm{cosx})}^2}\mathrm{dx}\\ \mathrm{Let}\hspace{0.17em}\hspace{0.17em}\mathrm{t}=\mathrm{sinx}-\mathrm{cosx}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=\mathrm{cosx}+\mathrm{sinx}\\ \mathrm{When}\hspace{0.17em}\hspace{0.17em}\mathrm{x}=0,\mathrm{\hspace{0.17em}}\mathrm{t}=-1\mathrm{and}\mathrm{when}\mathrm{x}=\frac{\mathrm{\pi }}{4},\mathrm{\hspace{0.17em}}\mathrm{t}=0\\ \therefore {\int }_0^{\frac{\mathrm{\pi }}{4}}\frac{\mathrm{sinx}+\mathrm{cosx}}{9+16\sin 2\mathrm{x}}\mathrm{dx}=\frac{1}{16}{\int }_{-1}^0\frac{1}{{\left(\frac{5}{4}\right)}^2-{\mathrm{t}}^2}\mathrm{dt}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\; \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{1}{16}\times \frac{1}{2\left(\frac{5}{4}\right)}{\left[\log \left(\frac{\frac{5}{4}+\mathrm{t}}{\frac{5}{4}-\mathrm{t}}\right)\right]}_{-1}^0\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\; \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{1}{40}\{\log \left(\frac{\frac{5}{4}+0}{\frac{5}{4}-0}\right)-\log \left(\frac{\frac{5}{4}-1}{\frac{5}{4}+1}\right)\}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\; \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{1}{40}\{\log \left(1\right)-\log \left(\frac{1}{9}\right)\}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\; \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{1}{40}\log \left(9\right)\end{array}$

Q.187

$$\begin{gathered} \mathrm{Evaluate}\mathrm{the}\mathrm{definite}\mathrm{integrals} \\ {\int }_0^{\frac{\mathrm{\pi }}{2}}\sin 2\mathrm{x}{\tan }^{–1}\left(\mathrm{sinx}\right)\mathrm{dx} \end{gathered}$$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}\sin 2\mathrm{x}{\tan }^{–1}\left(\mathrm{sinx}\right)\mathrm{dx}\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}2\mathrm{sinxcosx}{\tan }^{–1}\left(\mathrm{sinx}\right)\mathrm{dx}\\ \mathrm{When}\mathrm{x}=0,\mathrm{t}=0\text{and when x}=\frac{\mathrm{\pi }}{2},\mathrm{t}=1\\ \therefore \mathrm{I}={\int }_0^{1}2\mathrm{t}{\tan }^{–1}\left(\mathrm{t}\right)\mathrm{dt}\\ =2{\{{\tan }^{-1}\mathrm{t}\int \mathrm{t}\mathrm{dt}-\int (\frac{\mathrm{d}}{\mathrm{dt}}{\tan }^{-1}\mathrm{t}\int \mathrm{t}\mathrm{dt})\mathrm{dt}\}}_0^1\\ =2{\{{\tan }^{-1}\mathrm{t}.\frac{{\mathrm{t}}^2}{2}-\int (\frac{1}{1+{\mathrm{t}}^2}.\frac{{\mathrm{t}}^2}{2})\mathrm{dt}\}}_0^1\\ =2{\{\frac{{\mathrm{t}}^2}{2}.{\tan }^{-1}\mathrm{t}-\frac{1}{2}\int \left(\frac{1+{\mathrm{t}}^2-1}{1+{\mathrm{t}}^2}\right)\mathrm{dt}\}}_0^1\\ \mathrm{Let}\mathrm{t}=\mathrm{sinx}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=\mathrm{cosx}\\ =2{\{\frac{{\mathrm{t}}^2}{2}.{\tan }^{-1}\mathrm{t}-\frac{1}{2}\int (\frac{1+{\mathrm{t}}^2}{1+{\mathrm{t}}^2}-\frac{1}{1+{\mathrm{t}}^2})\mathrm{dt}\}}_0^1\\ =2{\{\frac{{\mathrm{t}}^2}{2}.{\tan }^{-1}\mathrm{t}-\frac{1}{2}\int (1-\frac{1}{1+{\mathrm{t}}^2})\mathrm{dt}\}}_0^1\\ =2{\{\frac{{\mathrm{t}}^2}{2}.{\tan }^{-1}\mathrm{t}-\frac{1}{2}(\mathrm{t}-{\tan }^{-1}\mathrm{t})\}}_0^1\\ =2\{\frac{1^2}{2}.{\tan }^{-1}1-\frac{1}{2}(1-{\tan }^{-1}1)\}-2\{\frac{0^2}{2}.{\tan }^{-1}0-\frac{1}{2}(0-{\tan }^{-1}0)\}\\ =2\{\frac{1}{2}.\frac{\mathrm{\pi }}{4}-\frac{1}{2}(1-\frac{\mathrm{\pi }}{4})\}-0\\ =\frac{\mathrm{\pi }}{4}-1+\frac{\mathrm{\pi }}{4}=\frac{\mathrm{\pi }}{2}-1\end{array}$

