NCERT Solutions for Class 12 Maths Chapter 7- Integrals Exercise 7.1

Mathematics is one of the oldest and most prominent scientific disciplines. It is of great use in our daily lives and is also a field equipped with many lucrative career options. Therefore, it is undoubtedly an indispensable academic subject. The prescribed NCERT academic curriculum for Class 12 Mathematics is composed of thirteen chapters organized in a logical sequence. Chapter 7 is titled Integrals. Exercise 7.1 Class 12th Maths is a challenging and comprehensive exercise. Exercise 7.1 Class 12th necessitates adequate conceptual clarity on the part of the students in order to be comprehended and solved effectively. To this end, the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.1 have been engineered in collaboration with renowned subject experts from Extramarks.

The students are advised to constantly practice the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.1 in order to retain important formulae. These NCERT Solutions would also be helping students to familiarise themselves with the logical sequence of steps in a calculation so that they can efficiently manage their time during examinations. The NCERT Solutions Class 12 for Integrals Exercise 7.1 have been designed in consideration of the latest updated CBSE syllabus by Extramarks. The content organization of the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.1 ensures that there are no discrepancies between the instruction students are receiving from their teachers and the learning resources available on the Extramarks’ website.

NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.1) Exercise 7.1

Students are often apprehensive of topics like Integrals and Calculus which are indispensable parts of the NCERT academic curriculum for Class 12. The NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.1 crafted painstakingly by Extramarks, are reliable, comprehensive, and detailed. The well-organized structure of the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.1 would be of help to students in case they are facing difficulties with regard to the clarity of the concepts at hand. Integrals are not only a prominent part of the academic curriculum for Class 12th, but they are also a crucial field of study for those who desire to pursue a career in Mathematics. The NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.1 available on the Extramarks’ website, along with adequate guidance from teachers, would ensure that students excelled in their understanding of Integrals.

List of Topics Covered Under NCERT Solutions for Class 12 Maths Chapter 7- Integrals (Include a Table)

A variety of topics associated with Integrals have been covered under the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.1. Exercise 7.1 starts with an introduction to Integrals. Further, this chapter describes the two major types of Integrals namely, the Definite and the Indefinite Integrals. Calculus can be defined as the theorem that makes up the connection between these two types of Integrals. A multitude of other associated themes have also been covered in great detail in this chapter. These include Integration as an inverse process of Differentiation, Geometrical Interpretation of Indefinite Integral, properties of Indefinite Integrals and comparison between Differentiation and Integration. The Extramarks’ website provides the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.1 which cover all of these topics in minute detail. These notes are easy-to-comprehend and logically organized to ensure maximum conceptual clarity.

Introduction to Integration 

The NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.1 provides a comprehensive and detailed introduction to the topic of Integration. Integration is a crucial theme within the larger topic of Integrals. It is a complex and diverse topic which has to be comprehensively dealt with to be adequately prepared for the examinations. Integration is one of the fundamental parts of the theme of Integrals, and it has to be adequately understood in order to excel in the examinations. Extramarks provides the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.1 with detailed solutions to problems involving the Application of integration formulae. These can be effectively used by students in order to ace the topic of Integration.

Access NCERT Solutions for Class 12 Mathematics Chapter 7- Integrals

Integrals are a very versatile and complex theme within the academic curriculum prescribed for Class 12 mathematics. It is not only a conceptually rich theme; it is also an introduction to a vast field of study for those interested in pursuing a career in Mathematics. In order to excel in this topic, it is important that students regularly practice comprehensive assessments coupled with reliable learning resources like the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.1. Regular practice would help towards enhancing the memory retention capabilities of students. Additionally, students would also be able to figure out how to appropriately approach a particular calculus or integral problem and solve it efficiently by following a logical sequence of steps. This would undoubtedly be of great assistance to them during the time of examinations.

Extramarks provides reliable, authentic, and easy-to-understand NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.1. This content available on the Extramarks’ website is curated by reputed subject experts and would act as a versatile learning resource for the students.

NCERT Solutions For Exercise 7.1 Maths Class 12

Exercise 7.1 Class 12th Maths is the first comprehensive assessment prescribed within the chapter titled Integrals in the NCERT book for Class 12. It consists of a variety of questions that introduce students to the topic of Integrals. It is an assessment aimed at preparing students for solving complex problems involving Integrals and Calculus efficiently. Integrals and Calculus are not only crucial topics within the prescribed NCERT academic curriculum for Class 12; they also possess immense potential for research and, consequently, are fields with many lucrative career options. Therefore, it is highly recommended that students should utilize the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.1 available on the Extramarks’ website to solve and practice assessments like Exercise 7.1 Class 12th.


The NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.1 begin with an explanation of what Integrals are. The two crucial types of integrals that have been explained thoroughly in this chapter are referred to as Definite and Indefinite Integrals. The theorem of Calculus is the connecting link between the two varieties of Integrals. Apart from this, themes such as integration as the inverse process of differentiation, constant of integration, geometrical interpretation of indefinite integral, properties of the indefinite integral etc have also been covered in the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.1.

