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

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NCERT Solutions helps students prepare for their board examinations and understand the fundamental concepts better. A step-by-step explanation is provided in NCERT Solutions that helps students understand the concept easily. Since there are no detailed solutions in the NCERT books, students who might have trouble understanding the concept might refer to the Extramarks website. Extramarks is a digital learning platform that students can access to refer to learning resources. Extramarks allows students to get acquainted with the preparation methods that work best for them. The Assessment Center, Smart Class Solutions, and Live Class Platform are just a few of the school-based tools that Extramarks provides to help students achieve their full potential through engaging instruction and customized curriculum.

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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.

The NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.3 explain Integration using some trigonometric functions, and a few specific identities are used to find the integral. The NCERT Solutions for Exercise 7.3 Class 12th are focused on covering various determinants of Integrals. Class 12 Maths Ex 7.3 is said to be a significant exercise, as the same pattern of questions can be observed in past years’ board examinations. Hence, students should refer to the NCERT Solutions For Class12 Maths Exercise 7.3 for improving their scores. Consistency is the key to success, so for scoring well in examinations, students must revise the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.3 consistently. Extramarks provides detailed NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.3 that help in understanding the concepts, thereby helping in improving grades.

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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The NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.3 emphasise learning various techniques for solving problems concerning Integrals in a way that the student gets the most out of it. The first and foremost step to understanding different formulas thoroughly is to understand the fundamental concepts from practising the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.3. The NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.3 are provided on the Extramarks website, so students can easily get acquainted with their problem areas. Extramarks provides various benefits; it connects teachers who have years of expertise with students who can learn from these teachers at their convenience. Students can easily access the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.3 through the Extramarks website. 7.3 Class 12th Maths focuses on various concepts through detailed and quick learning, and in this way, it helps students understand and improve their scores.

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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.∫23x2 dx

Ans.

We have, ∫23x2 dx∫abf(x) dx=(b−a)limn→∞1n[f(a)+f(a+h)+...+f(a+(n−1)h)],where,   h=b−anHere,  a=2,  b=3,  f(x)=x2,  h=3−2n=1n∴ ∫23x2 dx          =(3−2)limn→∞1n[f(2)+f(2+h)+...+f(2+(n−1)h)]          =limn→∞1n[(2)2+(2+h)2+...+{2+(n−1)h}2]          =limn→∞1n[4+(4+4h+h2)+...+{4+4(n−1)h+(n−1)2h2}]          =limn→∞1n[4n+4{1+2+...+(n−1)}h+{1+22+32+...+(n−1)2}h2]          =limn→∞1n[4n+4(n−1)n2h+(n−1)n(2n−2+1)6h2]          =limn→∞1n[4n+4(n−1)n2.1n+(n−1)n(2n−2+1)6.1n2]                                        [∵∑n=n(n+1)2,∑n2=n(n+1)(2n+1)6]          =limn→∞[4+4(1−1n)2+(1−1n)(2−1n)6]          =[4+4(1−0)2+(1−0)(2−0)6]          =4+2+13          =193

Q2.

Evaluate the following definite integrals as limit of sums.∫05(x+1) dx

Ans.

We have,  ∫05(x+1) dxBy definition∫abf(x) dx=(b−a)limn→∞1n[f(a)+f(a+h)+...+f(a+(n−1)h)],where,   h=b−anHere,  a=0,  b=5,  f(x)=x+1,  h=5−0n=5n∴∫05(x+1) dx=(b−a)limn→∞1n[f(0)+f(0+h)+...+f(0+(n−1)h)]                            =5limn→∞1n[(0+1)+(h+1)+...+{(n−1)h+1}]                            =5limn→∞1n[n+h+2h+...+(n−1)h]                            =5limn→∞1n[n+{1+2+...+(n−1)}h]                            =5limn→∞1n[n+(n−1)n2h][∵∑n=n(n+1)2]                            =5limn→∞[1+(n−1)2(5n)]                            =5limn→∞[1+52(1−1n)]                            =5[1+52(1−0)]                            =5×72  =352

Q3.

Evaluate the following definite integrals as limit of sums.∫abx dx

Ans.