Q.188

$$\begin{gathered} \mathrm{Evaluate}\mathrm{the}\mathrm{definite}\mathrm{integrals} \\ {\int }_0^{\mathrm{\pi }}\frac{\mathrm{xtanx}}{\mathrm{secx}+\mathrm{tanx}}\mathrm{dx} \end{gathered}$$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}={\int }_0^{\mathrm{\pi }}\frac{\mathrm{xtanx}}{\mathrm{secx}+\mathrm{tanx}}\mathrm{dx}\\ ={\int }_0^{\mathrm{\pi }}\frac{\mathrm{xsinx}}{1+\mathrm{sinx}}\mathrm{dx}...\left(\mathrm{i}\right)\\ ={\int }_0^{\mathrm{\pi }}\frac{(\mathrm{\pi }-\mathrm{x})\sin (\mathrm{\pi }-\mathrm{x})}{1+\sin (\mathrm{\pi }-\mathrm{x})}\mathrm{dx}[\because {\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_0^{\mathrm{a}}\mathrm{f}(\mathrm{a}-\mathrm{x})]\\ \mathrm{I}={\int }_0^{\mathrm{\pi }}\frac{(\mathrm{\pi }-\mathrm{x})\mathrm{sinx}}{1+\mathrm{sinx}}\mathrm{dx}...\left(\mathrm{ii}\right)\\ \mathrm{Adding}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\mathrm{we}\text{get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2I}=\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\frac{1+\mathrm{sinx}-1}{1+\mathrm{sinx}}\mathrm{dx}\\ =\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\frac{1+\mathrm{sinx}}{1+\mathrm{sinx}}\mathrm{dx}-\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\frac{(1-\mathrm{sinx})}{(1+\mathrm{sinx})(1-\mathrm{sinx})}\mathrm{dx}\\ =\mathrm{\pi }{\int }_0^{\mathrm{\pi }}1\mathrm{dx}-\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\frac{(1-\mathrm{sinx})}{1-{\sin }^2\mathrm{x}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2I}=\mathrm{\pi }{\left[\mathrm{x}\right]}_0^{\mathrm{\pi }}-\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\frac{(1-\mathrm{sinx})}{{\cos }^2\mathrm{x}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2I}=\mathrm{\pi }[\mathrm{\pi }-0]-\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\frac{1}{{\cos }^2\mathrm{x}}\mathrm{dx}+\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\frac{\mathrm{sinx}}{{\cos }^2\mathrm{x}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2I}={\mathrm{\pi }}^2-\mathrm{\pi }{\int }_0^{\mathrm{\pi }}{\sec }^2\mathrm{xdx}+\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\mathrm{tanxsecx}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2I}={\mathrm{\pi }}^2-\mathrm{\pi }{[\mathrm{tanx}+\mathrm{secx}]}_0^{\mathrm{\pi }}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2I}={\mathrm{\pi }}^2-\mathrm{\pi }[\mathrm{tan\pi }+\mathrm{sec\pi }-\tan 0-\sec 0]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2I}={\mathrm{\pi }}^2-\mathrm{\pi }[0-1-\tan 0-1]\\ ={\mathrm{\pi }}^2+2\mathrm{\pi }\\ \mathrm{{\rm I}}=\frac{\mathrm{\pi }}{2}(\mathrm{\pi }+2)\end{array}$

Q.189

$$\begin{gathered} \mathrm{Evaluate}\mathrm{the}\mathrm{definite}\mathrm{integrals} \\ {\int }_1^4[|\mathrm{x}-1|+|\mathrm{x}-2|+|\mathrm{x}-3|]\mathrm{dx} \end{gathered}$$

Ans.