These solutions have been created by Extramarks’ subject-matter experts as online learning resources. These solutions are readily and conveniently available for students.

Integral Calculus Is Again Classified Into Two Types

The Inverse process of Differentiation can be defined as Integration. During Integration, it is necessary to find the primitives through the use of the derivative of the function. Therefore, Integration can also be referred to as Anti-Differentiation. The symbol ‘C’ represents the constant. Definite Integrals are the first type, and Indefinite Integrals are the second type of Integral Calculus. Calculus can be explained as the association between the two types of Integrals. The NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.1 are engineered by Extramarks to provide detailed explanations of both the processes of Integration and Differentiation. The definitions and examples of Definite and Indefinite Integrals have also been provided within the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.1 prepared by Extramarks.

Integration As The Inverse Process of Differentiation

Extramarks provides the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.1 which cover the topic of Integration as the Inverse process of Differentiation. The NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.1 describe and define the processes of Integration and Differentiation in detail, along with suitable examples for the convenience of learning.

Constant of Integration

The Constant of Integration refers to the constant, symbolized by the letter ‘C’ that is added to the result of the Integration. The practical application of the constant of integration along with other formulae to attempt calculations and solve problems has been covered within the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.1. Students would be able to familiarise themselves with the functions and applications of the constant of Integration in the process of Integration.

Geometrical Interpretation Of Indefinite Integral

The geometrical interpretation of Indefinite Integral is a derivative formula that is arrived at through a comprehensive calculation attempted with the aid of a graph. The NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.1 comprise a detailed step-by-step explanation of the Mathematical Derivation of and format of the Geometrical Interpretation of Indefinite Integral. Extramarks provides high-quality graphs along with the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.1 to enhance the quality of learning.

Properties of Indefinite Integrals

The Extramarks’ website provides easy access to quality academic resources like the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.1 which comprehensively cover topics like Properties of Indefinite Integrals along with other nuances of Calculus.

Comparison Between Differentiation and Integration 

The NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.1 available on the Extramarks’ website, contains detailed descriptive explanations of Differentiation and Integration. Comprehensive comparisons between both have also been included in an easy-to-understand format for the convenience of students.

Why Should You Download Ex 7.1 Class 12 Maths Solutions

The NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.1 have been cautiously engineered in consideration of the latest updated CBSE academic syllabus. These solutions are easy to understand and retain. In addition, they follow a logical sequence of calculations. This has been done to ensure that there are no discrepancies between the pedagogy employed by the teachers in class and the online learning resources used by students. The formulas being applied in practical calculations have also been mentioned appropriately. This would prove to be extremely helpful for students who find it challenging to retain essential formulas. The format of the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.1 compiled by the Extramarks subject-matter experts would also help the students in acquiring the comprehension skills necessary for figuring out the type and number of formulae they would have to apply to solve a particular question.

Clarify Doubts

Integrals are undoubtedly one of the most versatile and complex topics within the prescribed NCERT academic curriculum for Class 12. It has to be understood at both the level of individual questions and the comprehensive working principles of the applied formulae. During classroom teaching, students may find it challenging to follow the guidance of the teachers. Some students may have challenges in following along at the same pace as their teachers. Classroom teaching also makes accommodating all doubts and adequately responding to them a bit cumbersome process. However, unclarified doubts can lead to a lack of conceptual clarity, which could be disadvantageous to students during examinations. Therefore, the Extramarks’ website is providing crucial learning resources like the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.1 to complement the guidance of teachers as well as the sincere efforts of students. The well-organized structure of the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.1 compiled by Extramarks can be utilized by students to clarify their doubts by themselves through self-learning and regular practice.

Convenience At Its Best

The NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.1 have been carefully crafted to ensure that students are able to appropriately follow the logical organization of calculations. It is important for students to not only retain crucial formulae but to also be able to practically apply these formulae for solving problems. This is a skill that can be attained through constant practice and revision. Therefore, Extramarks provides the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.1 as an easily accessible learning resource with quality content for the convenience of reference whenever required. Instead of searching high and low for answers to complex calculus problems, students can conveniently approach the Extramarks’ website to access comprehensive and well-structured solutions.

Preparing A Base

Integrals and Calculus are indispensable parts of the prescribed NCERT academic curriculum for Class 12. However, these are also vast areas of study with many attractive career options and unparalleled potential for academic research. Therefore, it is imperative for students to build a firm and reliable foundation in 12th grade itself in order to be able to reach new heights in the field in the future. The NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.1 would act as great guidelines for students and would help them in acquiring conceptual clarity with regard to the themes of Integrals and Calculus.

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Choose the correct answer

The value of 01tan12x11+xx2dx is equal toA 1B 0C1Dπ4


Let  I=01tan1(2x11+xx2)dx       =01tan1(x(1x)1+x(1x))dx       I=01{tan1xtan1(1x)}dx     ...(i)       =01{tan1(1x)tan1(11+x)}dx       [  0af(x)dx=0af(ax)dx]      =01{tan1(1x)tan1(x)}dx    ...(ii)Adding equation (i) and equation (ii), we get     2I=01{tan1xtan1(1x)}dx+01{tan1(1x)tan1(x)}dx       =0       I=0Thus, the correct answer is B.


