NCERT Solutions Class 12 Mathematics Chapter 7

NCERT Solutions for Class 12 Mathematics Chapter 7 Integrals introduces 'Integration,' the other side of differentiation. Differentiation provides us with the rate of change, or in geometric terms, the slope of the tangent, at every given point in the curve, while Integration gives us the area under the curve. Integrals are the inverse of differentiation; therefore, we can get the original function back by integrating the resultant derived from differentiating a function. NCERT solutions Class 12 Mathematics Chapter 7 explains what Integration is and how to use it, as well as the numerous methods for calculating it.

The integration method allows you to sum infinitesimally small parts infinitely many times, allowing you to calculate the area under the curve. With the help of examples and easily comprehensible methods provided by Extramarks, students find the concept intuitive and are able to compute integration quickly. Chapter 7 Mathematics Class 12 Integrals teaches students about integration, Its types (indefinite and definite) as well as their relationship to differentiation. It also outlines multiple approaches that can be used to achieve integration. Integration is a useful method for resolving scientific and technical issues. It can also be used to answer questions in economics, finance, and probability.

Key Topics Covered in NCERT Solutions Class 12 Mathematics Chapter 7

Integration, a process opposite of differentiation, consists of functions known as integrals that must satisfy a given differential equation. Students shall be responsible for identifying the functions whose differential will be given to them. Here are some concepts that will be covered in this paper.

  • Integration
  •  Indefinite integrals
  • Application of integrals
  • Integrals for class 12 
  • Application of integrals for class 12
  • Integral calculus

Indefinite integrals, geometrical interpretation of indefinite integrals, properties of indefinite integrals, standard integrals, and methods of integration including integration by substitution method, integration by partial fractions, and integration by parts are among the significant topics covered in NCERT Solutions Class 12 Mathematics Chapter 7. Definite integrals, the fundamental theorem of calculus, and the properties of definite integrals are some of the other essential topics covered in this chapter as well. The goal of curating these solutions is to promote fundamental knowledge of integral calculus to assist students in their studies.

List of NCERT Solutions Class 12 Mathematics Chapter 7 Exercises & Answer Solutions

One of the most important topics in calculus - integration, has a wide range of practical applications. This lesson contains several formulae and hence necessitates laser-sharp focus. Revising the solutions regularly is the greatest approach to remember these concepts and measure your understanding of integration. The following is an exercise-by-exercise detailed analysis of NCERT Solutions Class 12 Mathematics Chapter 7 Integrals to assist students in developing a strong understanding of this subject:

Chapter 7 Ex 7.1 - 22 Questions - Class 12 Mathematics

Chapter 7 Ex 7.2 - 39 Questions - Class 12 Mathematics

Chapter 7 Ex 7.3 - 24 Questions - Class 12 Mathematics

Chapter 7 Ex 7.4 - 25 Questions - Class 12 Mathematics

Chapter 7 Ex 7.5 - 23 Questions - Class 12 Mathematics

Chapter 7 Ex 7.6 - 24 Questions - Class 12 Mathematics

Chapter 7 Ex 7.7 - 11 Questions - Class 12 Mathematics

Chapter 7 Ex 7.8 - 6 Questions - Class 12 Mathematics

Chapter 7 Ex 7.9 - 22 Questions - Class 12 Mathematics

Chapter 7 Ex 7.10 - 10 Questions - Class 12 Mathematics

Chapter 7 Ex 7.11 - 21 Questions - Class 12 Mathematics

Chapter 7 Miscellaneous Exercise - 44 Questions - Class 12 Mathematics

Our subject experts create NCERT Solutions so as to assist students in understanding concepts more quickly and correctly. NCERT Solutions provide extensive, step-by-step explanations of textbook difficulties. NCERT  Solutions for all primary, secondary and higher secondary classes –

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NCERT Solutions Class 12 Mathematics Chapter 7 Formula List

NCERT solutions Class 12 Mathematics chapter 7 includes various important concepts necessary for mastering higher-level mathematics. Students must understand the derivation of the principles and formulas given. This will assist them in implementing the measures necessary to solve integration problems and limit their mistakes. Students should also keep a formula chart to quickly and efficiently review the formulae before exams. The following are some key formulae mentioned in NCERT Solutions for Class 12 Mathematics Chapter 7:

  • ∫ f(x) dx = F(x) + C
  • Power Rule: ∫ xn dx = (xn+1)/ (n+1)+ C. (Where n ≠ -1)
  • Exponential Rules: ∫ ex dx = ex + C
  • ∫ ax dx = ax /ln(a) + C
  • ∫ ln(x) dx = x ln(x) -x + C
  • Constant Multiplication Rule: ∫ a dx = ax + C, where a is the constant.
  • Reciprocal Rule: ∫ (1/x) dx = ln(x)+ C
  • Sum and Difference Rules:
  • ∫ [f(x) + g(x)] dx = ∫f(x) dx + ∫g(x) dx
  • ∫ [f(x) - g(x)] dx = ∫f(x) dx - ∫g(x) dx
  • ∫ k f(x) dx = k ∫f(x) dx, where k is any real number.
  • Integration by parts: ∫ f(x) g(x) dx = f(x) ∫ g(x) dx - ∫[d/dx f(x) * ∫ g(x) dx]dx
  • ∫ cos x dx = sin x + C
  • ∫ sin x dx = -cos x + C
  • ∫ sec 2x dx = tan x + C
  • ∫ cosec 2x dx = -cot x + C
  • ∫ sec x tan x dx = sec x + C
  • ∫ cosec x cot x dx = - cosec x + C

NCERT Class 12 Mathematics Syllabus CBSE

Term - 1

Unit Name

 

Chapter Name
 

Relations and Function

Relations and Functions

Inverse Trigonometric Functions

Algebra

 

Matrices

Determinants

Calculus

 

Continuity and Differentiability

Application of Derivatives 

Linear Programming Linear Programming

 Term - 2

Unit Name Chapter Name
 

Calculus

 

Integrals

Application of Integrals

Differential Equations

Vectors and Three-Dimensional Geometry  Vector Algebra

Three Dimensional Geometry 

Probability Probability 

 

NCERT Class 12 Mathematics Exam Pattern

Duration of Marks 3 hours 15 minutes
Marks for Internal 20 marks
Marks for Theory 80 marks
Total Number of Questions 38 Questions
Very short answer question 20 Questions
Short answer questions 7 Questions
Long Answer Questions (4 marks each) 7 Questions
Long Answer Questions (6 marks each) 4 Questions

 Key features NCERT Mathematics class 12th chapter 7

Students can study the following topics by learning the NCERT Solutions for Integrals:

As the inverse of differentiation, integration Substitution, partial fractions, and parts are used to integrate a number of functions. Simple integrals of the following categories and problems dependent on them are evaluated. Fundamental Theorem of Calculus, Definite Integrals as a Limit of a Sum (without proof). Evaluation of definite integrals and basic properties of definite integrals.

NCERT Exemplar Class 12 Mathematics 

All solutions and problems are given to help students prepare for their final exams. These example questions are a little more complex, and they cover each and every concept covered in each chapter of the Class 12 Mathematics subject. Students will fully understand all the concepts covered in each chapter by practising these NCERT exemplar for Mathematics Class 12. Exemplars provide the best solutions to challenges that students confront. To match the ideas taught in each class and provide the greatest practising materials or worksheets for students, all of these questions strictly follow updated 2022-23 CBSE guidelines.

Q.1

Find the integral of the function given below.cosxsinx1+sin2x

Ans.

We have,         cosxsinx1+sin2x=cosxsinxsin2x+cos2x+2sinxcosx                    =cosxsinx(sinx+cosx)2cosxsinx1+sin2xdx                    =cosxsinx(sinx+cosx)2dxLet t=sinx+cosxdtdx=cosxsinx                  dx=dtcosxsinxcosxsinx1+sin2xdx=cosxsinx(t)2dtcosxsinx                     =t2dt                     =1t+C                     =1sinx+cosx+C

.

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

Ans.

tan32xsec2xdx=tan22xtan2xsec2xdx                           =tan22xtan2xsec2xdx                           =sec22x1tan2xsec2xdx=sec22x.tan2xsec2xdxtan2xsec2xdxLet      t=sec2xdtdx=2sec2xtan2x                       dx=dt2sec2xtan2xSo,tan32xsec2xdx=t2.tan2xsec2x.dt2sec2xtan2x12sec2x+C                       =12t2dt12sec2x+C                       =12×t3312sec2x+C                       =16sec32x12sec2x+C

Q.3 Find the integral of the function tan4x.

Ans.

tan4xdx=tan2xtan2xdx             =(sec2x1)tan2xdx             =sec2xtan2xdxtan2xdx             =sec2xtan2xdx(sec2x1)dx             =sec2xtan2xdxsec2xdx+1dxLet t=tanxdtdx=sec2xdx=dtsec2xtan4xdx=sec2x.t2dtsec2xtanx+x+C             =t2dttanx+x+C             =t33tanx+x+C             =tan3x3tanx+x+C

Q.4

Find the integral of the function given below.sin3x+cos3xsin2xcos2x

Ans.