$\begin{array}{l}\mathrm{I}={\int }_1^4[|\mathrm{x}-1|+|\mathrm{x}-2|+|\mathrm{x}-3|]\mathrm{dx}\\ ={\mathrm{I}}_1+{\mathrm{I}}_2+{\mathrm{I}}_3...\left(\mathrm{i}\right)\\ {\mathrm{I}}_1={\int }_1^4|\mathrm{x}-1|\mathrm{dx}\\ ={\int }_1^4(\mathrm{x}-1)\mathrm{dx}[\because (\mathrm{x}-1)\ge 0\text{for 1}\le \text{x}\le \text{4}]\\ ={[\frac{{\mathrm{x}}^2}{2}-\mathrm{x}]}_1^4\\ =(\frac{4^2}{2}-4)-(\frac{1^2}{2}-1)\\ =8-4-\frac{1}{2}+1\\ =4\frac{1}{2}\\ {\mathrm{I}}_2={\int }_1^4|\mathrm{x}-2|\mathrm{dx}\\ \text{Since},(\mathrm{x}-2)\ge 0\mathrm{for}2\le \mathrm{x}\le 4\mathrm{and}(\mathrm{x}-2)\le 0\text{for 1}\le \mathrm{x}\le 2\\ ={\int }_1^2-(\mathrm{x}-2)\mathrm{dx}+{\int }_2^4(\mathrm{x}-2)\mathrm{dx}\\ ={[-\frac{{\mathrm{x}}^2}{2}+2\mathrm{x}]}_1^2+{[\frac{{\mathrm{x}}^2}{2}-2\mathrm{x}]}_2^4\\ =-\frac{2^2}{2}+2\left(2\right)+\frac{1^2}{2}-2\left(1\right)+\frac{4^2}{2}-2\left(4\right)-\frac{2^2}{2}+2\left(2\right)\\ =-2+4+\frac{1}{2}-2+8-8-2+4=\frac{5}{2}\\ {\mathrm{I}}_3={\int }_1^4|\mathrm{x}-3|\mathrm{dx}\\ \text{Since},(\mathrm{x}-3)\ge 0\mathrm{for}3\le \mathrm{x}\le 4\mathrm{and}(\mathrm{x}-3)\le 0\text{for 1}\le \mathrm{x}\le 3\\ ={\int }_1^3-(\mathrm{x}-3)\mathrm{dx}+{\int }_3^4(\mathrm{x}-3)\mathrm{dx}\\ ={[-\frac{{\mathrm{x}}^2}{2}+3\mathrm{x}]}_1^3+{[\frac{{\mathrm{x}}^2}{2}-3\mathrm{x}]}_3^4\\ =-\frac{3^2}{2}+3\left(3\right)+\frac{1^2}{2}-3\left(1\right)+\frac{4^2}{2}-3\left(4\right)-\frac{3^2}{2}+3\left(3\right)\\ =-\frac{9}{2}+9+\frac{1}{2}-3+8-12-\frac{9}{2}+9\\ =26\frac{1}{2}-24=2\frac{1}{2}\\ \text{From equation}\left(\text{i}\right)\text{, we have}\\ \mathrm{I}=4\frac{1}{2}+\frac{5}{2}+2\frac{1}{2}=9\frac{1}{2}\end{array}$