Choose the correct answers

If f(a+bx)=f(x),thenabxf(x)dx  is equal to(A)a+b2abf(bx)dx(B)a+b2abf(b+x)dx(C)ba2abf(x)dx(D)a+b2abf(x)dx


Let I=abxf(x)dx       =ab(b+ax)f(b+ax)dx           [  abf(x)dx=abf(b+ax)dx]       =ab(b+ax)f(x)dx                    [  f(b+ax)=f(x)]       =ab(b+a)f(x)dx  abxf(x)dx       I=(b+a)abf(x)dx  I     2I=(b+a)abf(x)dx         I=(b+a)2abf(x)dxTherefore, the correct answer is D.


Choose the correct answers

cos2x(sinx+cosx)2dx is equal to(A)1sinx+cosx+C(B)log|sinx+cosx|+C(C)log|sinxcosx|+C(D)1(sinx+cosx)2+C


Let  I=cos2x(sinx+cosx)2dx        =cos2xsin2x(sinx+cosx)2dx        =(cosxsinx)(cosx+sinx)(sinx+cosx)2dx        =(cosxsinx)(sinx+cosx)dxLet​ t=sinx+cosxdtdx=cosxsinx        I=1tdt       =logt+C       =log(sinx+cosx)+CTherefore, option B is correct.


Choose the correct answer

dxex+ex is equal to(A) tan1(ex)+C(B) tan1(ex)+C(C) log(exex)+C         (D) log(ex+ex)+C


Let  I=dxex+ex       =exdxe2x+1Let t=exdtdx=ex     I=dtt2+1      =tan1t+C      =tan1(ex)+CThus, the correct option is A.


Evaluate  01e23xdx as a limit of a sum.


We  have, 01e23xdxabf(x)dx=(ba)limn1n[f(a)+f(a+h)+...+f(a+(n1)h)],where, h=banHere,  a=0,  b=1,  f(x)=e23x,  h=10n=1n01e23xdx=(10)limn1n[f(0)+f(0+h)+...+f(0+(n1)h)]               =1limn1n[e2+e23h+...+e2(n1)h]Using the sum to n terms of a G.P., where a=e2, r=e3h,we have01e23xdx=limn1n[e2(e3nh1)(e3h1)]                    =limn1n[e2(e3n.1n1)(e3.1n1)]                    =limn1n[e2(e31)(e3n1)]                    =e2(e31)limn[e3n13n].(3)                     =e2(e31)3                    =13(1ee2)                 [limx0eh1h=1]01e23xdx=13(e21e)


Prove the following01sin1xdx=π21


Let   I=01sin1xdx         =01sin1x.1dx        =[sin1x1dx(ddxsin1x1dx)dx]        =[sin1x.x11x2.xdx]01        =[xsin1x+12.21x2]01        =[1sin11+1120sin10102]        =π21=R.H.S.


Prove the follwoing0π42tan3xdx=1log2


Let   I=0π42tan3xdx         =20π4tan2xtanxdx         =20π4(sec2x1)tanxdx        =20π4sec2xtanxdx20π4tanxdx        =2[tan2x2]0π42[logcosx]0π4        =[tan2π4tan20]+2[logcosπ4logcos0]        =[10]+2[log12log1]        =1+2[12log20]        =1log2=R.H.S.


Prove the following0π2sin3xdx=23


Let  I=0π2sin3xdx       =0π2sin2xsinxdx       =0π2(1cos2x)sinxdxLett=cosx    dtdx=sinxWhenx=0,t=1 and when x=π2,t=0       I=10(1t2)sinxdtsinx        =[tt33]10=[001+13]       =23=R.H.S.


Prove the following11x17cos4xdx=0


Let   I=11x17cos4xdxLet   f(x)=x17cos4xf(x)=(x)17cos4(x)         =x17cos4x=f(x)Therefore, f(x) is an odd function.And  aaf(x)dx=0       I=11x17cos4xdx            =0=R.H.S.


Prove the following01xexdx=1


Let  I=01xexdx       =x01exdx01(ddxx01exdx)dx       =[xex]0101(1.ex)dx       =[1e10.e0][ex]01        =e[e1e0]       =ee+1       =1=R.H.S.


Prove the following13dxx2(x+1)dx=23+log23


Let  I=13dxx2(x+1)dxLet  1x2(x+1)=Ax+Bx2+C(x+1)             1=Ax(x+1)+B(x+1)+Cx2             1=x2(A+C)+x(A+B)+BEquating the coefficients of x2, x and constant term, we get        A+C=0, A+B=0 and B=1Solving the above equations, we getA=1, C=1 and B=1     1x2(x+1)=1x+1x2+1(x+1)              I=131xdx+131x2dx+131(x+1)dx              I=[logx]13[1x]13+[log(x+1)]13              I=[log3log1][1311]+[log(3+1)log(1+1)]               =[log3log1][23]+[log4log2]               =[log30]+23+log42               =23+log2log3               =23+log23=R.H.S.


