NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.3) Exercise 7.3

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NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.3) Exercise 7.3

The NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.3 deals with the exercise on the topic of Definite and Indefinite Integrals and their properties. Integrals include two major concepts – the problem of finding the function whenever the derivatives are given, and the problem of finding the area bounded by the graph function under certain conditions. The NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.3 contain Integration which is called the inverse process of Differentiation. Instead of differentiating a function, the derivative function is given, and the question asks for its primitive. For a better understanding of Integrals, one should refer to the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.3. Practising the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.3 allows students to build a strong conceptual foundation for a better understanding of mathematical solutions. It covers all important theorems and formulae with detailed explanations.

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CBSE Class 12 Maths NCERT Solutions For Chapter 7 Exercise 7.3

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The NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.3 cover the topic of Integration while also using some trigonometric functions. Trigonometric functions are not just used while solving NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7, they can be used in a variety of topics. Therefore, mathematics is not just a subject that a student needs to learn formulas for, aiming for a good score, but it is also necessary for the future and daily planning of any individual.

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Q1. $Evaluate the following definite integrals as limit of sums . \int_{2}^{3} x^{2} dx$

Ans.

$\begin{array}{l} We have, \int_{2}^{3} x^{2} dx \\ \int_{a}^{b} f (x) dx = (b - a) \lim_{n \to \infty} \frac{1}{n} [f (a) + f (a + h) + . .. + f (a + (n - 1) h)], \\ where, h = \frac{b - a}{n} \\ Here, a = 2, b = 3, f (x) = x^{2}, h = \frac{3 - 2}{n} = \frac{1}{n} \\ ∴ \int_{2}^{3} x^{2} dx \\ = (3 - 2) \lim_{n \to \infty} \frac{1}{n} [f (2) + f (2 + h) + . .. + f (2 + (n - 1) h)] \\ = \lim_{n \to \infty} \frac{1}{n} [{(2)}^{2} + {(2 + h)}^{2} + . .. + {2 + (n - 1) h}^{2}] \\ = \lim_{n \to \infty} \frac{1}{n} [4 + (4 + 4 h + h^{2}) + . .. + {4 + 4 (n - 1) h + {(n - 1)}^{2} h^{2}}] \\ = \lim_{n \to \infty} \frac{1}{n} [4 n + 4 {1 + 2 + . .. + (n - 1)} h + {1 + 2^{2} + 3^{2} + . .. + {(n - 1)}^{2}} h^{2}] \\ = \lim_{n \to \infty} \frac{1}{n} [4 n + 4 \frac{(n - 1) n}{2} h + \frac{(n - 1) n (2 n - 2 + 1)}{6} h^{2}] \\ = \lim_{n \to \infty} \frac{1}{n} [4 n + 4 \frac{(n - 1) n}{2} . \frac{1}{n} + \frac{(n - 1) n (2 n - 2 + 1)}{6} . \frac{1}{n^{2}}] \\ [∵ \sum n = \frac{n (n + 1)}{2}, \sum n^{2} = \frac{n (n + 1) (2 n + 1)}{6}] \\ = \lim_{n \to \infty} [4 + 4 \frac{(1 - \frac{1}{n})}{2} + \frac{(1 - \frac{1}{n}) (2 - \frac{1}{n})}{6}] \\ = [4 + 4 \frac{(1 - 0)}{2} + \frac{(1 - 0) (2 - 0)}{6}] \\ = 4 + 2 + \frac{1}{3} \\ = \frac{19}{3} \end{array}$

Q2. $Evaluate the following definite integrals as limit of sums . \int_{0}^{5} (x + 1) dx$

Ans.