Q.190

$$\begin{gathered} \mathrm{Prove}\mathrm{the}\mathrm{following} \\ {\int }_1^3\frac{\mathrm{dx}}{{\mathrm{x}}^2(\mathrm{x}+1)}\mathrm{dx}=\frac{2}{3}+\log \frac{2}{3} \end{gathered}$$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_1^3\frac{\mathrm{dx}}{{\mathrm{x}}^2(\mathrm{x}+1)}\mathrm{dx}\\ \text{Let\hspace{0.17em}\hspace{0.17em}}\frac{1}{{\mathrm{x}}^2(\mathrm{x}+1)}=\frac{\mathrm{A}}{\mathrm{x}}+\frac{\mathrm{B}}{{\mathrm{x}}^2}+\frac{\mathrm{C}}{(\mathrm{x}+1)}\\ 1=\mathrm{Ax}(\mathrm{x}+1)+\mathrm{B}(\mathrm{x}+1)+{\mathrm{Cx}}^2\\ 1={\mathrm{x}}^2(\mathrm{A}+\mathrm{C})+\mathrm{x}(\mathrm{A}+\mathrm{B})+\mathrm{B}\\ {\text{Equating the coefficients of x}}^{\text{2}}\text{, x and constant term, we get}\\ \mathrm{A}+\mathrm{C}=0,\mathrm{A}+\mathrm{B}=0\mathrm{and}\mathrm{B}=1\\ \text{Solving the above equations, we get}\\ \text{A}=-\text{1, C}=-1\text{and B}=\text{1}\\ \therefore \frac{1}{{\mathrm{x}}^2(\mathrm{x}+1)}=\frac{-1}{\mathrm{x}}+\frac{1}{{\mathrm{x}}^2}+\frac{-1}{(\mathrm{x}+1)}\\ \mathrm{I}={\int }_1^3-\frac{1}{\mathrm{x}}\mathrm{dx}+{\int }_1^3\frac{1}{{\mathrm{x}}^2}\mathrm{dx}+{\int }_1^3\frac{1}{(\mathrm{x}+1)}\mathrm{dx}\\ \mathrm{I}=-{\left[\mathrm{logx}\right]}_1^3-{\left[\frac{1}{\mathrm{x}}\right]}_1^3+{\left[\log (\mathrm{x}+1)\right]}_1^3\\ \mathrm{I}=-[\log 3-\log 1]-[\frac{1}{3}-\frac{1}{1}]+[\log (3+1)-\log (1+1)]\\ =-[\log 3-\log 1]-[-\frac{2}{3}]+[\log 4-\log 2]\\ =-[\log 3-0]+\frac{2}{3}+\log \frac{4}{2}\\ =\frac{2}{3}+\log 2-\log 3\\ =\frac{2}{3}+\log \frac{2}{3}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.191

$$\begin{gathered} \mathrm{Prove}\mathrm{the}\mathrm{following} \\ {\int }_0^1{\mathrm{xe}}^{\mathrm{x}}\mathrm{dx}=1 \end{gathered}$$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_0^1{\mathrm{xe}}^{\mathrm{x}}\mathrm{dx}\\ =\mathrm{x}{\int }_0^1{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}-{\int }_0^1\left(\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{x}{\int }_0^1{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}\right)\mathrm{dx}\\ ={\left[{\mathrm{xe}}^{\mathrm{x}}\right]}_0^1-{\int }_0^1(1.{\mathrm{e}}^{\mathrm{x}})\mathrm{dx}\\ =[1{\mathrm{e}}^1-0.{\mathrm{e}}^0]-{\left[{\mathrm{e}}^{\mathrm{x}}\right]}_0^1\\ =\mathrm{e}-[{\mathrm{e}}^1-{\mathrm{e}}^0]\\ =\mathrm{e}-\mathrm{e}+1\\ =1=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.192

$$\begin{gathered} \mathrm{Prove}\mathrm{the}\mathrm{following} \\ {\int }_{-1}^1{\mathrm{x}}^{17}{\cos }^4\mathrm{x}\mathrm{dx}=0 \end{gathered}$$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_{-1}^1{\mathrm{x}}^{17}{\cos }^4\mathrm{x}\mathrm{dx}\\ \text{Let\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{17}{\cos }^4\mathrm{x}\\ \Rightarrow \mathrm{f}(-\mathrm{x})={(-\mathrm{x})}^{17}{\cos }^4(-\mathrm{x})\\ =-{\mathrm{x}}^{17}{\cos }^4\mathrm{x}=-\mathrm{f}\left(\mathrm{x}\right)\\ \text{Therefore, f(x) is an odd function.}\\ \text{And\hspace{0.17em}\hspace{0.17em}}{\int }_{-\mathrm{a}}^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=0\\ \therefore \mathrm{I}={\int }_{-1}^1{\mathrm{x}}^{17}{\cos }^4\mathrm{x}\mathrm{dx}\\ =0=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.193