We have,             sin3x+cos3xsin2x cos2x=sin3xsin2x cos2x+cos3xsin2x cos2x                                       =sinxcos2x+cosxsin2x                                       =tanxsecx+cotxcosecxsin3x+cos3xsin2x cos2xdx=tanxsecxdx+cotxcosecxdx                                       =secxcosecx+C

Q.5

Find the integral of the function given below.cos2x+2sin2xcos2x

Ans.

We have,    cos2x+2sin2xcos2x=cos2x+(1cos2x)cos2x                           =1cos2x                           =sec2x  cos2x+2sin2xcos2xdx=sec2xdx                           =tanx+C

Q.6

Find the integral of the function  1sinxcos3x.

Ans.

We have, 1sinxcos3x=sin2x+cos2xsinxcos3x                             =sin2xsinxcos3x+cos2xsinxcos3x                             =tanx  sec2x+1sinx  cosx                             =tanx  sec2x+sec2xtanx   1sinxcos3xdx=tanx  sec2x  dx+sec2xtanx  dxLet t=tanxdtdx=sec2xdx=dtsec2xSo,      1sinxcos3xdx=t  sec2x.dtsec2x+sec2xt.dtsec2x                          =t  dt+1t   dt                          =t22+logt+C      1sinxcos3xdx=tan2x2+logtanx+C

Q.7

Find the integral of the function given below.cos2x(cosx+sinx)2

Ans.

We have,                   cos2x(cosx+sinx)2=cos2x(cos2x+sin2x+2sinxcosx)                                   =cos2x(1+sin2x)       cos2x(cosx+sinx)2dx=cos2x(1+sin2x)dxLet  t=1+sin2xdtdx=2cos2xdx=dt2cos2x    cos2x(cosx+sinx)2dx=cos2xt.dt2cos2x                          =121tdt                          =12logt+C                          =12log1+sin2x+C                          =12log(cosx+sinx)2+C                             =12×2log(cosx+sinx)+CHence,cos2xcosx+sinx2dx=log(cosx+sinx)+C

Q.8

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

Ans.

We have, sin1cosxdxLet t=cosxsinx=1t2and  dtdx=sinx     dx=dtsinx=dt1t2sin1cosxdx=sin1t×dt1t2                    =sin1t1t2dt                    =u1t21t2du       Let u=sin1tdudx=11t2dx=1t2du                   =udu                   =u22+C                   =sin1t22+C   sin1cosxdx=sin1cosx22+C                   =π2cos1cosx22+C    sin1x+cos1x=π2                   =π2x22+C                   =12π24πx+x2+C                   =π28+πx2x22+C                   =x22+πx2+Cπ28                   =πx2x22+C, where C’= C+π28

Q.9

Find the integral of the function given below.1cos(xa)cos(xb)

Ans.

We have 1cosxacosxb           =1sinabsinabcosxacosxb           =1sinabsinxbx+acosxacosxb           =1sinabsinxbxacosxacosxb           =1sinabsinxbcosxacosxbsinxacosxacosxb           =1sinabsinxbcosxacosxacosxbcosxbsinxacosxacosxb           =1sinabsinxbcosxbsinxacosxa           =1sinabtanxbtanxa1cosxacosxbdx               =1sinab  tanxbdxtanxadx               =1sinablogcosxb+logcosxa+C               =1sinablogcosxacosxb+C

Q.10

Choosethecorrectanswersin2xcos2xsin2xcos2xdx is equal toAtanx+cotx+CBtanx+cosecx+CCtanx+cotx+CDtanx+secx+C

Ans.

sin2xcos2xsin2xcos2xdx=sin2xsin2xcos2xdxcos2xsin2xcos2xdx       =sec2xdxcosec2xdx       =tanx+cotx+CHence, the correct option is A.

Q.11

Choosethecorrectanswerex1+xcos2exxdxequalsA​ cotexx+CBtanxex+CCtanex+CDcotex+C

Ans.

We have ex(1+x)cos2(exx)dxLet t=exxdtdx=exddxx+xddxex         =ex+xex         =ex(1+x)dx=dtex(1+x)ex(1+x)cos2(exx)dx=ex(1+x)cos2tdtex(1+x)                    =1cos2tdt                   =sec2tdx                    =tant+C                    =tan(xex)+CHence, the correct option is B.

Q.12

Integrate the function given below.3x2x6+1

Ans.

Let    t=x3dtdx=3x2dx=dt3x23x2x6+1dx=3x2(x3)2+1dx                 =3x2t2+1.dt3x2                =1t2+1dt                 =tan1t+C               =tan1(x3)+C

Q.13

Integrate the function given below.11+4x2

Ans.

We have, 11+4x2=11+(2x)2Let  t=2xdtdx=2dx=dt211+4x2dx=11+(2x)2dx                       =11+t2.dt2                       =1211+t2dt                       =12log|t+t2+1|+C                       =12log|2x+(2x)2+1|+C                       =12log|2x+4x2+1|+C

Q.14

Integrate the function given below.1(2x)2+1

Ans.

We have, 1(2x)2+1Let   t=2xdtdx=1dx=dt1(2x)2+1dx=1t2+1(dt)                                       =11+t2.dt                                       =log|t+t2+1|+C   [  1x2+a2dx     =log|x+x2+a2|]                                      =log|(2x)+(2x)2+1|+C                                      =log|1(2x)+(2x)2+1|+C   [logx=log(1x)]                                     =log|1(2x)+44x+x2+1|+C                                      =log|1(2x)+x24x+5|+C

Q.15

Integrate the function given below. 1925x2

Ans.

We have, 1925x2=195x2Let  t=5xdtdx=5dx=15dt1925x2dx=15195x2dt                   =15132t2.dt                   =15sin1t3+C                  =15sin15x3+C

Q.16

Integrate the function 3x1+2x4

Ans.

We have, 3x1+2x4=3x1+(2x2)2Let   t=2x2dtdx=22xdx=dt22x3x1+2x4dx=3x1+(2x2)2dx                   =3x1+t2.dt22x                  =32211+t2dt                   =322tan1t+C=322tan1(2x2)+C

Q.17

Integrate the function x21x6

Ans.

We have, x21x6=x21(x3)2Let  t=x3dtdx=3x2dx=dt3x2x21x6dx=x21(x3)2dx               =x21t2.dt3x2               =1311t2dt               =13(12log|1+t1t|)+C               =16log|1+x31x3|+C

Q.18

Integrate the function x1x21

Ans.

We have, x1x21=x1x21Let  t=x21dtdx=2xdx=dt2xx1x21dx=xx21dx1x21dx                =xt.dt2x1x21dx               =12t12.dtlog|x+x21|+C               =12(2t)log|x+x21|+C               =tlog|x+x21|+C               =x21log|x+x21|+C

Q.19

Integrate the function x2x6+a6

Ans.

We have, x2x6+a6=x2a6+(x3)2Let  t=x3dtdx=3x2dx=dt3x2x2x6+a6dx=x2(a3)2+(x3)2dx                  =x2(a3)2+t2.dt3x2                  =131(a3)2+t2dt                  =13log|t+t2+(a3)2|+C                  =13log|x3+(x3)2+(a3)2|+C                  =13log|x3+x6+a6|+C

Q.20

Integrate the function sec2xtan2x+4

Ans.

We have, sec2xtan2x+4=sec2xtan2x+22Let  t=tanxdtdx=sec2xdx=dtsec2xsec2xtan2x+4dx=sec2xtan2x+22dx                     =sec2xt2+22.dtsec2x                     =1t2+22dt                     =log|t+t2+(2)2|+C                     =log|tanx+tan2x+4|+C

Q.21

Integrate the function 1x2+2x+2

Ans.

We have,1x2+2x+2=1(x+1)2+1Lett=x+1dtdx=1dx=dt1x2+2x+2dx=1(x+1)2+1dx                             =1t2+12.dt                          =log|t+t2+(1)2|+C                         =log|(x+1)+(x+1)2+1|+C                        =log|(x+1)+x2+2x+2|+C

Q.22

Integrate the function 19x2+6x+5

Ans.

We have, 19x2+6x+5=1(3x+1)2+22Lett=3x+1dtdx=3dx=dt319x2+6x+5dx=1(3x+1)2+22dx                =1t2+22.dt3                =131t2+22dt                =13(12tan1t2)+C               =16tan1(3x+1)2+C

Q.23

Integrate the function 176xx2

Ans.

We have, 176xx2=116(x+3)2Let  t=x+3dtdx=1dx=dt176xx2dx=116(x+3)2dx                      =142t2dt                      =sin1(t4)+C                      =sin1(x+34)+C

Q.24

Integrate the function 1(x1)(x2)

Ans.

We have, 1(x1)(x2)=1x23x+2                              =1(x32)214Lett=x32dtdx=1dx=dt1(x1)(x2)dx=1(x32)214dx                        =1t2(12)2dt                       =log|t+t2(12)2|+C                      =log|(x32)+(x32)2(12)2|+C                      =log|(x32)+x23x+2|+C

Q.25

Integrate the function 18+3xx2

Ans.

We have, 18+3xx2=18(x23x+9494)                               =18+94(x32)2                               =1414(x32)2Lett=x32dtdx=1dx=dt18+3xx2dx=1414(x32)2dx             =1(412)2t2dt             =sin1(t412)+C             =sin1(2t41)+C             =sin1{2(x32)41}+C             =sin1{(2x3)41}+C

Q.26

Integrate the function1(xa)(xb)

Ans.