$\begin{array}{l} We have, \int_{0}^{5} (x + 1) dx \\ By definition \\ \int_{a}^{b} f (x) dx = (b - a) \lim_{n \to \infty} \frac{1}{n} [f (a) + f (a + h) + . .. + f (a + (n - 1) h)], \\ where, h = \frac{b - a}{n} \\ Here, a = 0, b = 5, f (x) = x + 1, h = \frac{5 - 0}{n} = \frac{5}{n} \\ ∴ \int_{0}^{5} (x + 1) dx = (b - a) \lim_{n \to \infty} \frac{1}{n} [f (0) + f (0 + h) + . .. + f (0 + (n - 1) h)] \\ = 5 \lim_{n \to \infty} \frac{1}{n} [(0 + 1) + (h + 1) + . .. + {(n - 1) h + 1}] \\ = 5 \lim_{n \to \infty} \frac{1}{n} [n + h + 2 h + . .. + (n - 1) h] \\ = 5 \lim_{n \to \infty} \frac{1}{n} [n + {1 + 2 + . .. + (n - 1)} h] \\ = 5 \lim_{n \to \infty} \frac{1}{n} [n + \frac{(n - 1) n}{2} h] [∵ \sum n = \frac{n (n + 1)}{2}] \\ = 5 \lim_{n \to \infty} [1 + \frac{(n - 1)}{2} (\frac{5}{n})] \\ = 5 \lim_{n \to \infty} [1 + \frac{5}{2} (1 - \frac{1}{n})] \\ = 5 [1 + \frac{5}{2} (1 - 0)] \\ = 5 \times \frac{7}{2} = \frac{35}{2} \end{array}$

Q3. $Evaluate the following definite integrals as limit of sums . \int_{a}^{b} x dx$

Ans.

$\begin{array}{l} By definition \\ \int_{a}^{b} f (x) dx = (b - a) \lim_{n \to \infty} \frac{1}{n} [f (a) + f (a + h) + . .. + f (a + (n - 1) h)], \\ where, h = \frac{b - a}{n} \\ Here, f (x) = x, h = \frac{b - a}{n} \\ ∴ \int_{a}^{b} x dx = (b - a) \lim_{n \to \infty} \frac{1}{n} [f (a) + f (a + h) + . .. + f (a + (n - 1) h)] \\ = (b - a) \lim_{n \to \infty} \frac{1}{n} [a + a + h + . .. + a + (n - 1) h] \\ = (b - a) \lim_{n \to \infty} \frac{1}{n} [na + h + 2 h + . .. + (n - 1) h] \\ = (b - a) \lim_{n \to \infty} \frac{1}{n} [na + {1 + 2 + . .. + (n - 1)} h] \\ = (b - a) \lim_{n \to \infty} \frac{1}{n} [na + \frac{(n - 1) n}{2} h] [∵ \sum n = \frac{n (n + 1)}{2}] \\ = (b - a) \lim_{n \to \infty} [a + \frac{(n - 1)}{2} (\frac{b - a}{n})] \\ = (b - a) \lim_{n \to \infty} [a + \frac{(b - a)}{2} (1 - \frac{1}{n})] \\ = (b - a) [a + \frac{(b - a)}{2} (1 - 0)] \\ = (b - a) (\frac{2 a + b - a}{2}) \\ = (b - a) \frac{(b + a)}{2} = \frac{b^{2} - a^{2}}{2} \end{array}$

Q4. $(x + 3) \sqrt{3 - 4 x - x^{2}}$

Ans.

$\begin{array}{l} \int (x + 3) \sqrt{3 - 4 x - x^{2}} dx \\ Let x + 3 = A \frac{d}{dx} (3 - 4 x - x^{2}) + B \\ x + 3 = A (- 4 - 2 x) + B \\ \Rightarrow 3 = - 4 A + B & - 2 A = 1 \\ \Rightarrow A = \frac{1}{- 2} \\ & B = 3 + 4 A \\ = 3 + 4 (\frac{1}{- 2}) \\ = 3 - 2 = 1 \\ ∴ \int (x + 3) \sqrt{3 - 4 x - x^{2}} dx = \frac{1}{- 2} \int (- 4 - 2 x) \sqrt{3 - 4 x - x^{2}} dx \\ + 1 \int \sqrt{3 - 4 x - x^{2}} dx \\ = \frac{1}{- 2} \frac{{(3 - 4 x - x^{2})}^{\frac{3}{2}}}{\frac{3}{2}} + \int \sqrt{7 - {(x + 2)}^{2}} dx \\ = \frac{{(3 - 4 x - x^{2})}^{\frac{3}{2}}}{- 3} + \int \sqrt{{(\sqrt{7})}^{2} - {(x + 2)}^{2}} dx \\ = \frac{{(3 - 4 x - x^{2})}^{\frac{3}{2}}}{- 3} + \frac{1}{2} (x + 2) \sqrt{{(\sqrt{7})}^{2} - {(x + 2)}^{2}} + \frac{7}{2} \sin^{- 1} (\frac{x + 2}{\sqrt{7}}) + C \\ + \frac{{(\sqrt{7})}^{2}}{2} \sin^{- 1} (\frac{x + 2}{\sqrt{7}}) + C \\ = \frac{{(3 - 4 x - x^{2})}^{\frac{3}{2}}}{- 3} + \frac{1}{2} (x + 2) \sqrt{3 - 4 x - x^{2}} + \frac{7}{2} \sin^{- 1} (\frac{x + 2}{\sqrt{7}}) + C \end{array}$

Q5. $\int (x + 1) \sqrt{2 x^{2} + 3} dx$

Ans.