By definition∫abf(x) dx=(b−a)limn→∞1n[f(a)+f(a+h)+...+f(a+(n−1)h)],where,   h=b−anHere,  f(x)=x,  h=b−an∴∫abx dx=(b−a)limn→∞1n[f(a)+f(a+h)+...+f(a+(n−1)h)]          =(b−a)limn→∞1n[a+a+h+...+a+(n−1)h]          =(b−a)limn→∞1n[na+h+2h+...+(n−1)h]          =(b−a)limn→∞1n[na+{1+2+...+(n−1)}h]          =(b−a)limn→∞1n[na+(n−1)n2h]        [∵∑n=n(n+1)2]          =(b−a)limn→∞[a+(n−1)2(b−an)]          =(b−a)limn→∞[a+(b−a)2(1−1n)]          =(b−a)[a+(b−a)2(1−0)]          =(b−a)(2a+b−a2)          =(b−a)(b+a)2  =b2−a22

Q4.

(x+3)3–4x–x2

Ans.

∫(x+3)3−4x−x2 dxLet x+3=Addx(3−4x−x2)+B        x+3 =A(− 4−2x)+B⇒          3=− 4A+B   & −2A=1⇒    A=1−2 &     B=3+4A       =3+4(1−2)       =3−2=1∴∫(x+3)3−4x−x2 dx=1−2∫(− 4−2x)3−4x−x2 dx                                          +1∫3−4x−x2dx                              =1−2(3−4x−x2)3232+∫7−(x+2)2dx                              =(3−4x−x2)32−3+∫(7)2−(x+2)2dx                         =(3−4x−x2)32−3+12(x+2)(7)2−(x+2)2+72sin−1(x+27)+C                                                              +(7)22sin−1(x+27)+C                    =(3−4x−x2)32−3+12(x+2)3−4x−x2  +72sin−1(x+27)+C

Q5.

∫x+12x2+3dx

Ans.

∫(x+1)2x2+3 dxLet x+1=Addx(2x2+3)+B        x+1 =A(4x+0)+B⇒    x+1=4Ax   & B=1⇒    A=14 & B=1∴∫(x+1)2x2+3 dx=14∫4x2x2+3 dx+1∫2x2+3 dx                       =14×(2x2+3)3232+2 ∫x2+(32)2 dx                =16×(2x2+3)32+2 ∫x2+(32)2 dx                =16×(2x2+3)32+2(12xx2+(32)2        +  (32)22log|x+(32)2|)+C                =16×(2x2+3)32+22 x x2+32                                    +324log|x+x2+(32)2|+C                =(2x2+3)326−x2 2x2+3−324log|x+x2+32|+C

Q6.

∫xx+x2dx

Ans.

∫xx+x2dxLet x=Addx(x+x2)+B       x   =A(1+2x)+B⇒     1=2A & A+B=0⇒    A=12 & B=−12∴∫xx+x2dx=12∫(1+x)x+x2dx−12∫x+x2dx               =12×(x+x2)32(32)−12∫(x+12)2−(12)2dx               =13×(x+x2)32−12×(x+12)2(x+12)2−(12)2                              +12×(12)22log|x+(x+12)2−(12)2|+C               =(x+x2)323−18(2x+1)x+x2+116log|x+x+x2|+C

Q7.

Choose the correct answer∫x2−8x+7 dx is equal toA 12x−4x2−8x+7+9 logx−4+x2−8x+7+CB 14x+4x2−8x+7+9logx+4+x2−8x+7+CC 12x−4x2−8x+7−32 logx−4+x2−8x+7+CD 12x−4x2−8x+7−92logx−4+x2−8x+7+C

Ans.

We have,  x2−8x+7=(x−4)2−9Let I=∫(x−4)2−(3)2 dx          =∫t2−(3)2 dt     [Let t=x−4⇒dtdx=1]                            [∵∫x2−a2 dx=x2x2−a2−a22log|x+x2−a2|]        =t2t2−(3)2−(3)22log|t+t2−(3)2|+CPutting t=x−4, we get     =(x−4)2(x−4)2−(3)2−92log|(x−4)+(x−4)2−(3)2|+C     =(x−4)2x2−8x+7−92log|(x−4)+x2−8x+7|+CHence, the correct option is D.

Q8.

Choose the correct answer∫1+x2 dx is equal toA x21+x2+12logx+1+x2+CB 231+x232+CC 23x1+x232+CD x221+x2 +12 x2 logx+1+x2+C

Ans.

Let  I=∫1+x2 dx          =∫x2+12 dx                        [∵∫x2+a2 dx=x2x2+a2+a22log|x+x2+a2|]           =(x2x2+12+122log|x+x2+12|)+C           =x2x2+1+12log|x+x2+1|+C           =x21+x2+12log|x+1+x2|+CHence,​  the correct option is A.

Q9. 

∫1+x29dx

Ans.