Evaluate the definite integrals14[|x1|+|x2|+|x3|]dx


      I=14[|x1|+|x2|+|x3|]dx       =I1+I2+I3        ...(i)    I1=14|x1|dx       =14(x1)dx       [(x1)0 for 1x4]        =[x22x]14     =(4224)(1221)       =8412+1       =412      I2=14|x2|dxSince,(x2)0for2x4and  (x2)0 for 1x2        =12(x2)dx+24(x2)dx        =[x22+2x]12+[x222x]24        =222+2(2)+1222(1)+4222(4)222+2(2)        =2+4+122+882+4=52      I3=14|x3|dxSince,(x3)0for3x4  and  (x3)0 for 1x3       =13(x3)dx+34(x3)dx       =[x22+3x]13+[x223x]34       =322+3(3)+1223(1)+4223(4)322+3(3)       =92+9+123+81292+9       =261224=212From equation (i), we have      I=412+52+212=912


Evaluate the definite integrals0πxtanxsecx+tanxdx


Let  I=0πxtanxsecx+tanxdx       =0πxsinx1+sinxdx              ...(i)       =0π(πx)sin(πx)1+sin(πx)dx     [0af(x)dx=0af(ax)]       I=0π(πx)sinx1+sinxdx         ...(ii)Adding equation(i) and equation(ii),we get    2I=π0π1+sinx11+sinxdx     =π0π1+sinx1+sinxdxπ0π(1sinx)(1+sinx)(1sinx)dx     =π0π1dxπ0π(1sinx)1sin2xdx    2I=π[x]0ππ0π(1sinx)cos2xdx    2I=π[π0]π0π1cos2xdx+π0πsinxcos2xdx    2I=π2π0πsec2xdx+π0πtanxsecxdx    2I=π2π[tanx+secx]0π    2I=π2π[tanπ+secπtan0sec0]    2I=π2π[01tan01]      =π2+2π      Ι=π2(π+2)


Evaluate the definite integrals0π2sin2xtan1(sinx)dx


LetI=0π2sin2xtan1(sinx)dx         =0π22sinxcosxtan1(sinx)dxWhenx=0,t=0 and when x=π2,t=1  I=012ttan1(t)dt        =2{tan1ttdt(ddttan1ttdt)dt}01        =2{tan1t.t22(11+t2.t22)dt}01       =2{t22.tan1t12(1+t211+t2)dt}01Let  t=sinxdtdx=cosx       =2{t22.tan1t12(1+t21+t211+t2)dt}01       =2{t22.tan1t12(111+t2)dt}01       =2{t22.tan1t12(ttan1t)}01       =2{122.tan1112(1tan11)}2{022.tan1012(0tan10)}       =2{12.π412(1π4)}0       =π41+π4=π21


Evaluate the definite integrals0π4sinx+cosx9+16sin2xdx


0π4sinx+cosx9+16sin2xdx=0π4sinx+cosx9+16(11+sin2x)dx                    =0π4sinx+cosx9+1616(sin2x+cos2x2sinxcosx)dx                    =0π4sinx+cosx2516(sinxcosx)2dxLet  t=sinxcosxdtdx=cosx+sinxWhen  x=0,t=1 and when x=π4,t=00π4sinx+cosx9+16sin2xdx=116101(54)2t2dt                      =116×12(54)[log(54+t54t)]10                      =140{log(54+0540)log(54154+1)}                      =140{log(1)log(19)}                     =140log(9)


Evaluate the definite integrals01dx1+x+x


01dx1+x+x=011+xx(1+x)2(x)2dx                 =011+xx1+xxdx                 =011+xdx01xdx                 =01(1+x)12dx01x12dx                 =[(1+x)3232x3232]01                 =23[(1+x)32x32]01                 =23{(1+1)32132(1+0)32+032}                 =23{(2)3211}                 =23×(2)32=2523


Evaluate the definite integralsπ6π3sinx+cosxsin2xdx


Let  I=π6π3sinx+cosxsin2xdx       =π6π3sinx+cosx1(1sin2x)dx       =π6π3sinx+cosx1(sin2x+cos2x2sinxcosx)dx       =π6π3sinx+cosx1(sinxcosx)2dxLett=sinxcosxdtdx=cosx+sinxWhen x=π6,  t=sinπ6cosπ6=1232andwhenx=π3,t=sinπ3cosπ3=3212=(1232)I=(1232)(3212)11t2dtSince,11(t)2=11t2, therefore, 11t2 is an even function.Since,aaf(x)dx=20af(x)I=20(3212)11t2dt    =2[sin1t]0(3212)    =2sin1(312)