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

Q6. $\int x \sqrt{x + x^{2}} dx$

Ans.

$\begin{array}{l} \int x \sqrt{x + x^{2}} dx \\ Let x = A \frac{d}{dx} (x + x^{2}) + B \\ x = A (1 + 2 x) + B \\ \Rightarrow 1 = 2 A & A + B = 0 \\ \Rightarrow A = \frac{1}{2} & B = - \frac{1}{2} \\ ∴ \int x \sqrt{x + x^{2}} dx = \frac{1}{2} \int (1 + x) \sqrt{x + x^{2}} dx - \frac{1}{2} \int \sqrt{x + x^{2}} dx \\ = \frac{1}{2} \times \frac{{(x + x^{2})}^{\frac{3}{2}}}{(\frac{3}{2})} - \frac{1}{2} \int \sqrt{{(x + \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}} dx \\ = \frac{1}{3} \times {(x + x^{2})}^{\frac{3}{2}} - \frac{1}{2} \times \frac{(x + \frac{1}{2})}{2} \sqrt{{(x + \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}} \\ + \frac{1}{2} \times \frac{{(\frac{1}{2})}^{2}}{2} \log | x + \sqrt{{(x + \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}} | + C \\ = \frac{{(x + x^{2})}^{\frac{3}{2}}}{3} - \frac{1}{8} (2 x + 1) \sqrt{x + x^{2}} + \frac{1}{16} \log | x + \sqrt{x + x^{2}} | + C \end{array}$

Q7.

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

Ans.

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

Q8.

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

Ans.

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

Q9. $\int \sqrt{1 + \frac{x^{2}}{9}} dx$

Ans.

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

Q10. $\int \sqrt{x^{2} + 3 x} dx$

Ans.

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

Q11. $\int \sqrt{1 + 3 x - x^{2}} dx$

Ans.

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

Q12. $\int \sqrt{x^{2} + 4 x - 5} dx$

Ans.

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

Q13. $\int \sqrt{1 - 4 x - x^{2}} dx$

Ans.

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

Q14. $\int \sqrt{x^{2} + 4 x + 1} dx$

Ans.

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

Q15. $\int \sqrt{x^{2} + 4 x + 6} dx$

Ans.

$\begin{array}{l} We have, \sqrt{x^{2} + 4 x + 6} = \sqrt{{(x + 2)}^{2} + 2} \\ Let I = \int \sqrt{{(x + 2)}^{2} + {(\sqrt{2})}^{2}} dx \\ = \int \sqrt{t^{2} + {(\sqrt{2})}^{2}} dt [Let t = x + 2 \Rightarrow \frac{dt}{dx} = 1] \\ [∵ \int \sqrt{x^{2} + a^{2}} dx = \frac{x}{2} \sqrt{x^{2} + a^{2}} + \frac{a^{2}}{2} \log | x + \sqrt{x^{2} + a^{2}} |] \\ = \frac{t}{2} \sqrt{t^{2} + {(\sqrt{2})}^{2}} + \frac{{(\sqrt{2})}^{2}}{2} \log | t + \sqrt{t^{2} + {(\sqrt{2})}^{2}} | + C \\ Putting t = x + 2, we get \\ = \frac{(x + 2)}{2} \sqrt{{(x + 2)}^{2} + {(\sqrt{2})}^{2}} + \frac{2}{2} \log | (x + 2) + \sqrt{{(x + 2)}^{2} + {(\sqrt{2})}^{2}} | + C \\ = \frac{(x + 2)}{2} \sqrt{x^{2} + 4 x + 6} + \log | (x + 2) + \sqrt{x^{2} + 4 x + 6} | +</mo&g
t;C \end{array}$

Q16. $\int \sqrt{1 - 4 x^{2}} dx$

Ans.