We have, 1+x29=139+x2Let  ∫139+x2 dx=13∫32+x2 dx                                           =13∫x2+32 dx                                                        [∵∫x2+a2 dx=x2x2+a2+a22log|x+x2+a2|]                       =13(x2x2+32+322log|x+x2+32|)+C                       =x6x2+9+32log|x+x2+9|+C

Q10.

∫x2+3xdx

Ans.

We have,  x2+3x=(x2+3x+94)−94                                          =(x+32)2−(32)2Let I=∫(x+32)2−(32)2 dx         =∫t2−(32)2 dt     [Let t=x+32⇒dtdx=1]                                 [∵∫x2−a2 dx=x2x2−a2−a22log|x+x2−a2|]         =t2t2−(32)2−(32)22log|t+t2−(32)2|+CPutting t=x+32, we get     =(x+32)2(x+32)2−(32)2−98log|(x+32)+(x+32)2−(32)2|+C     =(2x+3)4x2+3x−98log|(x+32)+x2+3x|+C

Q11.

∫1+3x−x2dx

Ans.

We have,  1+3x−x2=1−(x2−3x+94−94)                                                   =(132)2−(x−32)2Let I=∫(132)2−(x−32)2 dx         =∫(132)2−t2 dt      [Let t=x−32⇒dtdx=1]                  [∵∫a2−x2 dx=x2a2−x2+a22sin−1xa]      =t2(132)2−t2+(132)22sin−1t(132)+CPutting t=x−32, we get      =(x−32)2(132)2−(x−32)2+(132)22sin−1(x−32)(132)+C      =(2x−3)41+3x−x2+138sin−1(2x−3)(13)+C

Q12.

∫x2+4x−5dx

Ans.

We have,  x2+4x−5=(x+2)2−9Let I=∫(x+2)2−(3)2 dx         =∫t2−(3)2 dt      [Let t=x+2⇒dtdx=1]                                 [∵∫x2−a2 dx=x2x2−a2−a22log|x+x2−a2|]      =t2t2−(3)2−(3)22log|t+t2−(3)2|+CPutting t=x+2, we get      =(x+2)2(x+2)2−(3)2−92log|(x+2)+(x+2)2−(3)2|+C      =(x+2)2x2+4x−5−92log|(x+2)+x2+4x−5|+C

Q13.

∫1−4x−x2dx

Ans.

We have,  1−4x−x2=5−(x2+4x+4)                                                    =(5)2−(x+2)2Let I=∫(5)2−(x+2)2 dx         =∫(5)2−t2 dt     [Let t=x+2⇒dtdx=1]                      [∵∫a2−x2 dx=x2a2−x2+a22sin−1xa]      =t2(5)2−t2+(5)22sin−1t(5)+CPutting t=x+2, we get      =(x+2)2(5)2−(x+2)2+(5)22sin−1(x+2)(5)+C      =(x+2)21−4x−x2+52sin−1(x+2)(5)+C

Q14.

∫x2+4x+1dx

Ans.

We have,  x2+4x+1=(x+2)2−3Let I=∫(x+2)2−(3)2 dx         =∫t2−(3)2 dt     [Let t=x+2⇒dtdx=1]                      [∵∫x2−a2 dx=x2x2−a2−a22log|x+x2−a2|]       =t2t2−(3)2−(3)22log|t+t2−(3)2|+CPutting t=x+2, we get       =(x+2)2(x+2)2−(3)2−32log|(x+2)+(x+2)2−(3)2|+C       =(x+2)2x2+4x+1−32log|(x+2)+x2+4x+1|+C

Q15.

∫x2+4x+6dx

Ans.

We have,  x2+4x+6=(x+2)2+2Let I=∫(x+2)2+(2)2 dx         =∫t2+(2)2 dt   [Let t=x+2⇒dtdx=1]                                      [∵∫x2+a2 dx=x2x2+a2+a22log|x+x2+a2|]       =t2t2+(2)2+(2)22log|t+t2+(2)2|+CPutting t=x+2, we get     =(x+2)2(x+2)2+(2)2+22log|(x+2)+(x+2)2+(2)2|+C     =(x+2)2x2+4x+6+log|(x+2)+x2+4x+6|+</mo&g
t;C

Q16.

∫1−4x2dx

Ans.

We have,  1−4x2Let I=∫1−4x2 dx          =∫12−(2x)2 dx[Let t=2x⇒dtdx=2⇒dx=dt2]            =∫12−t2 dt2                             [∵∫a2−x2 dx=12xa2−x2+a22sin−1xa+C]           =12(12t12−t2+a22sin−1t1)+C           =12(2x21−4x2+12sin−12x1)+C          =x21−4x2+14sin−12x+C

Q17.