We have, 1(xa)(xb)=1x2(a+b)x+ab                              =1(xa+b2)2(a+b2)2Lett=xa+b2dtdx=1dx=dt1(xa)(xb)dx=1(xa+b2)2(a+b2)2dx                        =1t2(a+b2)2dt                        =1t2(a+b2)2dt                       =log|t+t2(a+b2)2|+C                       =log|(xa+b2)+(xa+b2)2(a+b2)2|+C                      =log|(xa+b2)+x2(a+b)x+ab|+C

Q.27

Integrate the function 4x+12x2+x3

Ans.

We have,    4x+12x2+x3Here,  4x+1=Addx(2x2+x3)+B               =A(4x+1)+BEquating the coefficients of x and the contant terms fromboth sides, we get4=4A and A+B=1A=1 and B=11=0Now,​  let t=2x2+x3dtdx=4x+1     dx=dt4x+14x+12x2+x3dx=4x+12x2+x3.dt4x+1                           =1t.dt                        =2t+C                           =22x2+x3+C

Q.28

Integrate the function x+2x21

Ans.

We have,    x+2x21Here,  x+2=Addx(x21)+B                   =A(2x)+BEquating the coefficients of x and the contant terms fromboth sides, we get1=2A and B=2A=12 and B=2Therefore,  x+2=12(2x)+2x+2x21dx=12(2x)+2x21.dx                      =12(2x)x21dx+2x21dxNow,​         let t=x21dtdx=2x            dx=dt2xthen   x+2x21dx=12(2x)t.dt2x+2x21dx                      =1t.dt+2log|x+x21|                      =t+2log|x+x21|+C                     =x21+2log|x+x21|+C

Q.29

Integrate the function 5x21+2x+3x2

Ans.

We have,    5x21+2x+3x2Here,  5x2=Addx(1+2x+3x2)+B       =A(6x+2)+BEquating the coefficients of x and the contant terms fromboth sides, we get       5=6A and 2A+B=2A=56 and B=22(56)=253=1135x2=56(6x+2)1135x21+2x+3x2dx=56(6x+2)1132x2+x3dx         =56(6x+2)1+2x+3x2dx11311+2x+3x2dxNow,​  let t=1+2x+3x2dtdx=6x+2     dx=dt6x+2,​ then5x21+2x+3x2dx=56(6x+2)1+2x+3x2.dt6x+211311+2x+3x2.dx    =561t.dt11313{(x+13)2+(23)2}.dx    =56log|t|1191(x+13)2+(23)2.dx    =56log|1+2x+3x2|119×1(23)tan1(x+1323)+C    =56log|1+2x+3x2|1132tan1(3x+12)+C

Q.30

Integrate the function 6x+7(x5)(x4)

Ans.

We have,    6x+7x5x4=6x+7x29x+20Here,  6x+7=Addxx29x+20+B   6x+7=A2x9+BEquating the coefficients of x and the contant terms fromboth sides, we get       6=2A and 9A+B=7A=62=3 and B=7+93=346x+7=32x9+346x+7x5x4dx=6x+7x29x+20.dx             =32x9+34x29x+20.dx             =32x9x29x+20dx+34x29x+20dxNow,​  let t=x29x+20dtdx=2x9     dx=dt2x9,​ then6x+7x5x4dx=32x9t.dt2x9+341x29x+20.dx   =31t.dt+341x922122.dx   =3×2t+34logx92+x922122+C   =6x29x+20+34logx92+x29x+20+C

Q.31

Integrate the function x+24xx2

Ans.

We have,    x+24xx2Here,  x+2=Addx(4xx2)+B    x+2=A(42x)+BEquating the coefficients of x and the contant terms fromboth sides, we get        1=2A and 4A+B=2A=12 and B=24(12)=4x+2=12(42x)+4x+24xx2dx=12(42x)+44xx2dx    =12(42x)4xx2dx+44xx2dx   =12(42x)4xx2dx+44(44x+x2).dx   =12(42x)4xx2dx+422(2x)2dxNow,​  let t=4xx2dtdx=42x     dx=dt42x,​ thenx+24xx2dx=12(42x)tdt42x+422(2x)2dx         =12×2t4sin1(2x2)+C         =4xx24sin1(2x2)+C

Q.32

Integrate the function x+2x2+2x+3

Ans.

We have, x+2x2+2x+3x+2x2+2x+3dx=122(x+2)x2+2x+3dx                   =122x+4x2+2x+3dx                   =122x+2x2+2x+3dx+122x2+2x+3dxLet t=x2+2x+3dtdx=2x+2dx=dt2x+2,thenx+2x2+2x+3dx=122x+2tdt2x+2+122x2+2x+3dx                    =121tdt+1(x+1)2+(2)2dx                    =12.2t+log|(x+1)+(x+1)2+(2)2|+C                    =x2+2x+3+log|(x+1)+x2+2x+3|+C

Q.33

Integrate the function x+3x22x5

Ans.

We have,    x+3x22x5Here,  x+3=Addx(x22x5)+B    x+3=A(2x2)+BEquating the coefficients of x and the contant terms fromboth sides, we get       1=2A and 2A+B=3A=12 and B=3+2(12)=4x+3=12(2x2)+4x+3x22x5dx=12(2x2)+4x22x5dx        =12(2x2)x22x5dx+41x22x5dxLet t=x22x5dtdx=2x2dx=dt2x2,thenx+3x22x5dx=12(2x2)tdt2x2+41(x1)26dx        =121tdt+41(x1)2(6)2dx        =12log|t|+4×126log(x16x1+6)+C        =12log|x22x5|+26log(x16x1+6)+C

Q.34

Integrate the function 5x+3x2+4x+10

Ans.

We have, 5x+3x2+4x+10Here,  5x+3=Addx(x2+4x+10)+B    5x+3=A(2x+4)+BEquating the coefficients of x and the contant terms fromboth sides, we get         5=2A and 4A+B=3   A=52 and B=34(52)=75x+3=52(2x+4)75x+3x2+4x+10dx=52(2x+4)7x2+4x+10dx   =52(2x+4)x2+4x+10dx7x2+4x+10dx   =52(2x+4)x2+4x+10dx7(x+2)2+6dx   =52(2x+4)x2+4x+10dx71(x+2)2+(6)2dxNow,​  let t=x2+4x+10dtdx=2x+4     dx=dt2x+4,​ then5x+3x2+4x+10dx=52(2x+4)tdt2x+471(x+2)2+(6)2dx         =52×2t7log|(x+2)+(x+2)2+(6)2|+C         =5x2+4x+107log|(x+2)+x2+4x+10|+C

Q.35

Choosethecorrectanswerdxx2+2x+2A​ xtan1x+1+CBtan1x+1+CCx+1tan1x+CDtan1x+C

Ans.

We have, dxx2+2x+2=dx(x+1)2+12       =tan1(x+11)+C       =tan1(x+1)+CHence,​ the correct option is B.

Q.36

Choosethecorrectanswerdx9x4x2equalsA​ 19sin19x88+CB12sin18x99+CC13sin19x88+CD12sin19x89+C

Ans.

dx9x4x2=dx4(x294x)=dx4{x294x+(98)2(98)2}=12dx(98)2(x98)2=12sin1(x9898)+C=12sin1(8x99)+CHence, the correct option is B.

Q.37

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

Ans.

Let  x(x+1)(x+2)=Ax+1+Bx+2    x=A(x+2)+B(x+1)    x=(A+B)x+2A+B     A+B=1 and 2A+B=0   A=1 and B=2So,  x(x+1)(x+2)=1x+1+2x+2x(x+1)(x+2)dx=1x+1dx+2x+2dx=log(x+1)1+log(x+2)2+C=log(x+2)2(x+1)+C

Q.38

Integrate the rational function 1x29

Ans.

Let  1x29=1(x+3)(x3)     =Ax+3+Bx3           1=A(x3)+B(x+3)             1=(A+B)x+3(A+B)       A+B=0 and 3(A+B)=1        A=16 and B=16So,  1(x+3)(x3)=16(x+3)+16(x3)1(x+3)(x3)dx=16(x+3)dx+16(x3)dx                      =16log|x+3|+16log|x3|+C                      =16log|x3x+3|+C

Q.39

Integrate the rational function 3x1(x1)(x2)(x3)

Ans.

Let  3x1x1x2x3           =Ax1+Bx2+Cx3       3x1=Ax2x3+Bx1x3+Cx1x2...iSubstituting x=1,2and3 respectively in equationi, we get  A=1, B=5 and C=4         3x1x1x2x3=1x1+5x2+4x33x1x1x2x3dx=1x1dx5x2dx+4x3dx    =logx15logx2+4logx3+C

Q.40

Integrate the rational function x(x1)(x2)(x3)

Ans.