Evaluate the definite integrals0π2cos2x.dxcos2x+4sin2x


Let  I=0π2cos2x.dxcos2x+4sin2x       =0π2cos2x.dxcos2x+4(1cos2x)       =0π2cos2x.dx43cos2x       =130π23cos2x.dx43cos2x       =130π2(43cos2x4).dx43cos2x       =130π2(43cos2x).dx43cos2x+430π2dx43cos2x       =130π2dx+430π2sec2xdx4sec2x3       =13[x]0π2+430π2sec2xdx4(tan2x+1)3       =13[π20]+430π2sec2xdx4tan2x+43       =π6+430π2sec2xdx1+(2tanx)2Let  t=2tanxdtdx=2sec2xWhen x=0, t=0 and when x=π2,​ t=      I=π6+43011+t2.dt2       =π6+43×120dt1+t2      =π6+23[tan1t]0       =π6+23[tan1tan10]       =π6+23(π20)       =π6+π3       =π6


Evaluate the definite integrals0π4sinxcosxcos4x+sin4xdx


0π4sinxcosxcos4x+sin4xdx=0π4sinxcosxcos4x(1+sin4xcos4x)dx                       =0π4tanxsec2x(1+tan4x)dx                      =0π4tanxsec2x1+(tan2x)2dxLet  t=tan2xdtdx=2tanxsec2xWhen  x=0,t=0  and  when  x=π4,t=10π4sinxcosxcos4x+sin4xdx=01tanxsec2x1+t2.dt2tanxsec2x                         =120111+t2.dt                         =12[tan1t]01                         =12[tan11tan10]                         =12×π4                         =π8


Evaluate the definite integralsπ2πex(1sinx1cosx)dx


Let  I=π2πex(1sinx1cosx)dx       =π2πex(12sinx2cosx22sin2x2)dx       =π2πex(12sin2x2sinx2cosx2sin2x2)dx       =π2πex(12cosec2x2cotx2)dxLet  f(x)=cotx2f(x)=(12cosec2x2)=12cosec2x2     I=π2πex{f(x)+f(x)}dx       =[exf(x)]π2π=[excotx2]π2π       =[eπcotπ2eπ2cot(π4)]       =[eπ.0eπ2.1]       =eπ2


Integrate the functionsx2+1[log(x2+1)2logx]x4


x2+1[log(x2+1)2logx]x4=1x4x2+1{log(x2+1)2logx}                                =1x3x2+1x2{log(x2+1)logx2}                                 =1x31+1x2log(x2+1x2)                                 =1x31+1x2log(1+1x2)Let  t=1+1x2dtdx=02x3dx=x32dtx2+1[log(x2+1)2logx]x4dx=1x31+1x2logt×x32dt                                        =12tlogtdt                                        =12{logtt12dt(ddtlogtt12dt)dt}                                        =12{logt.t32321t.t3232dt}                                        =12{23logt.t3223t12dt}                                        =12{23t32logt23.t3232}+C                                        =12{23t32logt49t32}+C                                        =13t32(logt23)+C                                        =13(1+1x2)32{log(1+1x2)23}+C


Integrate the functionstan11x1+x


tan11x1+xdxPutting​ x=cosθdx=sinθtan11x1+xdx=tan11cosθ1+cosθ×sinθ                     =tan12sin2θ22cos2θ2×sinθ                     =tan1tanθ2×sinθ                     =θ2sinθ                     =12{θsinθ(dθsinθ)}                     =12{θ(cosθ)1(cosθ)}                     =12{θ(cosθ)+sinθ}+C                     =12θcosθ121cos2θ+C                     =12(xcos1x1x2)+C


Integrate the functionsx2+x+1(x+1)2(x+2)


Let  x2+x+1(x+1)2(x+2)=A(x+1)+B(x+1)2+C(x+2)   ..(i)              x2+x+1=A(x+1)(x+2)+B(x+2)+C(x+1)2              x2+x+1=A(x2+3x+2)+B(x+2)+C(x2+2x+1)              x2+x+1=x2(A+C)+x(3A+B+2C)+(2A+2B+C)Comparing coefficients of x2, x and constant from both sides,we getA+C=1, 3A+B+2C=1  and 2A+2B+C=1Solving the above equations, we getA=2,  B=1,  C=3So, from equation(i), we have         x2+x+1(x+1)2(x+2)=2(x+1)+1(x+1)2+3(x+2)  x2+x+1(x+1)2(x+2)dx=2(x+1)dx+1(x+1)2dx+3(x+2)dx     =2log|x+1|1(x+1)+3log|x+2|+C     =2log|x+1|+3log|x+2|1(x+1)+C


Integrate the functions2+sin2x(1+cos2x)ex


Let        I=2+sin2x(1+cos2x)exdx=2+2sinxcosx2cos2xexdx        =1+sinxcosxcos2xexdx        =(sec2x+tanx)dxLetf(x)=tanxf(x)=sec2x     I=[f(x)+t(x)]exdx           =exf(x)+C           =extanx+C