∫4−x2dx

Ans.

We have,  4−x2Let I=∫4−x2 dx           =∫22−x2 dx                         [∵∫a2−x2 dx=12xa2−x2+a22sin−1xa+C]           =12x22−x2+222sin−1x2+C          =x24−x2+2sin−1x2+C

Q18.

Choose the correct answer∫ex secx1+tanxdx equalsA​ excosx+CB exsecx+CC exsinx+CD extanx+C

Ans.

Let  I=∫ex secx(1+tanx)dx          =∫ex(secx+secxtanx) dxLet​ f(x)=secx  and f’(x)=secxtanxThus, the given integrand is of the form ex[f(x)+f‘(x)].Therefore,       I=∫ex(secx+secxtanx) dx        =exsecx+C       [∵ex{f(x)+f‘(x)}=exf(x)]Hence, the correct option is B.

Q19.

Choose the correct answer∫x2ex3 dx equalsA​ 13ex3+CB 13ex2+CC 12ex3+CD 12ex2+C

Ans.

We  have  ∫x2ex3 dxLet​ t=x3⇒dtdx=3x2∴∫x2ex3 dx=∫x2et dt3x2                          =13∫et dt                          =13ex3+CHence, the correct option is A.

Q20.

Integrate the functionssin–1(2x1+x2)

Ans.

We have, sin–1(2x1+x2)Let  I=∫sin–1(2x1+x2) dxLet x=tanθ⇒dxdθ=sec2θ∴     I=∫sin–1(2x1+x2) dx          =∫sin–1(2tanθ1+tan2θ) sec2θ dθ          =∫sin–1(sin2θ)sec2θ dθ          =∫2θ sec2θ dθ         =2[θ∫sec2θ dθ−(∫ddθ θ∫sec2θ dθ) dθ]         =2[θtanθ−∫1tanθ dθ]         =2θtanθ−2log|secθ|+C          =2θtanθ−2log|1+tan2θ|+C       I=2xtan−1x−2log|1+x2|+C.

Q21. Integrate the functions e2x sinx

Ans.

We have, e2xsinxLet  I=∫e2xsinx dxTaking sinx as first function and e2x as second function, then Integrating by parts, we get      I=∫e2xsinx dx        =sinx∫e2x dx−∫(ddxsinx∫e2x dx) dx        =sinx(e2x2)−∫{cosx(e2x2)} dx        =e2x2sinx−12∫cosx.e2x dx        =e2x2sinx−12{cosx∫e2x dx−∫(ddxcosx∫e2x dx) dx}        =e2x2sinx−12{cosx(e2x2)−∫{−sinx(e2x2)} dx}        =e2x2sinx−12cosx(e2x2)−14∫e2xsinx dx+C     I =e2xsinx2−14cosx e2x−14I+CI+14I=e2xsinx2−14cosx e2x+C      54I=e2xsinx2−14cosx e2x+C         I=45e2x4(2sinx−cosx )+C

Q22.

Integrate the functions(x−3)ex(x−1)3

Ans.

Let  I=∫(x−3)ex(x−1)3 dx           =∫ex{x−1−2(1−x)3} dx           =∫ex{x−1(1−x)3−2(1−x)3} dx           =∫ex{1(1−x)2−2(1−x)3} dxLet​ f(x)=1(1−x)2  and f’(x)=−2(1−x)3Thus, the given integrand is of the form ex[f(x)+f‘(x)].Therefore,       I=∫(x−3)ex(x−1)3 dx        =∫ex{1(1−x)2−2(1−x)3} dx       =ex(1−x)2+C            [∵ex{f(x)+f‘(x)}=exf(x)]

Q23.

Integrate the functionse x(1x−1x2)

Ans.

Let  I=∫ex(1x−1x2) dxLet​ f(x)=1x  and f’(x)=−1x2Thus, the given integrand is of the form ex[f(x)+f‘(x)].Therefore,   I=∫ex(1x−1x2) dx                            =exx+C           [∵ex{f(x)+f‘(x)}=exf(x)]

Q24.

Integrate the functions ex(1+sinx1+cosx)

Ans.

We have, ex(1+sinx1+cosx)=ex(sin2x2+cos2x2+2sinx2cosx22cos2x2)                           =12ex(sinx2+cosx2cosx2)2                           =12ex(tanx2+1)2                           =12ex(1+tan2x2+2tanx2)                           =12ex(sec2x2+2tanx2)                          =ex(12sec2x2+tanx2)                          =ex(tanx2+12sec2x2)Let f(x)=tanx2  and f’(x)=12sec2x2Thus, the given integrand is of the form ex[f(x)+f‘(x)].Therefore,∫ex(1+sinx1+cosx) dx=∫ex(tanx2+12sec2x2) dx                                           =extanx2+C    [∵ex{f(x)+f‘(x)}=exf(x)]

Q25.