Let     x(x1)(x2)(x3)           =A(x1)+B(x2)+C(x3)              x=A(x2)(x3)+B(x1)(x3)+C(x1)(x2)...(i)Substituting x=1,2and3 respectively in equation(i), we get  A=12, B=2 and C=32         x(x1)(x2)(x3)=12(x1)+2x2+32(x3)x(x1)(x2)(x3)dx=121x1dx21x2dx+321(x3)dx     =12log|x1|2log|x2|+32log|x3|+C

Q.41

Integrate the rational function 2xx2+3x+2

Ans.

Let  2xx2+3x+2=2x(x+1)(x+2)              =Ax+1+Bx+2                 2x=A(x+2)+B(x+1)...(i)Substituting x=1and2 respectively in equation(i), we get   A=2, B=4        2x(x+1)(x+2)=2x+1+4x+22x(x+1)(x+2)dx=2x+1dx+4x+2dx                        =2log|x+1|+4log|x+2|+C

Q.42

Integrate the rational function 1x2x(12x)

Ans.

      1x2x12x=12+122xx12x     ...iLet   2xx12x=Ax+B12x        2x=A12x+BxSubstituting x=0and12 respectively in equationi, we get  A=2, B=3     2xx12x=2x+312xFrom​ equation i,  we​​ have1x2x12xdx=12dx+122xdx+12312xdx        =12x+122logx34log12x+C        =12x+logx34log12x+C

Q.43

Integrate the rational function x(x2+1)(x1)

Ans.

Let  x(x2+1)(x1)=Ax+Bx2+1+Cx1    ...(i)                         x=(Ax+B)(x1)+C(x2+1)          =Ax2Ax+BxB+Cx2+C          =(A+C)x2+(A+B)x+(B+C)Equating coefficients of x2, x and constant term from both sides,we get     A+C=0, A+B=1 and  B+C=0On​ solving these equations, we getA=12,   B=12,  C=12So, from equation(i), we get         x(x2+1)(x1)=12(x+1)x2+1+12x1x(x2+1)(x1)dx=12(x)x2+1dx+121x2+1dx+121x1dx                       =14log|x2+1|+12tan1x+12log|x1|+C                       =12log|x1|14log|x2+1|+12tan1x+C

Q.44

Integrate the rational function x(x1)2(x+2)

Ans.

Let     x(x1)2(x+2)=A(x1)+B(x1)2+C(x+2)               x=A(x1)(x+2)+B(x+2)+C(x1)2      ...(i)Substituting x=1 in equation(i), we get              B=13Equating coefficients of x2 and constant term, we get      A+C=02A+2B+C=0On solving, we getA=29 and C=29        x(x1)2(x+2)=29(x1)+13(x1)229(x+2)x(x1)2(x+2)dx=291x1dx+131(x1)2dx291(x+2)dx                        =29log|x1|+131(x1)29log|x+2|+C                        =29log|x1x+2|131(x1)+C

Q.45

Integrate the rational function 3x+5x3x2x+1

Ans.

Given,  3x+5x3x2x+1=3x+5x2(x1)1(x1)      =3x+5(x1)(x21)       =3x+5(x1)(x1)(x+1)      =3x+5(x1)2(x+1)Let,     3x+5(x1)2(x+1)=A(x1)+B(x1)2+C(x+1)         3x+5=A(x1)(x+1)+B(x+1)+C(x1)2                        =A(x21)+B(x+1)+C(x22x+1)      ...(i)Substituting x=1 in equation (i), we get     B=82=4Equating coefficients of x2, x and constant term,we getA+C=0 and B2C=3On solving these equations, we getA=12 and C=12    3x+5(x1)2(x+1)=12(x1)+4(x1)2+12(x+1)3x+5(x1)2(x+1)dx=121(x1)dx+41(x1)2dx+121(x+1)dx     =12log|x1|+4(1x1)+12log|x+1|+C    =12log|x+1x1|4x1+C

Q.46

Integrate the rational function 2x3(x21)(2x+3)

Ans.

2x3(x21)(2x+3)=2x3(x1)(x+1)(2x+3)            =A(x+1)+B(x1)+C(2x+3)             2x3=A(x1)(2x+3)+B(x1)(2x+3)+C(x1)(x+1)    ...(i)Substituting x=1,1 and32 respectively in equation(i), we getA=52,  B=110 and C=245  2x3(x1)(x+1)(2x+3)=52(x+1)110(x1)245(2x+3)  2x3(x21)(2x+3)dx=521(x+1)dx1101(x1)dx                                          2451(2x+3)dx                          =52log|(x+1)|110log|(x1)|                                                245×2log|(2x+3)|+D                        =52log|(x+1)|110log|(x1)|                                                125log|(2x+3)|+D

Q.47

Integrate the rational function 5x(x+1)(x24)

Ans.

5xx+1x24=5xx+1x2x+2                 =Ax+1+Bx+2+Cx2      5x=Ax2x+2+Bx+1x2+Cx+1x+2      ...iSubstituting x=1,2 and  2 respectively in equationi, we getA=53,  B=52 and C=565xx+1x2x+2=53x+152x+2+56x22x3x212x+3dx=531x+1dx521x+2dx                                         +561x2dx                         =53logx+152logx+2                                       +56logx2+D

Q.48

Integrate the rational function x3+x+1x21

Ans.

    x3+x+1x21=x+2x+1x21Let,   2x+1x21=Ax+1+Bx12x+1=A(x1)+B(x+1)   ...(i)Substituting x=1 and 1 respectively in equation  (i), we get2(1)+1=A(11)+B(1+1)             1=A(2)A=12and B=32      x3+x+1x21=x+12(x+1)+32(x1)x3+x+1x21dx=xdx+121(x+1)dx+321x1dx          =x22+12log|x+1|+32log|x1|+C

Q.49

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

Ans.

We have, 2(1x)(1+x2)Let   2(1x)(1+x2)=A1x+Bx+C1+x2                    2=A(1+x2)+(Bx+C)(1x)                    2=x2(AB)+x(BC)+(A+C)Equating the coefficients of x2, x and constant term, we getAB=0BC=0A+C=2On solving these equations, we getA=1, B=1 and C=12(1x)(1+x2)=11x+x+11+x22(1x)(1+x2)dx=11xdx+x+11+x2dx                         =11xdx+x1+x2dx+11+x2dx                         =log|1x|+12log|1+x2|+tan1x+C

Q.50

Integrate the rational function3x1(x+2)2

Ans.

We have, 3x1(x+2)2Let  3x1(x+2)2=A(x+2)+B(x+2)2      3x1=A(x+2)+BEquating the coefficients of x2, x and constant term, we getA=32A+B=1Solving, the above equations, we getA=3 and B=7         3x1(x+2)2=3(x+2)7(x+2)23x1(x+2)2dx=3(x+2)dx7(x+2)2dx                 =3log|x+2|7(1x+2)+C                 =3log|x+2|+7x+2+C

Q.51

Integrate the rational function 1x41

Ans.

We have, 1x41=1(x+1)(x1)(x2+1)Let  1x41=A(x+1)+B(x1)+Cx+D(x2+1)         1=A(x1)(x2+1)+B(x+1)(x2+1)+(x21)(Cx+D)         1=A(x3+xx21)+B(x3+x+x2+1)+C(x3x)+D(x21)         1=x3(A+B+C)+x2(A+B+D)+x(A+BC)+(A+BD)Equating the coefficients of x2, x and constant term, we get     A+B+C=0A+B+D=0     A+BC=0A+BD=1Solving, the above equations, we getA=14,  B=14,C=0 and D=12    1x41=14(x+1)+14(x1)+12(x2+1)1x41dx=141x+1dx+141x1dx121x2+1dx                =14log|(x+1)|+14log|(x1)|12tan1x+C                =14log|x1x+1|12tan1x+C

Q.52

Integrate the rational function 1x(xn1)

Ans.

We have, 1x(xn1)=1x(xn1)×xn1xn1                        =xn1xn(xn1)Let   t=xndtdx=nxn1dtnxn1=dx1x(xn1)dx=xn1xn(xn1)dx                  =1n1t(t1)dtLet 1t(t1)=At+Bt1                     1=A(t1)+Bt      =t(A+B)ASubstituting​ t= 0 and 1 respectively, we getA=1  and   B=1          1t(t1)=1t+1t11x(xn1)dx=1n(1t1t1)dt      =1n1tdt1n1t1dt      =1nlog|t|1nlog|t1|+C      =1nlog|xn|1nlog|xn1|+C     =1nlog|xnxn1|+C

Q.53

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

Ans.

We have, cosx(1sinx)(2sinx)Let  t=sinxdtdx=cosxdtcosx=dxcosx(1sinx)(2sinx)dx=cosx(1t)(2t)dtcosx                              =1(1t)(2t)dtLet  1(1t)(2t)=A1t+B2t                  1=A(2t)+B(1t)     ...(1)Substituting t=1 and 2 respectively in eqution (1),we getA=1 and B=11(1t)(2t)=11t12tSo,cosx(1sinx)(2sinx)dx=1(1t)(2t)dt                             =11tdt12tdt                            =log|1t|+log|2t|+C                           =log|2t1t|+C                            =log|2sinx1sinx|+C

Q.54

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=1dx(2(x2+3)dx+61(x2+4)dx)                       =x+23tan1(x3)62tan1(x2)+C                      =x+23tan1(x3)3tan1(x2)+C

Q.55

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

Ans.