Integrate the functions 1x1+x


1x1+xLet  x=cos2t    dxdt=2costsint1x1+xdx=1cos2t1+cos2t×2costsintdt                            =1cost1+cost×2costsintdt                           =2sin2(t2)2cos2(t2)×2costsintdt                          =sin(t2)cos(t2)×2cost{2sin(t2)cos(t2)}dt                         =4sin(t2)cos(t2)×cost{sin(t2)cos(t2)}dt                        =4sin2(t2)costdt                       =4sin2(t2)(2cos2(t2)1)dt                      =4{2sin2(t2)cos2(t2)sin2(t2)}dt                     =24sin2(t2)cos2(t2)dt+4sin2(t2)dt                    =2sin2tdt+2(1cost)dt                   =(1cos2t)dt+2(1cost)dt                   =(tsin2t2)+2(tsint)+C                  =sin2t2t+2t2sint+C                  =t+2sintcost22sint+C                  =t+sintcost2sint+C                  =t+1cos2tcost21cos2t+C                  =cos1(x)+1xx21x+C                  =cos1(x)+xx221x+C                  =21x+xx2+cos1(x)+C


Integrate the functionssin-1x cos1xsin-1x + cos1x,    x[0, 1]


sin1xcos1xsin1x+cos1x,    x[0,1]                      =sin1xcos1xπ2        [sin1x+cos1x=π2]                     =2π(π2cos1xcos1x)                     =2π(π22cos1x)sin1xcos1xsin1x+cos1xdx                     =2π(π22cos1x)dx                     =1dx4πcos1xdx                     =x4πcos1t.2tdt       [Let  t=xdtdx=12x=12t]                     =x8πtcos1tdt                     =x8π[cos1ttdt{ddtcos1ttdt}dt]                     =x8π[cos1t.t22{11t2.t22}dt]                     =x8π[cos1t.t22+12{t21t2}dt]                     =x8π[cos1t.t2212{1t211t2}dt]                     =x8π[12t2cos1t12{1t21t211t2}dt]                     =x8π[12t2cos1t12{1t211t2}dt]                     =x8π.12t2cos1t+8π.12(t21t2+12sin1t             sin1t)+C                     =x4πxcos1x+4π(x21x12sin1x)+C                     =x4πx(π2sin1x)+2πxx22πsin1x+C                     =x2x+2πxx2+4πxsin1x2πsin1x+C                     =x+2πxx2+2π(2x1)sin1x+C




1sin3xsin(x+α)dx=1sin3x(sinxcosα+cosxsinα)dx                         =1sin4x(cosα+cotxsinα)dx                        =cosec2x(cosα+cotxsinα)dx[Let  t=cosα+cotxsinαdtdx=cosec2x.sinα]                       =cosec2xtdtcosec2x.sinα                       =1sinαt12dt                       =1sinα[t1212]+C                       =2sinα[cosα+cotxsinα]+C                       =2sinα[cosα+cosxsinxsinα]+C                       =2sinα[sinxcosα+cosxsinαsinx]+C                       =2sinα[sin(x+α)sinx]+C




f(ax + b)[f(ax + b)]ndxLett=f(ax + b)dtdx=af(ax + b)f(ax + b)[f(ax + b)]ndx=f(ax + b)tndtaf(ax + b)                                                                       =1atndx                                                                      =1atn+1n+1+C                                                                     =1a[f(ax + b)]n+1n+1+C




     e3logx(x4+1)1=elogx3(x4+1)1                                            =x3x4+1          [elogx=x]e3logx(x4+1)1dx=x3x4+1dx   =x3tdt4x3      [Let  t=x4+1dtdx=4x3]     =1tdt    =logt+C         =log(x4+1)+C




       cos3x elogsinx=cos3xsinx      [  elogx=x]cos3x elogsinxdx=cos3xsinxdx                                                 =t3sinxdtsinx     [Let  t=cosxdtdx=sinx]                                                =t3dt                                               =t44+C                                               =t3dt                                              =cos4x4+C


Integrate the functions 1(x2+1)(x2+4)


Let  1(x2+1)(x2+4)=Ax+B(x2+1)+Cx+D(x2+4)       1=(Ax+B)(x2+4)+(Cx+D)(x2+1)       1=x3(A+C)+x2(B+D)+x(4A+C)+(4B+D)Comparing coefficients of x2, x and constant both sides, we getA+C=0,  B+D=0,4A+C=0and4B+D=1Solving the above equations, we getA=0,B=13,C=0 and D=13    1(x2+1)(x2+4)=13(x2+1)13(x2+4)1(x2+1)(x2+4)dx=131(x2+1)dx131(x2+4)dx                                                    =13tan1(x)13×2tan1(x2)+C                                                    =13tan1(x)16tan1(x2)+C


Integrate the functions ex(1+ex)(2+ex)


Given,  ex(1+ex)(2+ex)dxLet  t=exdtdx=exdx=dtexex(1+ex)(2+ex)dx=ex(1+t)(2+t)dtex                                                =(12+t11+t)dt                                               =12+tdt11+tdt                                               =log|2+t|log|1+t|+C                                               =log|2+t1+t|+C                                              =log|2+ex1+ex|+C