Integrate the functions xex(1+x)2

Ans.

Let  I=∫xex(1+x)2 dx           =∫ex{x(1+x)2} dx          =∫ex{1+x−1(1+x)2} dx         =∫ex{1+x(1+x)2−1(1+x)2} dx          =∫ex{1(1+x)−1(1+x)2} dxLet​ f(x)=1(1+x)  and f’(x)=−1(1+x)2Thus, the given integrand is of the form ex[f(x)+f‘(x)].Therefore,       I=∫xex(1+x)2 dx        =∫ex{1(1+x)−1(1+x)2} dx       =ex1+x+C            [∵ex{f(x)+f‘(x)}=exf(x)]

Q26. Integrate the functions ex (sinx + cosx)

Ans.

Let  I=∫ex(sinx+cosx) dxHere, f(x)=sinx and f’(x)=cosxThus, the given integrand is of the form ex[f(x)+f‘(x)].Therefore,        I=∫ex(sinx+cosx) dx        =exsinx+C       [∵∫ex{f(x)+f‘(x)} dx=ex f(x)+C]

Q27. Integrate the functions (x2+1) logx

Ans.

Let  I=∫(x2+1)logx dxTaking logx as a first function and (x2+1) as second function and integrating by parts, we get      =logx∫(x2+1) dx−∫{(ddxlogx)∫(x2+1) dx} dx      =logx(x33+x)−∫1x.(x33+x) dx      =(x33+x)logx−∫(x23+1) dx+C      =(x33+x)logx−x39−x+C

Q28. Integrate the functions x (logx)2

Ans.

Let  I=∫x(logx)2 dxTaking (logx)2 as a first function and x as second function and integrating by parts, we get      =(logx)2∫x dx−∫{(ddx(logx)2)∫x dx} dx      =(logx)2(x22)−∫2logxx.(x22) dx      =x22(logx)2−∫xlogx dx+C      =x22(logx)2−[logx∫x dx−∫{(ddxlogx)∫x dx} dx]      =x22(logx)2−{logx(x22)−∫1x.(x22) dx}      =x22(logx)2−{x22logx−∫x2 dx}      =x22(logx)2−x22logx+x24+C

Q29. Integrate the functions tan-1x

Ans.

Let  I=∫tan−1x dx=∫tan−1x .1dxTaking tan−1x as a first function and 1 as second function and integrating by parts, we get         =tan−1x∫1 dx−∫{(ddxtan−1x)∫1 dx} dx         =tan−1x(x)−∫11+x2.(x) dx         =x tan−1x−∫x1+x2 dx+C         =x tan−1x−12log|1+x2|+C 

Q30. Integrate the functions x sec2x

Ans.

Let  I=∫xsec2x dxTaking algebraic function i.e. x as a first function and sec2x assecond function and integrating by parts, we get         =x∫sec2x dx−∫{(ddxx)∫sec2x dx} dx        =x(tanx)−∫1.tanx dx         =xtanx−∫tanx dx+C         =xtanx+log|cosx|+C

Q31.

Integrate the functionsxcos–1x1−x2

Ans.

Let  I=∫xcos–1x1−x2 dx          =−12∫cos−1x−2x1−x2 dxTaking cos−1x as first function and −2x1−x2 as second function and integrating by parts, we get        I=−12{cos−1x∫−2x1−x2 dx−(∫ddxcos−1x∫−2x1−x2 dx)dx}          =−12{cos−1x∫−2xt dt−2x−(∫ddxcos−1x∫−2xt dt−2x) dx}                                              [Let t=1−x2⇒dtdx=−2x⇒dx=dt−2x]        =−12[cos−1x.2t−(∫−11−x22t) dx]        =−12[cos−1x.21−x2−(∫−11−x221−x2) dx]        =−12(2cos−1x1−x2+2∫1 dx)       =−12(2cos−1x1−x2+2x)+C       =−(1−x2cos−1x+x)+C

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

Ans.