We have,  2x(x2+1)(x2+3)Let  t=x2dtdx=2xdx=dt2x2x(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 and3 respectively, we getA=12  and  B=121(t+1)(t+3)=12(t+1)12(t+3)1(t+1)(t+3)dx=12(t+1)dt12(t+3)dx                     =12log|t+1|12log|t+3|+C                      =12log|t+1t+3|+C                      =12log|x2+1x2+3|+C

Q.56

Integrate the rational function 1x(x41)

Ans.

We have, 1x(x41)=1x(x41)×x3x3                        =x3x4(x41)Let  t=x4dtdx=4x41dt4x3=dx1x(x41)dx=x3x4(x41)dx                 =x3t(t1)dt4x3                 =141t(t1)dtLet 1t(t1)=At+Bt1             1=A(t1)+BtSubstituting​ t= 0 and 1 respectively, we getA=1  and   B=1   1t(t1)=1t+1t11x(x41)dx=14(1t+1t1)dt                     =141tdt+141t1dt                    =14log|t|+14log|t1|+C                    =14log|x4|+14log|x41|+C                    =14log|x41x4|+C

Q.57

Integrate the rational function 1(ex1)

Ans.

1(ex1)dxLet  t=exdtdx=ex     dx=dtex=dtt1(ex1)dx=1(t1)dtt         =1t(t1)dtLet  1t(t1)=At+Bt1      =A(t1)+Btt(t1)          1=A(t1)+Bt     ...(i)Putting t=0​ and 1 respectively in equation (i),we getA=1 and B=1    1t(t1)=1t+1t11t(t1)dt=1tdt+1t1dt                 =log|t|+log|t1|+C                 =log|t1t|+C1(ex1)dx=log|ex1ex|+C

Q.58

Choosethecorrectanswerxdxx1x2dxequalsA​ logx12x2+CBlogx22x1+CClogx1x22+CDlogx1x2+C

Ans.

We have, xdx(x1)(x2)dxLet x(x1)(x2)=Ax1+Bx2                        =A(x2)+B(x1)(x1)(x2)                 x=A(x2)+B(x1)    ...(i)Substituting x=1  and  2, in equation(i),​ we getA=1 and B=2         x(x1)(x2)=1x1+2x2x(x1)(x2)dx=1x1dx+2x2dx                       = log|x1|+2log|x2|+C                      =log|(x2)2(x1)|+CHence, the option (B) is correct.

Q.59

Choosethecorrectanswerdxxx2+1equalsA​ logx12logx2+1+CBlogx+12logx2+1+CClogx+12logx2+1+CD12logx+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=1xdxx(x2+1)dx                  = log|x|2log|x2+1|+CThus, the correct option is (A).

Q.60

Integrate the function x sin x.

Ans.

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

Q.61

Integrate the functions
x sin 3x

Ans.

Let  I=xsin3xdxTaking algebraic function i.e. x as a first function and sin3x as second function and integrating by parts, we get      =xsin3xdx{(ddxx)sin3xdx}dx      =x(13cos3x)1.(13cosx)dx      =13xcos3x+19sin3x+C

Q.62

Integrate the functions
x2 ex

Ans.

Let  I=x2exdxTaking algebraic function i.e. x2 as a first function and ex as second function and integrating by parts, we get     =x2exdx{(ddxx2)exdx}dx     =x2(ex)2x.(ex)dx     =x2ex2[xexdx{(ddxx)exdx}dx]+C     =x2ex2[xex1.exdx]+C     =x2ex2xex+2ex+C     =ex(x22x+2)+C

Q.63

Integrate the functions
x logx

Ans.

Let  I=xlogxdxTaking logx as a first function and x as second function and integrating by parts, we get =logxxdx{(ddxlogx)xdx}dx =logx(x22)1x.(x22)dx   =x22logxx2dx+C      =x22logxx24+C

Q.64

Integrate the functions
x log 2x

Ans.

Let  I=xlog2xdxTaking log2x as a first function and x as second function and integrating by parts, we get     =log2xxdx{(ddxlog2x)xdx}dx     =log2x(x22)12x×2.(x22)dx     =x22log2xx2dx+C     =x22log2xx24+C

Q.65

Integrate the functions
x sin-1x

Ans.

Let  I=xsin1xdxLet t=sin1xx=sintcost=1sin2t=1x2Then, dtdx=ddxsin1x     =11x2     =1cost        dx=costdtSo,          I=tsintcostdt     =12tsin2tdtTaking algebraic function i.e. t as a first function and sin2t as second function and integrating by parts, we get      =12[tsin2tdt{(ddtt)sin2tdt}dt]      =12[t(12cos2t)1.(12co2t)dx]      =12[12tcos2t+14sin2t+C]      =14t(12sin2t)+18×2sintcost+C      =14t(2sin2t1)+14sintcost+C      =14(2x21)sin1x+14x1x2+C

Q.66

Integrate the functions
x tan-1x

Ans.

Let  I=xtan1xdxTaking tan1x as a first function and x as second function and integrating by parts, we get=tan1xxdx{(ddxtan1x)xdx}dx=tan1x(x22)11+x2.(x22)dx=x22tan1x12x21+x2dx+C=x22tan1x121+x211+x2dx+C=x22tan1x121+x21+x2dx+1211+x2dx+C=x22tan1x121.dx+12tan1x+C=x22tan1x12xdx+12tan1x+C

Q.67

Integrate the functions
x cos-1x

Ans.

Let  I=xcos1xdxLet t=cos1xx=cost  sint=1cos2t=1x2Then, dtdx=ddxcos1x                =11x2               =1sint         dx=sintdtSo,           I=tcost(sint)dt              =12tsin2tdtTaking algebraic function i.e. t as a first function and sin2t as second function and integrating by parts, we get     =12[tsin2tdt{(ddtt)sin2tdt}dt]     =12[t(12cos2t)1.(12co2t)dx]     =12[12tcos2t+14sin2t+C]     =14t(12sin2t)18×2sintcost+C     =14t(12sin2t)14sintcost+C     =14(12x2)sin1x14x1x2+C

Q.68

Integrate the functions
(sin–1x)2

Ans.

Let  I=(sin1x)2dx       =(sin1x)2.1dxTaking(sin1x)2 as first function and 1 as second function and integrating by parts, we get        I=(sin1x)21dx{ddx(sin1x)21dx}dx      =(sin1x)2x{2sin1xddx(sin1x).x}dx      =(sin1x)2x2xsin1x1x2dx      =(sin1x)2x+sin1x2x1x2dx      =(sin1x)2x+sin1x2x1x2dx(ddxsin1x2x1x2dx)dx      =(sin1x)2x+sin1x2xtdt2x{ddxsin1x2xtdt2x}dx                               [Let t=1x2dtdx=2xdx=dt2x]      =(sin1x)2x+sin1x.2t{11x22t}dx      =(sin1x)2x+2sin1x.1x211x221x2dx      =(sin1x)2x+2sin1x.1x221dx      =x(sin1x)2+21x2sin1x2x+C

Q.69

Integrate the functions
x2 logx

Ans.

Let  I=x2logxdxTaking logx as a first function and x2 as second function and integrating by parts, we get     =logxx2dx{(ddxlogx)x2dx}dx      =logx(x33)1x.(x33)dx      =x33logxx23dx+C      =x33logxx39+C

Q.70

Integrate the functionsxcos1x1x2

Ans.

Let  I=xcos1x1x2dx       =12cos1x2x1x2dxTakingcos1x as first function and 2x1x2 as second function and integrating by parts, we get        I=12{cos1x2x1x2dx(ddxcos1x2x1x2dx)dx}       =12{cos1x2xtdt2x(ddxcos1x2xtdt2x)dx}                         [Let t=1x2dtdx=2xdx=dt2x]       =12[cos1x.2t(11x22t)dx]       =12[cos1x.21x2(11x221x2)dx]        =12(2cos1x1x2+21dx)       =12(2cos1x1x2+2x)+C       =(1x2cos1x+x)+C

Q.71

Integrate the functions x sec2x

Ans.

Let  I=xsec2xdxTaking algebraic function i.e. x as a first function and sec2x assecond function and integrating by parts, we get       =xsec2xdx{(ddxx)sec2xdx}dx        =x(tanx)1.tanxdx       =xtanxtanxdx+C       =xtanx+log|cosx|+C

Q.72

Integrate the functions tan-1x

Ans.

Let  I=tan1xdx=tan1x.1dxTaking tan1x as a first function and 1 as second function and integrating by parts, we get       =tan1x1dx{(ddxtan1x)1dx}dx        =tan1x(x)11+x2.(x)dx        =xtan1xx1+x2dx+C        =xtan1x12log|1+x2|+C

Q.73

Integrate the functions x (logx)2

Ans.

Let  I=x(logx)2dxTaking (logx)2 as a first function and x as second function and integrating by parts, we get      =(logx)2xdx{(ddx(logx)2)xdx}dx      =(logx)2(x22)2logxx.(x22)dx      =x22(logx)2xlogxdx+C      =x22(logx)2[logxxdx{(ddxlogx)xdx}dx]      =x22(logx)2{logx(x22)1x.(x22)dx}      =x22(logx)2{x22logxx2dx}      =x22(logx)2x22logx+x24+C

Q.74

Integrate the functions (x2+1) logx

Ans.