Integrate the functions x31x8



Since,x31x8=x31(x4)2Let  t=x4dtdx=4x3dx=dt4x3x31x8dx=x31t2dt4x3                                  =1411t2dt                                  =14sin1t+C                                  =14sin1x4+C



Integrate the functions 1cos(x+a)cos(x+b)



Since,x31x8=x31(x4)2Let  t=x4dtdx=4x3dx=dt4x3x31x8dx=x31t2dt4x3                                  =1411t2dt                                  =14sin1t+C                                  =14sin1x4+C

Sr34 (1701534)



Integrate the functions 1cos(x+a)cos(x+b)

Marks 1

Since,1cos(x+a)cos(x+b)             =1sin(ba)(sin(ba)cos(x+a)cos(x+b))             =1cos(ba){sin(x+bax)cos(x+a)cos(x+b)}             =1sin(ba)[sin{(x+b)(x+a)}cos(x+a)cos(x+b)]             =1sin(ba){sin(x+b)cos(x+a)cos(x+b)sin(x+a)cos(x+a)cos(x+b)}             =1sin(ba){tan(x+b)tan(x+a)}1cos(x+a)cos(x+b)dx             =1sin(ba){tan(x+b)tan(x+a)}dx             =1sin(ba){logcos(x+b)+logcos(x+a)}+C             =1sin(ba)log|cos(x+a)cos(x+b)|+C


Integrate the functionssin8xcos8x12sin2xcos2x


Here,  sin8xcos8x12sin2xcos2x=(sin4x)2(cos4x)212sin2xcos2x                                =(sin4x+cos4x)(sin4xcos4x)(sin2x+cos2x)22sin2xcos2x                               =(sin4x+cos4x)(sin2xcos2x)(sin2x+cos2x)sin4x+cos4x+2sin2xcos2x2sin2xcos2x                               =(sin4x+cos4x)(sin2xcos2x)(sin2x+cos2x)(sin4x+cos4x)                               =(sin2xcos2x)                               =cos2xsin8xcos8x12sin2xcos2xdx                            =cos2xdx                            =(sin2x2)+C


Integrate the functions cosx4sin2x


Given,  cosx4sin2xdxLet     t=sinxdtdx=cosxcosx4sin2xdx=cosx22t2dtcosx                                   =sin1(t2)+C                                   =sin1(sinx2)+C


Integrate the functions e5logxe4logxe3logxe2logx


Here,  e5logxe4logxe3logxe2logx=e4logx(elogx1)e2logx(elogx1)                                                =e2logx                                                =elogx2  e5logxe4logxe3logxe2logxdx=elogx2dx                                                  =x2dx                                                  =x33+C


Integrate the functions sinxsin(xa)


We have,  sinxsin(xa)dxLet  t=xadtdx=1sinxsin(xa)dx=sin(t+a)sintdt    =sintcosa+sinacostsintdt    =(cosa+sinacott)dt    =cosa(t)+sina(logsint)+C    =cosa(xa)+sina{logsin(xa)}+C    =xcosaacosa+sinalogsin(xa)+C    =xcosa+sinalogsin(xa)+Cacosa    =xcosa+sinalogsin(xa)+C1               [Let  C1=Cacosa]


Integrate the functions 5x(x+1)(x2+9)


Let   5x(x+1)(x2+9)=Ax+1+Bx+Cx2+9   ...(i)          5x(x+1)(x2+9)=A(x2+9)+(Bx+C)(x+1)(x+1)(x2+9)                           5x=A(x2+9)+(Bx+C)(x+1)                            5x=x2(A+B)+x(B+C)+(C+9A)Comparing coefficients of x2, x and constant terms, we getA+B=0,B+C=5  and  C+9A=0Solving the above equations, we getA=12,   B=12,  C=92From equation (i), we have        5x(x+1)(x2+9)=12(x+1)+x+92(x2+9)5x(x+1)(x2+9)dx=121x+1dx+12xx2+9dx+921x2+33dx                                        =12log|x+1|+12×12log|x2+9|+92×13tan1(x3)+C                                        =12log|x+1|+14log(x2+9)+32tan1(x3)+C


Integrate the functions 1x12+x13


We have,1x12+x13dx=1x13(x1213+1)dx                           =1x13(1+x16)dxPutting x=t6dxdt=6t5dx=6t5dt  =1t2(1+t)6t5dt  =6t3(1+t)dt=6{(t2t+1)1t+1}dx=6[t33t22+tlog(t+1)]+C  =6[(x16)33(x16)22+(x16)log{(x16)+1}]+C=6[x3x32+(x16)log{(x16)+1}]+C=6x3x3+6x166log(1+x16)+C


Integrate the functions 1x2(x4+1)34


We have, 1x2(x4+1)34dx=1x2{x4(1+1x4)}34dx                                     =1x2.x3(1+1x4)34dx                                     =1x5t34x5dt4Let  t=1+1x4dtdx=4x5dx=x5dt4                                    =14t34dt                                    =14[t1414]+C                                    =(1+1x4)14+C