Let  I=∫(sin−1x)2 dx          =∫(sin−1x)2.1 dxTaking (sin−1x)2 as first function and 1 as second function and integrating by parts, we get        I=(sin−1x)2∫1 dx−∫{ddx(sin−1x)2∫1 dx} dx      =(sin−1x)2x−∫{2sin−1xddx(sin−1x).x} dx      =(sin−1x)2x−∫2xsin−1x1−x2 dx      =(sin−1x)2x+∫sin−1x−2x1−x2 dx      =(sin−1x)2x+sin−1x∫−2x1−x2 dx−(∫ddxsin−1x∫−2x1−x2 dx) dx      =(sin−1x)2x+sin−1x∫−2xt dt−2x−{∫ddxsin−1x∫−2xt dt−2x} dx                                                        [Let t=1−x2⇒dtdx=−2x⇒dx=dt−2x]      =(sin−1x)2x+sin−1x.2t−{∫11−x22t} dx      =(sin−1x)2x+2sin−1x.1−x2−∫11−x221−x2dx      =(sin−1x)2x+2sin−1x.1−x2−2∫1 dx      =x(sin−1x)2+21−x2sin−1x−2x+C

Q33. Integrate the functions
x cos-1x

Ans.

Let  I=∫xcos−1x dxLet t=cos−1x⇒x=cost  sint=1−cos2t=1−x2Then,  dtdx=ddxcos−1x                       =−11−x2                      =−1sint⇒         dx=−sint dtSo,             I=∫tcost(− sint) dt                      =−12∫tsin2t dtTaking algebraic function i.e. t as a first function and sin2t as second function and integrating by parts, we get     =−12[t∫sin2t dt−∫{(ddtt)∫sin2t dt} dt]     =−12[t(−12cos2t)−∫1.(−12co2t) dx]     =−12[−12tcos2t+14sin2t+C]     =14t(1−2sin2t)−18×2sintcost+C     =14t(1−2sin2t)−14sintcost+C     =14(1−2x2)sin−1x−14x1−x2+C

Q34. Integrate the functions
x tan-1x

Ans.

Let  I=∫xtan−1x dxTaking tan−1x as a first function and x as second function and integrating by parts, we get=tan−1x∫x dx−∫{(ddxtan−1x)∫x dx} dx=tan−1x(x22)−∫11+x2.(x22) dx=x22tan−1x−12∫x21+x2 dx+C=x22tan−1x−12∫1+x2−11+x2 dx+C=x22tan−1x−12∫1+x21+x2 dx+12∫11+x2 dx+C=x22tan−1x−12∫1. dx+12tan−1x+C=x22tan−1x−12x dx+12tan−1x+C

Q35. Integrate the functions
x sin-1x

Ans.

Let  I=∫xsin−1x dxLet t=sin−1x⇒x=sintcost=1−sin2t=1−x2Then,  dtdx=ddxsin−1x                    =11−x2                   =1cost⇒        dx=cost dtSo,          I=∫tsint cost dt               =12∫tsin2t dtTaking algebraic function i.e. t as a first function and sin2t as second function and integrating by parts, we get      =12[t∫sin2t dt−∫{(ddtt)∫sin2t dt} dt]      =12[t(−12cos2t)−∫1.(−12co2t) dx]      =12[−12tcos2t+14sin2t+C]      =−14t(1−2sin2t)+18×2sintcost+C      =14t(2sin2t−1)+14sintcost+C      =14(2x2−1)sin−1x+14x1−x2+C

Q36. Integrate the functions
x2 logx

Ans.

Let  I=∫x2logx dxTaking logx as a first function and x2 as second function and integrating by parts, we get     =logx∫x2 dx−∫{(ddxlogx)∫x2 dx} dx      =logx(x33)−∫1x.(x33) dx      =x33logx−∫x23 dx+C      =x33logx−x39+C

Q37. Integrate the functions
x log 2x

Ans.

Let  I=∫xlog2x dxTaking log2x as a first function and x as second function and integrating by parts, we get     =log2x∫x dx−∫{(ddxlog2x)∫x dx} dx     =log2x(x22)−∫12x×2.(x22) dx     =x22log2x−∫x2 dx+C     =x22log2x−x24+C

Q38. Integrate the functions
x logx

Ans.

Let  I=∫xlogx dxTaking logx as a first function and x as second function and integrating by parts, we get =logx∫x dx−∫{(ddxlogx)∫x dx} dx =logx(x22)−∫1x.(x22) dx   =x22logx−∫x2 dx+C      =x22logx−x24+C

Q39. Integrate the functions
x2 ex

Ans.

Let  I=∫x2ex< /msup> dxTaking algebraic function i.e. x2 as a first function and ex as second function and integrating by parts, we get     =x2∫ex dx−∫{(ddxx2)∫ex dx} dx     =x2(ex)−∫2x.(ex) dx     =x2ex−2[x∫ex dx−∫{(ddxx)∫ex dx} dx]+C     =x2ex−2[xex−∫1.ex dx]+C     =x2ex−2xex+2ex+C     =ex(x2−2x+2)+C

Q40. Integrate the functions
x sin 3x

Ans.