$$\begin{gathered} \mathrm{Prove}\mathrm{the}\mathrm{following} \\ {\int }_0^{\frac{\mathrm{\pi }}{2}}{\sin }^3\mathrm{x}\mathrm{dx}=\frac{2}{3} \end{gathered}$$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}{\sin }^3\mathrm{x}\mathrm{dx}\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}{\sin }^2\mathrm{xsinx}\mathrm{dx}\\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}(1-{\cos }^2\mathrm{x})\mathrm{sinx}\mathrm{dx}\\ \mathrm{Let}\mathrm{t}=\mathrm{cosx}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=-\mathrm{sinx}\\ \mathrm{When}\mathrm{x}=0,\mathrm{t}=1\text{and when x}=\frac{\mathrm{\pi }}{2},\mathrm{t}=0\\ \therefore \mathrm{I}={\int }_1^0(1-{\mathrm{t}}^2)\mathrm{sinx}\frac{\mathrm{dt}}{-\mathrm{sinx}}\\ =-{[\mathrm{t}-\frac{{\mathrm{t}}^3}{3}]}_1^0=-[0-0-1+\frac{1}{3}]\\ =\frac{2}{3}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.194

$$\begin{gathered} \mathrm{Prove}\mathrm{the}\mathrm{follwoing} \\ {\int }_0^{\frac{\mathrm{\pi }}{4}}2{\tan }^3\mathrm{x}\mathrm{dx}=1-\log 2 \end{gathered}$$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{4}}2{\tan }^3\mathrm{x}\mathrm{dx}\\ =2{\int }_0^{\frac{\mathrm{\pi }}{4}}{\tan }^2\mathrm{xtanx}\mathrm{dx}\\ =2{\int }_0^{\frac{\mathrm{\pi }}{4}}({\sec }^2\mathrm{x}-1)\mathrm{tanx}\mathrm{dx}\\ =2{\int }_0^{\frac{\mathrm{\pi }}{4}}{\sec }^2\mathrm{xtanx}\mathrm{dx}-2{\int }_0^{\frac{\mathrm{\pi }}{4}}\mathrm{tanx}\mathrm{dx}\\ =2{\left[\frac{{\tan }^2\mathrm{x}}{2}\right]}_0^{\frac{\mathrm{\pi }}{4}}-2{[-\mathrm{logcosx}]}_0^{\frac{\mathrm{\pi }}{4}}\\ =[{\tan }^2\frac{\mathrm{\pi }}{4}-{\tan }^{2}0]+2[\mathrm{logcos}\frac{\mathrm{\pi }}{4}-\mathrm{logcos}0]\\ =[1-0]+2[\log \frac{1}{\sqrt{2}}-\log 1]\\ =1+2[-\frac{1}{2}\log 2-0]\\ =1-\log 2=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.195

$$\begin{gathered} \mathrm{Prove}\mathrm{the}\mathrm{following} \\ {\int }_0^1{\sin }^{–1}\mathrm{x}\mathrm{dx}=\frac{\mathrm{\pi }}{2}-1 \end{gathered}$$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_0^1{\sin }^{–1}\mathrm{x}\mathrm{dx}\\ ={\int }_0^1{\sin }^{–1}\mathrm{x}.1\mathrm{dx}\\ =[{\sin }^{–1}\mathrm{x}\int 1\mathrm{dx}-\int (\frac{\mathrm{d}}{\mathrm{dx}}{\sin }^{-1}\mathrm{x}\int 1\mathrm{dx})\mathrm{dx}]\\ ={[{\sin }^{–1}\mathrm{x}.\mathrm{x}-\int \frac{1}{\sqrt{1-{\mathrm{x}}^2}}.\mathrm{x}\mathrm{dx}]}_0^1\\ ={[{\mathrm{xsin}}^{–1}\mathrm{x}+\frac{1}{2}.2\sqrt{1-{\mathrm{x}}^2}]}_0^1\\ =[1{\sin }^{–1}1+\sqrt{1-1^2}-0{\sin }^{-1}0-\sqrt{1-0^2}]\\ =\frac{\mathrm{\pi }}{2}-1=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.196

$\text{Evaluate\hspace{0.17em}\hspace{0.17em}}{\int }_0^1{\mathrm{e}}^{2–3\mathrm{x}}\mathrm{dx}\text{as a limit of a sum}.$

Ans.