Let  I=(x2+1)logxdxTaking 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)logxx39x+C

Q.75

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=exf(x)+C]

Q.76

Integrate the functions xex(1+x)2

Ans.

Let  I=xex(1+x)2dx       =ex{x(1+x)2}dx       =ex{1+x1(1+x)2}dx        =ex{1+x(1+x)21(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)2dx       =ex{1(1+x)1(1+x)2}dx      =ex1+x+C            [ex{f(x)+f(x)}=exf(x)]

Q.77

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)]

Q.78

Integrate the functionse x(1x1x2)

Ans.

Let  I=ex(1x1x2)dxLet​ f(x)=1x  and f’(x)=1x2Thus, the given integrand is of the form ex[f(x)+f(x)].Therefore,   I=ex(1x1x2)dx                =exx+C           [ex{f(x)+f(x)}=exf(x)]

Q.79

Integrate the functions(x3)ex(x1)3

Ans.

Let  I=(x3)ex(x1)3dx       =ex{x12(1x)3}dx       =ex{x1(1x)32(1x)3}dx       =ex{1(1x)22(1x)3}dxLet​ f(x)=1(1x)2  and f’(x)=2(1x)3Thus, the given integrand is of the form ex[f(x)+f(x)].Therefore,       I=(x3)ex(x1)3dx       =ex{1(1x)22(1x)3}dx      =ex(1x)2+C            [ex{f(x)+f(x)}=exf(x)]

Q.80

Integrate the functions e2x sinx

Ans.

We have, e2xsinxLet  I=e2xsinxdxTaking sinx as first function and e2x as second function, then Integrating by parts, we get      I=e2xsinxdx        =sinxe2xdx(ddxsinxe2xdx)dx        =sinx(e2x2){cosx(e2x2)}dx        =e2x2sinx12cosx.e2xdx        =e2x2sinx12{cosxe2xdx(ddxcosxe2xdx)dx}        =e2x2sinx12{cosx(e2x2){sinx(e2x2)}dx}        =e2x2sinx12cosx(e2x2)14e2xsinxdx+C     I=e2xsinx214cosxe2x14I+CI+14I=e2xsinx214cosxe2x+C      54I=e2xsinx214cosxe2x+C         I=45e2x4(2sinxcosx)+C

Q.81

Integrate the functionssin1(2x1+x2)

Ans.

We have, sin1(2x1+x2)Let  I=sin1(2x1+x2)dxLetx=tanθdx=sec2θ     I=sin1(2x1+x2)dx      =sin1(2tanθ1+tan2θ)sec2θ     =sin1(sin2θ)sec2θ       =2θsec2θ       =2[θsec2θ(dθsec2θ)]      =2[θtanθ1tanθ]       =2θtanθ2log|secθ|+C      =2θtanθ2log|1+tan2θ|+C       I=2xtan1x2log|1+x2|+C.

Q.82

Choosethecorrectanswerx2ex3dxequalsA​ 13ex3+CB13ex2+CC12ex3+CD12ex2+C

Ans.

We  have  x2ex3dxLet​ t=x3dtdx=3x2x2ex3dx=x2etdt3x2              =13etdt              =13ex3+CHence, the correct option is A.

Q.83

Choosethecorrectanswerexsecx1+tanxdxequalsA​ excosx+CBexsecx+CCexsinx+CDextanx+C

Ans.

Let  I=exsecx(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.

Q.84

4x2dx

Ans.

We have, 4x2Let I=4x2dx       =22x2dx                    [a2x2dx=12xa2x2+a22sin1xa+C]       =12x22x2+222sin1x2+C      =x24x2+2sin1x2+C

Q.85

14x2dx

Ans.

We have, 14x2Let I=14x2dx         =12(2x)2dx[Let t=2xdtdx=2dx=dt2]        =12t2dt2                     [a2x2dx=12xa2x2+a22sin1xa+C]       =12(12t12t2+a22sin1t1)+C       =12(2x214x2+12sin12x1)+C      =x214x2+14sin12x+C

Q.86

x2+4x+6dx

Ans.

We have, x2+4x+6=(x+2)2+2Let I=(x+2)2+(2)2dx       =t2+(2)2dt   [Let t=x+2dtdx=1]                      [x2+a2dx=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|+C

Q.87

x2+4x+1dx

Ans.

We have, x2+4x+1=(x+2)23Let I=(x+2)2(3)2dx        =t2(3)2dt     [Let t=x+2dtdx=1]                      [x2a2dx=x2x2a2a22log|x+x2a2|]       =t2t2(3)2(3)22log|t+t2(3)2|+CPutting t=x+2, we get       =(x+2)2(x+2)2(3)232log|(x+2)+(x+2)2(3)2|+C       =(x+2)2x2+4x+132log|(x+2)+x2+4x+1|+C

Q.88

14xx2dx

Ans.

We have, 14xx2=5(x2+4x+4)                           =(5)2(x+2)2Let I=(5)2(x+2)2dx       =(5)2t2dt     [Let t=x+2dtdx=1]                      [a2x2dx=x2a2x2+a22sin1xa]      =t2(5)2t2+(5)22sin1t(5)+CPutting t=x+2,weget      =(x+2)2(5)2(x+2)2+(5)22sin1(x+2)(5)+C      =(x+2)214xx2+52sin1(x+2)(5)+C

Q.89

x2+4x5dx

Ans.

We have, x2+4x5=(x+2)29Let I=(x+2)2(3)2dx       =t2(3)2dt      [Let t=x+2dtdx=1]                   [x2a2dx=x2x2a2a22log|x+x2a2|]      =t2t2(3)2(3)22log|t+t2(3)2|+CPutting t=x+2, we get      =(x+2)2(x+2)2(3)292log|(x+2)+(x+2)2(3)2|+C      =(x+2)2x2+4x592log|(x+2)+x2+4x5|+C

Q.90

1+3xx2dx

Ans.

We have, 1+3xx2=1(x23x+9494)                           =(132)2(x32)2Let I=(132)2(x32)2dx      =(132)2t2dt      [Let t=x32dtdx=1]                  [a2x2dx=x2a2x2+a22sin1xa]      =t2(132)2t2+(132)22sin1t(132)+CPutting t=x32, we get      =(x32)2(132)2(x32)2+(132)22sin1(x32)(132)+C      =(2x3)41+3xx2+138sin1(2x3)(13)+C

Q.91

x2+3xdx

Ans.

We have, x2+3x=(x2+3x+94)94                       =(x+32)2(32)2Let I=(x+32)2(32)2dx       =t2(32)2dt     [Let t=x+32dtdx=1]                       [x2a2dx=x2x2a2a22log|x+x2a2|]      =t2t2(32)2(32)22log|t+t2(32)2|+CPutting t=x+32, we get     =(x+32)2(x+32)2(32)298log|(x+32)+(x+32)2(32)2|+C     =(2x+3)4x2+3x98log|(x+32)+x2+3x|+C

Q.92

1+x29dx

Ans.

Wehave,1+x29=139+x2Let  139+x2dx=1332+x2dx                         =13x2+32dx                            [x2+a2dx=x2x2+a2+a22log|x+x2+a2|]                       =13(x2x2+32+322log|x+x2+32|)+C                       =x6x2+9+32log|x+x2+9|+C

Q.93

Choosethecorrectanswer1+x2dx is equal toAx21+x2+12logx+1+x2+CB231+x232+CC23x1+x232+CDx221+x2+12 x2logx+1+x2+C

Ans.

Let  I=1+x2dx       =x2+12dx             [x2+a2dx=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.

Q.94

Choosethecorrectanswerx28x+7dx is equal toA12x4x28x+7+9logx4+x28x+7+CB14x+4x28x+7+9logx+4+x28x+7+CC12x4x28x+732logx4+x28x+7+CD12x4x28x+792logx4+x28x+7+C

Ans.

We have, x28x+7=(x4)29Let I=(x4)2(3)2dx       =t2(3)2dt     [Let t=x4dtdx=1]                     [x2a2dx=x2x2a2a22log|x+x2a2|]      =t2t2(3)2(3)22log|t+t2(3)2|+CPutting t=x4,weget     =(x4)2(x4)2(3)292log|(x4)+(x4)2(3)2|+C     =(x4)2x28x+792log|(x4)+x28x+7|+CHence, the correct option is D.

Q.95

xx+x2dx

Ans.

xx+x2dxLet x=Addx(x+x2)+B       x   =A(1+2x)+B     1=2A & A+B=0    A=12 & B=12xx+x2dx=12(1+x)x+x2dx12x+x2dx               =12×(x+x2)32(32)12(x+12)2(12)2dx               =13×(x+x2)3212×(x+12)2(x+12)2(12)2                              +12×(12)22log|x+(x+12)2(12)2|+C               =(x+x2)32318(2x+1)x+x2+116log|x+x+x2|+C

Q.96

x+12x2+3dx

Ans.