Let  I=∫xsin3x dxTaking algebraic function i.e. x as a first function and sin3x as second function and integrating by parts, we get      =x∫sin3x dx−∫{(ddxx)∫sin3x dx} dx      =x(−13cos3x)−∫1.(−13cosx) dx      =−13xcos3x+19sin3x+C

Q41. Integrate the function x sin x.

Ans.

Let  I=∫xsinx dxTaking algebraic function i.e. x as a first function and sinx as second function and integrating by parts, we get=x∫sinx dx−∫{(ddxx)∫sinx dx} dx=x(−cosx)−∫1.(−cosx) dx=−xcosx+sinx+C

Q42.

Choose the correct answer∫dxxx2+1equalsA​ logx−12logx2+1+CB logx+12logx2+1+CC −logx+12logx2+1+CD 12logx+logx2+1+C

Ans.

We have,  ∫dxx(x2+1)Let 1x(x2+1)=Ax+Bx+C(x2+1)                          =A(x2+1)+(Bx+C)xx(x2+1)⇒               1=A(x2+1)+(Bx2+Cx)⇒               1=x2(A+B)+Cx+AEquating the coefficients of x2, x and constant term, we getA+B=0,   C=0 and A=1On solving these equations, we getA=1,  B=−1 and C=0⇒             1x(x2+1)=1x+−x(x2+1)∴∫1x(x2+1) dx=∫1x dx−∫x(x2+1) dx                                       = log|x|−2log|x2+1|+CThus, the correct option is (A).

Q43.

Choose the correct answer∫xdxx−1x−2 dx equalsA​ logx−12x−2+CB logx−22x−1+CC logx−1x−22+CD logx−1x−2+C

Ans.

We have,  ∫xdx(x−1)(x−2) dxLet x(x−1)(x−2)=Ax−1+Bx−2                                    =A(x−2)+B(x−1)(x−1)(x−2)⇒                         x=A(x−2)+B(x−1)    ...(i)Substituting x=1  and  2, in equation(i),​ we getA=−1 and B=2⇒                x(x−1)(x−2)=−1x−1+2x−2∴∫x(x−1)(x−2) dx=∫−1x−1 dx+∫2x−2 dx                                                 =− log|x−1|+2log|x−2|+C                                                =log|(x−2)2(x−1)|+CHence, the option (B) is correct.

Q44.

Integrate the rational function 1(ex−1)

Ans.

∫1(ex−1) dxLet  t=ex⇒dtdx=ex     dx=dtex=dtt∴∫1(ex−1) dx=∫1(t−1) dtt                                  =∫1t(t−1)dtLet  1t(t−1)=At+Bt−1                       =A(t−1)+Btt(t−1)⇒              1=A(t−1)+Bt     ...(i)Putting t=0​ and 1 respectively in equation (i), we getA=−1 and B=1∴             1t(t−1)=−1t+1t−1⇒∫1t(t−1) dt=∫−1t dt+∫1t−1 dt                                    =−log|t|+log|t−1|+C                                    =log|t−1t|+C∴∫1(ex−1) dx=log|ex−1ex|+C

Q45.

Integrate the rational function 1x(x4−1)

Ans.

We have, 1x(x4−1)=1x(x4−1)×x3x3                                         =x3x4(x4−1)Let  t=x4⇒dtdx=4x4−1⇒dt4x3=dx∴∫1x(x4−1) dx=∫x3x4(x4−1) dx                                     =∫x3t(t−1) dt4x3                                     =14∫1t(t−1) dtLet 1t(t−1)=At+Bt−1                     1=A(t−1)+BtSubstituting​ t= 0 and 1 respectively, we getA=−1  and   B=1∴                 1t(t−1)=−1t+1t−1⇒∫1x(x4−1) dx=14∫(−1t+1t−1) dt                                       =−14∫1t dt+14∫1t−1 dt                                         =−14log|t|+14log|t−1|+C                                         =−14log|x4|+14log|x4−1|+C                                         =14log|x4−1x4|+C

Q46.

Integrate the rational function 2x(x2+1)(x2+3)

Ans.