$\begin{array}{l}\text{We\hspace{0.17em} have,}{\int }_0^1{\mathrm{e}}^{2–3\mathrm{x}}\mathrm{dx}\\ \because {\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=(\mathrm{b}-\mathrm{a})\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[\mathrm{f}\left(\mathrm{a}\right)+\mathrm{f}(\mathrm{a}+\mathrm{h})+...+\mathrm{f}(\mathrm{a}+(\mathrm{n}-1)\mathrm{h})],\\ \text{where},\text{h}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}\\ \text{Here},\mathrm{a}=0,\mathrm{b}=1,\mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{2-3\mathrm{x}},\mathrm{h}=\frac{1-0}{\mathrm{n}}=\frac{1}{\mathrm{n}}\\ \therefore {\int }_0^1{\mathrm{e}}^{2-3\mathrm{x}}\mathrm{dx}=(1-0)\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[\mathrm{f}\left(0\right)+\mathrm{f}(0+\mathrm{h})+...+\mathrm{f}(0+(\mathrm{n}-1)\mathrm{h})]\\ =1\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}[{\mathrm{e}}^2+{\mathrm{e}}^{2-3\mathrm{h}}+...+{\mathrm{e}}^{2-(\mathrm{n}-1)\mathrm{h}}]\\ \text{Using the sum to n terms of a G.P., where a}={\mathrm{e}}^2,\text{r}={\mathrm{e}}^{-3\mathrm{h}},\\ \text{we have}\\ \therefore {\int }_0^1{\mathrm{e}}^{2–3\mathrm{x}}\mathrm{dx}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}\left[\frac{{\mathrm{e}}^2({\mathrm{e}}^{-3\mathrm{nh}}-1)}{({\mathrm{e}}^{-3\mathrm{h}}-1)}\right]\\ =\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}\left[\frac{{\mathrm{e}}^2({\mathrm{e}}^{-3\mathrm{n}.\frac{1}{\mathrm{n}}}-1)}{({\mathrm{e}}^{-3.\frac{1}{\mathrm{n}}}-1)}\right]\\ =\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}\left[\frac{{\mathrm{e}}^2({\mathrm{e}}^{-3}-1)}{({\mathrm{e}}^{-\frac{3}{\mathrm{n}}}-1)}\right]\\ =\frac{{\mathrm{e}}^2({\mathrm{e}}^{-3}-1)}{\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\left[\frac{{\mathrm{e}}^{-\frac{3}{\mathrm{n}}}-1}{-\frac{3}{\mathrm{n}}}\right].(-3)}\\ =\frac{{\mathrm{e}}^2({\mathrm{e}}^{-3}-1)}{-3}\\ =-\frac{1}{3}(\frac{1}{\mathrm{e}}-{\mathrm{e}}^2)[\because \underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{h}}-1}{\mathrm{h}}=1]\\ \therefore {\int }_0^1{\mathrm{e}}^{2–3\mathrm{x}}\mathrm{dx}=\frac{1}{3}({\mathrm{e}}^2-\frac{1}{\mathrm{e}})\end{array}$