(x+1)2x2+3dxLet x+1=Addx(2x2+3)+B        x+1=A(4x+0)+B    x+1=4Ax   & B=1    A=14 & B=1(x+1)2x2+3dx=144x2x2+3dx+12x2+3dx                       =14×(2x2+3)3232+2x2+(32)2dx                =16×(2x2+3)32+2x2+(32)2dx                =16×(2x2+3)32+2(12xx2+(32)2        +  (32)22log|x+(32)2|)+C                =16×(2x2+3)32+22xx2+32                                    +324log|x+x2+(32)2|+C                =(2x2+3)326x22x2+3324log|x+x2+32|+C

Q.97

(x+3)34xx2

Ans.

(x+3)34xx2dxLet x+3=Addx(34xx2)+B        x+3=A(42x)+B          3=4A+B   & 2A=1    A=12 &     B=3+4A       =3+4(12)       =32=1(x+3)34xx2dx=12(42x)34xx2dx                                          +134xx2dx                              =12(34xx2)3232+7(x+2)2dx                              =(34xx2)323+(7)2(x+2)2dx           =(34xx2)323+12(x+2)(7)2(x+2)2+72sin1(x+27)+C                                   +(7)22sin1(x+27)+C         =(34xx2)323+12(x+2)34xx2  +72sin1(x+27)+C

Q.98

Evaluate the following definite integrals as limit of sums.abxdx

Ans.

By definitionabf(x)dx=(ba)limn1n[f(a)+f(a+h)+...+f(a+(n1)h)],where, h=banHere,  f(x)=x,  h=banabxdx=(ba)limn1n[f(a)+f(a+h)+...+f(a+(n1)h)]          =(ba)limn1n[a+a+h+...+a+(n1)h]          =(ba)limn1n[na+h+2h+...+(n1)h]          =(ba)limn1n[na+{1+2+...+(n1)}h]          =(ba)limn1n[na+(n1)n2h]        [n=n(n+1)2]          =(ba)limn[a+(n1)2(ban)]          =(ba)limn[a+(ba)2(11n)]          =(ba)[a+(ba)2(10)]          =(ba)(2a+ba2)          =(ba)(b+a)2  =b2a22

Q.99

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

Ans.

Wehave,  05(x+1)dxBy definitionabf(x)dx=(ba)limn1n[f(a)+f(a+h)+...+f(a+(n1)h)],where, h=banHere,  a=0,  b=5,  f(x)=x+1,  h=50n=5n05(x+1)dx=(ba)limn1n[f(0)+f(0+h)+...+f(0+(n1)h)]                            =5limn1n[(0+1)+(h+1)+...+{(n1)h+1}]                            =5limn1n[n+h+2h+...+(n1)h]                            =5limn1n[n+{1+2+...+(n1)}h]                            =5limn1n[n+(n1)n2h][n=n(n+1)2]                            =5limn[1+(n1)2(5n)]                            =5limn[1+52(11n)]                            =5[1+52(10)]                            =5×72  =352

Q.100

Evaluate the following definite integrals as limit of sums.23x2dx

Ans.

We have, 23x2dxabf(x)dx=(ba)limn1n[f(a)+f(a+h)+...+f(a+(n1)h)],where, h=banHere,  a=2,  b=3,  f(x)=x2,  h=32n=1n23x2dx          =(32)limn1n[f(2)+f(2+h)+...+f(2+(n1)h)]          =limn1n[(2)2+(2+h)2+...+{2+(n1)h}2]          =limn1n[4+(4+4h+h2)+...+{4+4(n1)h+(n1)2h2}]          =limn1n[4n+4{1+2+...+(n1)}h+{1+22+32+...+(n1)2}h2]          =limn1n[4n+4(n1)n2h+(n1)n(2n2+1)6h2]          =limn1n[4n+4(n1)n2.1n+(n1)n(2n2+1)6.1n2]                                        [n=n(n+1)2,n2=n(n+1)(2n+1)6]          =limn[4+4(11n)2+(11n)(21n)6]          =[4+4(10)2+(10)(20)6]          =4+2+13          =193

Q.101

Evaluate the following definite integrals as limit of sums.14(x2x)dx

Ans.

We have, 14(x2x)dx=14x2dx14xdx                           =I1I2    (Let)abf(x)dx=(ba)limn1n[f(a)+f(a+h)+...+f(a+(n1)h)]where, h=banHere,  a=1,  b=4,  f(x)=x2,  h=41n=3n   I1=14x2dx       =(41)limn1n[f(1)+f(1+h)+...+f(1+(n1)h)]           =3limn1n[(1)2+(1+h)2+...+{1+(n1)h}2]           =3limn1n[1+(1+2h+h2)+...+{1+2(n1)h+(n1)2h2}]            =3limn1n[n+{1+2+...+(n1)}2h+{1+22+32+...+(n1)2}h2]           =3limn1n[n+(n1)n22h+(n1)n(2n2+1)6h2]           =3limn1n[n+(n1)n.3n+(n1)n(2n2+1)6.9n2]                   n=nn+12,n2=nn+12n+16           =3limn[1+(11n)×3+(11n)(21n)6×9]

           =3[1+3(10)+3(10)(20)2]           =3(1+3+3)=21For​   I2=14xdx14xdx=(41)limn1n[f(1)+f(1+h)+...+f(1+(n1)h)]                   =3limn1n[1+1+h+...+1+(n1)h]                   =3limn1n[n+h+2h+...+(n1)h]                   =3limn1n[n+{1+2+...+(n1)}h]                   =3limn1n[n+(n1)n2h][n=n(n+1)2]                   =3limn[1+(n1)2(3n)]                    =3limn[1+32(11n)]                   =3[1+32(10)]                    =3(52)=152So,  14(x2x)dx                    =I1I2=21152=272

Q.102

Evaluate the following definite integrals as limit of sums.11exdx

Ans.

We  have, 11exdxabf(x)dx=(ba)limn1n[f(a)+f(a+h)+...+f(a+(n1)h)],where, h=banHere,  a=1,  b=1,  f(x)=ex,  h=1(1)n=2n11exdx=(1+1)limn1n[f(1)+f(1+h)+...+f(1+(n1)h)]           =2limn1n[e1+e1+h+...+e1+(n1)h]Using the sum to n terms of a G.P., where a=e1, r=eh,we have11exdx=2limn1n[e1(enh1)(eh1)]           =2limn1n[e1(en.2n1)(e2n1)]           =2limn1n[e1(e21)(e2n1)]           =2e1(e21)limn[e2n12n].2           =2e1(e21)2           =e1e           limx0eh1h=1

Q.103

Evaluate the following definite integrals as limit of sums.04(x+e2x)dx

Ans.

We have, 04(x+e2x)dx=04xdx+04e2xdx                           =I1+I2   (Let)abf(x)dx=(ba)limn1n[f(a)+f(a+h)+...+f(a+(n1)h)]where, h=banHere,  a=0,  b=4,  f(x)=x,  h=40n=4n        I1=04xdx04xdx=(40)limn1n[f(0)+f(0+h)+...+f(0+(n1)h)]             =4limn1n[0+h+...+(n1)h]             =4limn1n[h+2h+...+(n1)h]             =4limn1n[{1+2+...+(n1)}h]             =4limn1n[(n1)n2h][n=n(n+1)2]             =4limn[(n1)2(4n)]             =4limn[42(11n)]             =4[2(10)]             =4(2)=8        I2=04e2xdx             =(40)limn1n[f(0)+f(0+h)+...+f(0+(n1)h)]             =4limn1n[e0+e2h+...+e2(n1)h]             =4limn1n[1+e2h+...+e2(n1)h]Using the sum to n terms of a G.P., where a=1, r=e2h,we have

04e2xdx=4limn1n[1.(e2nh1)(e2h1)]           =4limn1n[(e2n.4n1)(e8n1)]           =4limn1n[(e81)(e8n1)]           =4(e81)limn[e8n18n].8           =4(e81)8         [limx0eh1h=1]           =e812So,  04(x+e2x)dx          =I1+I2          =8+e812          =15+e82

Q.104

11(x+1)dx

Ans.

Let  I=11(x+1)dx(x+1)dx=x22+x=F(x)Therefore, by the second fundamental theorem, we have      I=F(1)F(1)      =(122+1){(1)22+(1)}      =32(12)      =2

Q.105

231xdx

Ans.

Let  I=231xdx1xdx=logx=F(x)Therefore, by the second fundamental theorem, we have      I=F(3)F(2)      =log3log2=log32

Q.106

12(4x35x2+6x+9)dx

Ans.

Let  I=12(4x35x2+6x+9)dx(4x35x2+6x+9)dx      =4x3dx5x2dx+6xdx+9dx      =x453x3+3x2+9x=F(x)Therefore, by the second fundamental theorem, we have      I=F(2)F(1)       ={2453(2)3+3(2)2+9(2)}{(1)453(1)3+3(1)2+9(1)}      =(16403+12+18)(153+3+9)      =(46403)(1353)      =4640313+53      =33353      =643

Q.107

0π4sin2xdx

Ans.

Let  I=0π4sin2xdxsin2xdx=12cos2x=F(x)Therefore, by the second fundamental theorem, we have      I=F(π4)F(0)       =12cos2(π4)(12cos0)      =12cosπ2+12cos0      =0+12(1)=12

Q.108

0π2cos2xdx

Ans.