We have,    2x(x2+1)(x2+3)Let  t=x2⇒dtdx=2x⇒dx=dt2x∫2x(x2+1)(x2+3) dx=∫2x(t+1)(t+3) dt2x                                                =∫1(t+1)(t+3) dtLet  1(t+1)(t+3)=At+1+Bt+3                                1=A(t+3)+B(t+1)Substituting t=−1 and−3 respectively, we getA=12  and  B=−12∴1(t+1)(t+3)=12(t+1)−12(t+3)⇒∫1(t+1)(t+3) dx=∫12(t+1) dt−∫12(t+3) dx                                               =12log|t+1|−12log|t+3|+C                                               =12log|t+1t+3|+C                                               =12log|x2+1x2+3|+C

Q47.

Integrate the rational function (x2+1)(x2+2)(x2+3)(x2+4)

Ans.

We have,    (x2+1)(x2+2)(x2+3)(x2+4)=1+(4x2+10)(x2+3)(x2+4)Let , (4x2+10)(x2+3)(x2+4)=Ax+B(x2+3)+Cx+D(x2+4)             4x2+10 =(Ax+B)(x2+4)+(Cx+D)(x2+3)                                =x3(A+C)+x2(B+D)+x(4A+3C)+(4B+3D)Comparing the coefficients of x3, x2, x and constant terms,we get   A+C=0B+D=44A+3C=04B+3D=10On​ solving these equations, we getA=0,B=−2,C=0 and D=6∴(4x2+10)(x2+3)(x2+4)=−2(x2+3)+6(x2+4)Then,∫(x2+1)(x2+2)(x2+3)(x2+4) dx=∫1 dx−(∫−2(x2+3) dx+6∫1(x2+4) dx)                                                 =x+23tan−1(x3)−62tan−1(x2)+C                                                =x+23tan−1(x3)−3tan−1(x2)+C

Q48.

Integrate the rational function cosx(1−sinx)(2−sinx)

Ans.

We have, cosx(1−sinx)(2−sinx)Let  t=sinx⇒dtdx=cosx⇒dtcosx=dx∴∫cosx(1−sinx)(2−sinx) dx=∫cosx(1−t)(2−t) dtcosx                                                               =∫1(1−t)(2−t) dtLet  1(1−t)(2−t)=A1−t+B2−t                                1=A(2−t)+B(1−t)     ...(1)Substituting t=1 and 2 respectively in eqution (1), we getA=1 and B=−1∴1(1−t)(2−t)=11−t−12−tSo,∫cosx(1−sinx)(2−sinx) dx=∫1(1−t)(2−t) dt                                                 =∫11−t dt−∫12−t dt                                                  =−log|1−t|+log|2−t|+C                                                 =log|2−t1−t|+C                                                  =log|2−sinx1−sinx|+C

Q49.

Integrate the rational function 1x(xn−1)

Ans.

We have, 1x(xn−1)=1x(xn−1)×xn−1xn−1                                          =xn−1xn(xn−1)Let    t=xn⇒dtdx=nxn−1⇒dtnxn−1=dx∴∫1x(xn−1) dx=∫xn−1xn(xn−1) dx                                       =1n∫1t(t−1) dtLet 1t(t−1)=At+Bt−1                       1=A(t−1)+Bt                      =t(A+B)−ASubstituting​ t= 0 and 1 respectively, we getA=1  and   B=−1∴          1t(t−1)=1t+−1t−1⇒∫1x(xn−1) dx=1n∫(1t−1t−1) dt                                      =1n∫1t dt−1n∫1t−1 dt                                     =1nlog|t|−1nlog|t−1|+C                                     =1nlog|xn|−1nlog|xn−1|+C                                    =1nlog|xnxn−1|+C

Q50.

Integrate the rational function 1x4−1

Ans.

We have, 1x4−1=1(x+1)(x−1)(x2+1)Let  1x4−1=A(x+1)+B(x−1)+Cx+D(x2+1)                   1=A(x−1)(x2+1)+B(x+1)(x2+1)+(x2−1)(Cx+D)                   1=A(x3+x−x2−1)+B(x3+x+x2+1)+C(x3−x)+D(x2−1)                   1=x3(A+B+C)+x2(−A+B+D)+x(A+B−C)+(−A+B−D)Equating the coefficients of x2, x and constant term, we get       A+B+C=0 −A+B+D=0       A+B−C=0 −A+B−D=1Solving, the above equations, we getA=−14,  B=14, C=0 and D=−12∴    1x4−1=−14(x+1)+14(x−1)+−12(x2+1)⇒∫1x4−1 dx=−14∫1x+1 dx+14∫1x−1 dx−12∫1x2+1 dx                                 =−14log|(x+1)|+14log|(x−1)|−12tan−1x+C                                 =14log|x−1x+1|−12tan−1x+C

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