Integrate the functions 1xaxx2


We have, 1xaxx2dx=1xx2(ax1)dx                                   =1x2(ax1)dx                                  =1x2t×x2adt  [Let t=ax1dtdx=ax2dx=x2adt]                                 =1at12dt=1a[t1212]+C                                =2a(ax1)+C                                =2a(axx)+C


Integrate the functions 1x+a+x+b


We have 1x+a+x+b=1x+a+x+b×x+ax+bx+ax+b                                                 =x+ax+b(x+a)(x+b)                                                 =x+ax+b(ab)       1x+a+x+bdx=1(ab)(x+ax+b)dx                                                   =1(ab){(x+a)12(x+b)12}dx                                                    =1(ab){(x+a)3232(x+b)3232}+C           1x+a+x+bdx=23(ab){(x+a)32(x+b)32}+C


Integrate the functions 1xx3


We  have1xx3=1x(1x)(1+x)             =Ax+B(1x)+C(1+x)         1=A(1x)(1+x)+Bx(1+x)+Cx(1x)   ...(i)Substituting x = 0, 1 and –1 respectively in equation (i), we getA=1,   B=12 and C=12   1xx3=1x+12(1x)12(1+x)1xx3dx=1xdx+1211xdx1211+xdx      =log|x|12log|1x|12log|1+x|+D      =12(2log|x|log|(1x)(1+x)|)+D      =12log|x21x2|+D


Choose the correct answerThe value of0π2log4 + 3sinx4 + 3cosxdx isA2B34C0D2


Let  I=0π2log(4+3sinx4+3cosx)dx   ...(i)          =0π2log{4+3sin(π2x)4+3cos(π2x)}dx         =0π2log(4+3cosx4+3sinx)dx...(ii)Adding equation (i) and equation (ii), we get     2I=0π2{log(4+3sinx4+3cosx)+log(4+3cosx4+3sinx)}dx        =0π2{log(4+3sinx4+3cosx×4+3cosx4+3sinx)}dx       =0π2log(1)dx      I=0Therefore, the correction option is C.


Choose the correct answerThe value of π2π2x3+ xcosx+tan5x+1dxisA0B2CπD1


Let   I=π2π2(x3+xcosx+tan5x+1)dx           =π2π2x3dx+π2π2xcosxdx+π2π2tan5xdx+π2π21dxSince, if f(x) is even function then, aaf(x)dx=20af(x)dxand if f(x) is odd function then,   aaf(x)dx=0    I=0+0+0+20π21dx          =2[x]0π2          =2(π2)          =πTherefore, the correct option is C.


Show that0af(x)g(x)dx=20af(x)dx, if f and g are defined as f(x)=f(ax) and g(x)+g(ax)=4


Let  I=0af(x)g(x)dx...(i)         =0af(ax)g(ax)dx          [    0af(x)dx=0af(ax)dx]         =0af(x)g(ax)dx  ...(ii)       [f(x)=f(ax)]Adding equation (i) and (ii), we get     2I=0af(x)g(x)dx+0af(x)g(ax)dx        =0af(x){g(x)+g(ax)}dx        =0af(x)(4)dx          [  g(x)+g(ax)=4]       I=20af(x)dx          Hence​ proved.




Let  I=04|x1|dxSince,​ (x1)0 on [0,2] and (x1)0 on [1,4].  I=01(x1)dx+14(x1)dx         =[x22+x]01+[x22x]14         =[122+1+00]+[4224122+1]         =12+1+8412+1         =5  04|x1|dx=5




Let  I=0axx+axdx      ...(i)           =0aaxax+a(ax)dx           =0aaxax+xdx      ...(ii)Adding equation (i) and equation (ii), we get   2I=0ax+axx+axdx       =0a1dx       =[x]0a     I=a2a




Let  I=0πlog(1+cosx)dx          ...(i)         =0πlog{1+cos(πx)}dx       =0πlog(1cosx)dx         ...(ii)Adding equation (i)​  and equation (ii),​ we get  2I=0πlog(1+cosx)dx+0πlog(1cosx)dx       =0πlog{(1+cosx)(1cosx)}dx       =0πlog(1cos2x)dx       =0πlogsin2xdx  2I=20πlogsinxdx    I=0πlogsinxdx       ...(iii)    I=20π2logsinxdx      ...(iv)           [sin(πx)=sinx]    I=20π2logsin(π2x)dx     =20π2logcosxdx      ...(v)Adding equation (iv)​  2I=20π2(logsinx+logcosx)dx  I=0π2(logsinx+logcosx)dx   =0π2log(2sinxcosx2)dx  =0π2logsin2xdx0π2log2dxLet t=2xdtdx=2When x=0t=0 and when x=π2,​ t=π.Ι=0πlogsintdt2log2[x]0π2Ι=120πlogsinxdxlog2[π20][abf(x)dx=abf(t)dt]I=12Iπ2log2I12I=π2log2    12I=π2log2I=πlog2

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