Let  I=0π2cos2xdxcos2xdx=12sin2x=F(x)Therefore, by the second fundamental theorem, we have      I=F(π2)F(0)      =12sin2(π2)(12sin0)      =12sinπ+12sin0      =0+0=0

Q.109

45exdx

Ans.

Let  I=45exdxexdx=ex=F(x)Therefore, by the second fundamental theorem, we have      I=F(5)F(4)      =e5e4      =e4(e1)

Q.110

0π4tanxdx

Ans.

Let  I=0π4tanxdxtanxdx=log|cosx|=F(x)Therefore, by the second fundamental theorem, we have      I=F(π4)F(0)      =log|cosπ4|(log|cos0|)      =log|cosπ4|+log|cos0|      =log|12|+log1      =log2+0=12log2

Q.111

π6π4cosecxdx

Ans.

Let  I=π6π4cosecxdxcosecxdx=log|cosecxcotx|=F(x)Therefore, by the second fundamental theorem, we have      I=F(π4)F(π6)     =log|cosecπ4cotπ4|log|cosecπ6cotπ6|     =log|21|log|23|     =log(2123)

Q.112

0111x2dx

Ans.

Let  I=0111x2dx11x2dx=sin1x=F(x)Therefore, by the second fundamental theorem, we have      I=F(1)F(0)      =sin1(1)sin1(0)      =π20=π2

Q.113

011(1+x2)dx

Ans.

Let  I=0111+x2dx11+x2dx=tan1x=F(x)Therefore, by the second fundamental theorem, we have      I=F(1)F(0)       =tan1(1)tan1(0)       =π40=π4

Q.114

23dx(x21)

Ans.

Let  I=23dx(x21)1x21dx=12log|x1x+1|=F(x)Therefore, by the second fundamental theorem, we have      I=F(3)F(2)      =12log|313+1|12log|212+1|       =12log|24|12log|13|      =12log|24×31|      =12log|32|

Q.115

0π2cos2xdx

Ans.

Let  I=0π2cos2xdx=0π21+cos2x2dx1+cos2x2dx=12(x+sin2x2)=F(x)Therefore, by the second fundamental theorem, we have      I=F(π2)F(0)      =12{π2+sin2.(π2)2}12{0+sin2(0)2}     =12(π2+sinπ2)12{0+0}      =12(π2)=π4

Q.116

23xdx(x2+1)

Ans.

Let  I=23xdx(x2+1)x(x2+1)dx=12log|x2+1|=F(x)Therefore, by the second fundamental theorem, we have      I=F(3)F(2)      =12log|32+1|12log|22+1|     =12log|10|12log|5|      =12log|105|      =12log|2|

Q.117

012x+3(5x2+1)dx

Ans.

Let            I=012x+3(5x2+1)dx                   =150110x(5x2+1)dx+013(5x2+1)dx2x+3(5x2+1)dx=1510x(5x2+1)dx+31(5x2+1)dx                   =15log|5x2+1|+31(5x)2+1dx                   =15log|5x2+1|+3tan15x.15                   =15log|5x2+1|+35tan1(5x)=F(x)Therefore, by the second fundamental theorem, we have                I=F(1)F(0)                 =15log|5(1)2+1|+35tan1{5(1)}[15log|5(0)2+1|    +35tan1{5(0)}]                 =15log(6)+35tan1(5)15log|1|35tan1(0)                 =15log(6)+35tan1(5)

Q.118

01xex2dx

Ans.

Let       I=01xex2dxxex2dx=xetdt2x        [Let​ t=x2dtdx=2xdx=dt2x]    =12et    =12ex2=F(x)Therefore, by the second fundamental theorem, we have             I=F(1)F(0)       =12e1212e0=12(e1)

Q.119

125x2x2+4x+3  dx

Ans.

Let                 I=125x2x2+4x+3dx5x2x2+4x+3dx=(520x+15x2+4x+3)dx                       =5dx20x+15x2+4x+3dx  ...(i)Let     20x+15=Addx(x2+4x+3)+B                      =A(2x+4)+B                      =2Ax+4A+BEquating the coefficients of x and constant term, we get2A=20 and 4A+B=15A=10  and  B=25        20x+15=10(2x+4)25From equation(i), we have  5x2x2+4x+3dx=5dx10(2x+4)25x2+4x+3dx                    =5dx10(2x+4)x2+4x+3dx+251x2+4x+3dx                    =5x10log(x2+4x+3)+251(x+2)21dx                    =5x10log(x2+4x+3)+25×12log|x+21x+2+1|+C                    =5x10log(x2+4x+3)+252log|x+1x+3|125x2x2+4x+3dx=[5x10log(x2+4x+3)+252log|x+1x+3|]12              =[5(2)10log(22+4×2+3)+252log|2+12+3|]             [5(1)10log(12+4×1+3)+252log|1+11+3|]125x2x2+4x+3dx=[5x10log(x2+4x+3)+252log|x+1x+3|]12               =[5(2)10log(22+4×2+3)+252log|2+12+3|]                   [5(1)10log(12+4×1+3)+252log|1+11+3|]               =(1010log15+252log35)(510log8+252log24)           =510(log15log8)+252(log35log24)                  =510(log5+log3log4log2)                   +252(log3log5log2+log4)                  =5(10+252)log5+(10+252)log4+(10+252)log3                   +(10252)log2               =5452log5+452log4+52log352log2125x2x2+4x+3dx=5452log54+52log32  =552(9log54log32)

Q.120

0π4(2sec2x+x3+2)dx

Ans.

Let  I=0π4(2sec2x+x3+2)dx(2sec2x+x3+2)dx      =2sec2xdx+x3dx+2dx      =2tanx+x44+2x=F(x)Therefore, by the second fundamental theorem, we have       I=F(π4)F(0)      ={2tan(π4)+(π4)44+2(π4)}{2tan(0)+(0)44+2(0)}      =2[tanπ4tan0]+14[(π4)40]+2[π40]      =2[10]+14[(π4)4]+2[π4]      =2+π41024+π2

Q.121

0π(sin2x2cos2x2)dx

Ans.

Let                    I=0π(sin2x2cos2x2)dx(sin2x2cos2x2)dx=(cos2x2sin2x2)dx                          =cosxdx      [cos2θsin2θ=cos2θ]                         =sinx=F(x)Therefore, by the second fundamental theorem, we have                         I=F(π)F(0)                           =sinπ(sin0)  =0

Q.122

026x+3x2+4dx

Ans.

            Let  I=026x+3x2+4dx6x+3x2+4dx=6xx2+4dx+3x2+4dx                  =32xx2+4dx+31x2+22dx                  =3[log(x2+4)]+32[tan1x2]=F(x)Therefore, by the second fundamental theorem, we have                 I=F(2)F(0)                   =3[log(22+4)32tan122][log(02+4)+32tan102]                  =3[log8log4]32[tan1(1)tan1(0)]                  =3log8432[π40]026x+3x2+4dx=3log23π8

Q.123

01(xex+sinπx4)dx

Ans.

Let   I=01(xex+sinπx4)dx(xex+sinπx4)dx=xexdx+sinπx4dx          =xexdx(ddxxexdx)dx1(π4)cosπx4+C          =xexexdx4πcosπx4          =xexex4πcosπx4          =ex(x1)4πcosπx4=F(x)Therefore, by the second fundamental theorem, we have        I=F(1)F(0)         ={e1(11)4πcosπ(1)4}{e0(01)4πcosπ(0)4}         ={4πcosπ4}{14π×1}         ={4π×12}{14π}         =42π+1+4π         =1+4π22π

Q.124

Choosethecorrectanswer13dx1+x2 equalsAπ3B2π3Cπ6Dπ12

Ans.

Let            I=13dx1+x211+x2dx=tan1xTherefore, by the second fundamental theorem, we have                I=F(3)F(1)                 =tan1(3)tan1(1)                 =π3π4                 =4π3π12                 =π12Hence, the correct option is D.

Q.125

Choosethecorrectanswer023dx4+9x2 equalsAπ6Bπ12Cπ24Dπ4

Ans.

Let               I=023dx4+9x214+9x2dx=122+(3x)2dx                   =12tan1(3x2)×13     =16tan1(3x2)=F(x)Therefore, by the second fundamental theorem, we have               I=F(23)F(0)     =16tan1(32×23)16tan1(32×0)     =16tan1(1)16tan1(0)     =16×π4=π24Hence, the correct option is C.

Q.126

Evaluate the integrals 01xx2+1dx

Ans.

01xx2+1dxLet  t=x2+1dtdx=2xWhen x=0, t=1 and when x=1, t=201xx2+1dx=12121tdt       =12[log|t|]12       =12[log2log1]       =12log2

Q.127

Evaluate the integrals 0π2sinϕcos5ϕ

Ans.

Let  I=0π2sinϕcos5ϕ dϕ          =0π2sinϕcos4ϕcosϕ          =0π2sinϕ(cos2ϕ)2cosϕ           =0π2sinϕ(1sin2ϕ)2cosϕ          =0π2sinϕ(12sin2ϕ+sin4ϕ)cosϕ Let t=sinϕdt=cosϕWhen ϕ=0,t=0 and when ϕ=π2,t=1     I=01t(12t2+t4)dt=01(t122t