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

Q17. $\int \sqrt{4 - x^{2}} dx$

Ans.

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

Q18.

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

Ans.

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

Q19.

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

Ans.

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

Q20.

$\begin{array}{l} Integrate the functions \\ \sin^{- 1} (\frac{2 x}{1 + x^{2}}) \end{array}$

Ans.

$\begin{array}{l} We have, \sin^{- 1} (\frac{2 x}{1 + x^{2}}) \\ Let I = \int \sin^{- 1} (\frac{2 x}{1 + x^{2}}) dx \\ Let x = tanθ \Rightarrow \frac{dx}{dθ} = \sec^{2} θ \\ ∴ I = \int \sin^{- 1} (\frac{2 x}{1 + x^{2}}) dx \\ = \int \sin^{- 1} (\frac{2 tanθ}{1 + \tan^{2} θ}) \sec^{2} θ dθ \\ = \int \sin^{- 1} (\sin 2 θ) \sec^{2} θ dθ \\ = \int 2 θ \sec^{2} θ dθ \\ = 2 [θ \int \sec^{2} θ dθ - (\int \frac{d}{dθ} θ \int \sec^{2} θ dθ) dθ] \\ = 2 [θtanθ - \int 1 tanθ dθ] \\ = 2 θtanθ - 2 \log | secθ | + C \\ = 2 θtanθ - 2 \log | \sqrt{1 + \tan^{2} θ} | + C \\ I = 2 {xtan}^{- 1} x - 2 \log | \sqrt{1 + x^{2}} | + C . \end{array}$

Q21. Integrate the functions e^2x sinx

Ans.

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

Q22.

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

Ans.

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

Q23. $Integrate the functionse^{x} (\frac{1}{x} - \frac{1}{x^{2}})$

Ans.

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

Q24. $Integrate the functions e^{x} (\frac{1 + sinx}{1 + cosx})$

Ans.

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

Q25. $Integrate the functions \frac{{xe}^{x}}{{(1 + x)}^{2}}$

Ans.

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

Q26. Integrate the functions e^x(sinx + cosx)

Ans.

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

Q27. Integrate the functions (x²+1) logx

Ans.

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

Q28. Integrate the functions x (logx)²

Ans.

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

Q29. Integrate the functions tan^-1x

Ans.

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

Q30. Integrate the functions x sec²x

Ans.

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

Q31.

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

Ans.

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

Q32. Integrate the functions
(sin^–1x)²

Ans.

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

Q33. Integrate the functions
x cos^-1x

Ans.

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

Q34. Integrate the functions
x tan^-1x

Ans.

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

Q35. Integrate the functions
x sin^-1x

Ans.

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

Q36. Integrate the functions
x² logx

Ans.

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

Q37. Integrate the functions
x log 2x

Ans.

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

Q38. Integrate the functions
x logx

Ans.

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

Q39. Integrate the functions
x² e^x

Ans.

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

Q40. Integrate the functions
x sin 3x

Ans.

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

Q41. Integrate the function x sin x.

Ans.

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

Q42.

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

Ans.

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

Q43.

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

Ans.

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

Q44. $Integrate the rational function \frac{1}{(e^{x} - 1)}$

Ans.

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

Q45. $Integrate the rational function \frac{1}{x (x^{4} - 1)}$

Ans.

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

Q46. $Integrate the rational function \frac{2 x}{(x^{2} + 1) (x^{2} + 3)}$

Ans.

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

Q47. $Integrate the rational function \frac{(x^{2} + 1) (x^{2} + 2)}{(x^{2} + 3) (x^{2} + 4)}$

Ans.

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

Q48. $Integrate the rational function \frac{cosx}{(1 - sinx) (2 - sinx)}$

Ans.

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

Q49. $Integrate the rational function \frac{1}{x (x^{n} - 1)}$

Ans.

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

Q50. $Integrate the rational function \frac{1}{x^{4} - 1}$

Ans.

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

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NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.3) Exercise 7.3

NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.3) Exercise 7.3

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NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.3) Exercise 7.3

NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.3) Exercise 7.3

Access NCERT Solutions for Class 12 Mathematics Chapter 7- Integrals

CBSE Class 12 Maths NCERT Solutions For Chapter 7 Exercise 7.3

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