Q.197

$$\begin{gathered} \mathrm{Choose}\mathrm{the}\mathrm{correct}\mathrm{answer} \\ \begin{array}{l}\int \frac{\mathrm{dx}}{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{–\mathrm{x}}}\text{\hspace{0.17em}is equal to}\\ \left(\text{A}\right){\text{\hspace{0.17em}tan}}^{–1}\left({\mathrm{e}}^{\mathrm{x}}\right)+\mathrm{C}\\ \left(\text{B}\right){\text{\hspace{0.17em}tan}}^{–1}\left({\mathrm{e}}^{–\mathrm{x}}\right)+\mathrm{C}\\ \left(\text{C}\right)\text{\hspace{0.17em}log}({\mathrm{e}}^{\mathrm{x}}–{\mathrm{e}}^{–\mathrm{x}})+\mathrm{C}\\ \left(\text{D}\right)\text{\hspace{0.17em}log}({\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{–\mathrm{x}})+\mathrm{C}\end{array} \end{gathered}$$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int \frac{\mathrm{dx}}{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{–\mathrm{x}}}\\ =\int \frac{{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}}{{\mathrm{e}}^{2\mathrm{x}}+1}\\ \text{Let\hspace{0.17em}}\mathrm{t}={\mathrm{e}}^{\mathrm{x}}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}={\mathrm{e}}^{\mathrm{x}}\\ \therefore \mathrm{I}=\int \frac{\mathrm{dt}}{{\mathrm{t}}^2+1}\\ ={\tan }^{-1}\mathrm{t}+\mathrm{C}\\ ={\tan }^{-1}\left({\mathrm{e}}^{\mathrm{x}}\right)+\mathrm{C}\\ \text{Thus, the correct option is A.}\end{array}$

Q.198

$$\begin{gathered} \mathrm{Choose}\mathrm{the}\mathrm{correct}\mathrm{answers} \\ \begin{array}{l}\int \frac{\cos 2\mathrm{x}}{{(\mathrm{sinx}+\mathrm{cosx})}^2}\mathrm{dx}\text{\hspace{0.17em}is equal to}\\ \left(\text{A}\right)\frac{-1}{\mathrm{sinx}+\mathrm{cosx}}+\mathrm{C}\\ \left(\text{B}\right)\log |\mathrm{sinx}+\mathrm{cosx}|+\mathrm{C}\\ \left(\text{C}\right)\log |\mathrm{sinx}-\mathrm{cosx}|+\mathrm{C}\\ \left(\text{D}\right)\frac{1}{{(\mathrm{sinx}+\mathrm{cosx})}^2}+\mathrm{C}\end{array} \end{gathered}$$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int \frac{\cos 2\mathrm{x}}{{(\mathrm{sinx}+\mathrm{cosx})}^2}\mathrm{dx}\\ =\int \frac{{\cos }^2\mathrm{x}-{\sin }^2\mathrm{x}}{{(\mathrm{sinx}+\mathrm{cosx})}^2}\mathrm{dx}\\ =\int \frac{(\mathrm{cosx}-\mathrm{sinx})(\mathrm{cosx}+\mathrm{sinx})}{{(\mathrm{sinx}+\mathrm{cosx})}^2}\mathrm{dx}\\ =\int \frac{(\mathrm{cosx}-\mathrm{sinx})}{(\mathrm{sinx}+\mathrm{cosx})}\mathrm{dx}\\ \text{Let t}=\mathrm{sinx}+\mathrm{cosx}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=\mathrm{cosx}-\mathrm{sinx}\\ \mathrm{I}=\int \frac{1}{\mathrm{t}}\mathrm{dt}\\ =\mathrm{logt}+\mathrm{C}\\ =\log (\mathrm{sinx}+\mathrm{cosx})+\mathrm{C}\\ \text{Therefore, option B is correct.}\end{array}$

Q.199

$$\begin{gathered} \mathrm{Choose}\mathrm{the}\mathrm{correct}\mathrm{answers} \\ \begin{array}{l}\text{If}\mathrm{f}(\mathrm{a}+\mathrm{b}-\mathrm{x})=\mathrm{f}\left(\mathrm{x}\right),\text{then}{\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{x}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\text{\hspace{0.17em} is equal to}\\ \left(\text{A}\right)\frac{\mathrm{a}+\mathrm{b}}{2}{\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}(\mathrm{b}-\mathrm{x})\mathrm{dx}\\ \left(\text{B}\right)\frac{\mathrm{a}+\mathrm{b}}{2}{\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}(\mathrm{b}+\mathrm{x})\mathrm{dx}\\ \left(\text{C}\right)\frac{\mathrm{b}-\mathrm{a}}{2}{\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\\ \left(\text{D}\right)\frac{\mathrm{a}+\mathrm{b}}{2}{\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\end{array} \end{gathered}$$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}}\mathrm{I}={\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{x}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\\ ={\int }_{\mathrm{a}}^{\mathrm{b}}(\mathrm{b}+\mathrm{a}-\mathrm{x})\mathrm{f}(\mathrm{b}+\mathrm{a}-\mathrm{x})\mathrm{dx}[\because {\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}(\mathrm{b}+\mathrm{a}-\mathrm{x})\mathrm{dx}]\\ ={\int }_{\mathrm{a}}^{\mathrm{b}}(\mathrm{b}+\mathrm{a}-\mathrm{x})\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}[\because \mathrm{f}(\mathrm{b}+\mathrm{a}-\mathrm{x})=\mathrm{f}\left(\mathrm{x}\right)]\\ ={\int }_{\mathrm{a}}^{\mathrm{b}}(\mathrm{b}+\mathrm{a})\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}-{\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{xf}\left(\mathrm{x}\right)\mathrm{dx}\\ \mathrm{I}=(\mathrm{b}+\mathrm{a}){\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}-\mathrm{I}\\ 2\mathrm{I}=(\mathrm{b}+\mathrm{a}){\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\\ \mathrm{I}=\frac{(\mathrm{b}+\mathrm{a})}{2}{\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\\ \text{Therefore, the correct answer is D.}\end{array}$

Q.200

$$\begin{gathered} \mathrm{Choose}\mathrm{the}\mathrm{correct}\mathrm{answers} \\ \begin{array}{l}\text{The value of}{\int }_0^1{\text{tan}}^{-1}\left(\frac{2\mathrm{x}-1}{1+\mathrm{x}-{\mathrm{x}}^2}\right)\text{dx is equal to}\\ \left(\text{A}\right)\text{\hspace{0.33em}1}\\ \left(\text{B}\right)\text{\hspace{0.33em}0}\\ \left(\text{C}\right)-\text{1}\\ \left(\text{D}\right)\frac{\mathrm{\pi }}{\text{4}}\end{array} \end{gathered}$$

Ans.

$\begin{array}{l}\text{Let\hspace{0.17em}\hspace{0.17em}}\mathrm{I}={\int }_0^1{\text{tan}}^{–1}\left(\frac{2\mathrm{x}-1}{1+\mathrm{x}-{\mathrm{x}}^2}\right)\mathrm{dx}\\ ={\int }_0^1{\text{tan}}^{–1}\left(\frac{\mathrm{x}-(1-\mathrm{x})}{1+\mathrm{x}(1-\mathrm{x})}\right)\mathrm{dx}\\ \mathrm{I}={\int }_0^1\{{\text{tan}}^{–1}\mathrm{x}-{\text{tan}}^{-1}(1-\mathrm{x})\}\mathrm{dx}...\left(\mathrm{i}\right)\\ ={\int }_0^1\{{\text{tan}}^{–1}(1-\mathrm{x})-{\text{tan}}^{-1}(1-1+\mathrm{x})\}\mathrm{dx}[\because {\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_0^{\mathrm{a}}\mathrm{f}(\mathrm{a}-\mathrm{x})\mathrm{dx}]\\ ={\int }_0^1\{{\text{tan}}^{–1}(1-\mathrm{x})-{\text{tan}}^{-1}\left(\mathrm{x}\right)\}\mathrm{dx}...\left(\mathrm{ii}\right)\\ \text{Adding equation}\left(\text{i}\right)\text{and equation}\left(\text{ii}\right)\text{, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2I}={\int }_0^1\{{\text{tan}}^{–1}\mathrm{x}-{\text{tan}}^{-1}(1-\mathrm{x})\}\mathrm{dx}+{\int }_0^1\{{\text{tan}}^{–1}(1-\mathrm{x})-{\text{tan}}^{-1}\left(\mathrm{x}\right)\}\mathrm{dx}\\ =0\\ \mathrm{I}=0\\ \text{Thus, the correct answer is B.}\end{array}$