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.

Introduction

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.

Why Extramarks?

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Q.1

Choose the correct answer

The value of ∫01tan−12x−11+x−x2dx is equal toA 1B 0C−1D π4

Ans.

Let  I=∫01tan–1(2x−11+x−x2) dx       =∫01tan–1(x−(1−x)1+x(1−x)) dx       I=∫01{tan–1x−tan−1(1−x)} dx     ...(i)       =∫01{tan–1(1−x)−tan−1(1−1+x)} dx       [∵  ∫0af(x)dx=∫0af(a−x)dx]      =∫01{tan–1(1−x)−tan−1(x)} dx    ...(ii)Adding equation (i) and equation (ii), we get     2I=∫01{tan–1x−tan−1(1−x)} dx+∫01{tan–1(1−x)−tan−1(x)} dx       =0       I=0Thus, the correct answer is B.

Q.2

Choose the correct answers

If f(a+b−x)=f(x),then∫abx f(x)dx  is equal to(A) a + b2∫abf(b−x)dx(B) a + b2 ∫abf(b+x)dx(C) b−a2∫abf(x)dx(D) a + b2 ∫abf(x)dx

Ans.

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

Q.3

Choose the correct answers

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

Ans.

Let  I=∫ cos2x(sinx+cosx)2 dx        =∫ cos2x−sin2x(sinx+cosx)2 dx        =∫ (cosx−sinx)(cosx+sinx)(sinx+cosx)2 dx        =∫ (cosx−sinx)(sinx+cosx) dxLet​ t=sinx+cosx ⇒ dtdx=cosx−sinx        I=∫1t dt       =logt+C       =log(sinx+cosx)+CTherefore, option B is correct.

Q.4

Choose the correct answer

∫dxex+e–x is equal to(A) tan–1(ex)+C(B) tan–1(e–x)+C(C) log(ex–e–x)+C         (D) log(ex+e–x)+C

Ans.

Let  I=∫dxex+e–x       =∫exdxe2x+1Let t=ex ⇒ dtdx=ex∴     I=∫dtt2+1      =tan−1t+C      =tan−1(ex)+CThus, the correct option is A.

Q.5

Evaluate  ∫01e2–3xdx as a limit of a sum.

Ans.

We  have, ∫01e2–3xdx∵∫abf(x) dx=(b−a)limn→∞1n[f(a)+f(a+h)+...+f(a+(n−1)h)],where, h=b−anHere,  a=0,  b=1,  f(x)=e2−3x,  h=1−0n=1n∴∫01e2−3x dx=(1−0)limn→∞1n[f(0)+f(0+h)+...+f(0+(n−1)h)]               =1limn→∞1n[e2+e2−3h+...+e2−(n−1)h]Using the sum to n terms of a G.P., where a=e2, r=e−3h, we have∴∫01e2–3x dx=limn→∞1n[e2(e−3nh−1)(e−3h−1)]                    =limn→∞1n[e2(e−3n.1n−1)(e−3.1n−1)]                    =limn→∞1n[e2(e−3−1)(e−3n−1)]                    =e2(e−3−1)limn→∞[e−3n−1−3n].(−3)                     =e2(e−3−1)−3                    =−13(1e−e2)                 [∵limx→0eh−1h=1]∴∫01e2–3x dx=13(e2−1e)

Q.6

Prove the following∫01sin–1x dx=π2−1

Ans.

Let   I=∫01sin–1x dx         =∫01sin–1x.1 dx        =[sin–1x∫1 dx−∫(ddxsin−1x∫1 dx) dx]        =[sin–1x.x−∫11−x2.x dx]01        =[xsin–1x+12.21−x2]01        =[1sin–11+1−12−0sin−10−1−02]        =π2−1=R.H.S.

Q.7

Prove the follwoing∫0π42 tan3x dx=1−log2

Ans.

Let   I=∫0π42tan3x dx         =2∫0π4tan2xtanx dx         =2∫0π4(sec2x−1)tanx dx        =2∫0π4sec2xtanx dx−2∫0π4tanx dx        =2[tan2x2]0π4−2[−logcosx]0π4        =[tan2π4−tan20]+2[logcosπ4−logcos0]        =[1−0]+2[log12−log1]        =1+2[−12log2−0]        =1−log2=R.H.S.

Q.8

Prove the following∫0π2sin3x dx=23

Ans.

Let  I=∫0π2sin3x dx       =∫0π2sin2xsinx dx       =∫0π2(1−cos2x)sinx dxLet t=cosx  ⇒  dtdx=−sinxWhen x=0, t=1 and when x=π2, t=0∴       I=∫10(1−t2)sinx dt−sinx        =−[t−t33]10=−[0−0−1+13]       =23=R.H.S.

Q.9

Prove the following∫−11x17cos4x dx=0

Ans.

Let   I=∫−11x17cos4x dxLet   f(x)=x17cos4x⇒f(−x)=(−x)17cos4(−x)         =−x17cos4x=−f(x)Therefore, f(x) is an odd function.And  ∫−aaf(x)dx=0∴       I=∫−11x17cos4x dx            =0=R.H.S.

Q.10

Prove the following∫01xex dx=1

Ans.

Let  I=∫01xex dx       =x∫01exdx−∫01(ddxx∫01exdx)dx       =[xex]01−∫01(1.ex)dx       =[1e1−0.e0]−[ex]01        =e−[e1−e0]       =e−e+1       =1=R.H.S.

Q.11

Prove the following∫13dxx2(x+1) dx = 23 + log 23

Ans.

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=∫13−1x dx+∫131x2 dx+∫131(x+1)dx              I=−[logx]13−[1x]13+[log(x+1)]13              I=−[log3−log1]−[13−11]+[log(3+1)−log(1+1)]               =−[log3−log1]−[−23]+[log4−log2]               =−[log3−0]+23+log42               =23+log2−log3               =23+log23=R.H.S.

Q.12

Evaluate the definite integrals∫14[|x−1|+|x−2|+|x−3|]dx

Ans.

      I=∫14[|x−1|+|x−2|+|x−3|]dx       =I1+I2+I3        ...(i)    I1=∫14|x−1|dx       =∫14(x−1)dx       [∵(x−1)≥0 for 1≤x≤4]        =[x22−x]14     =(422−4)−(122−1)       =8−4−12+1       =412      I2=∫14|x−2|dxSince, (x−2)≥0 for 2≤x≤4 and  (x−2)≤0 for 1≤x≤2        =∫12−(x−2)dx+∫24(x−2)dx        =[−x22+2x]12+[x22−2x]24        =−222+2(2)+122−2(1)+422−2(4)−222+2(2)        =−2+4+12−2+8−8−2+4=52      I3=∫14|x−3|dxSince, (x−3)≥0 for 3≤x≤4  and  (x−3)≤0 for 1≤x≤3       =∫13−(x−3)dx+∫34(x−3)dx       =[−x22+3x]13+[x22−3x]34       =−322+3(3)+122−3(1)+422−3(4)−322+3(3)       =−92+9+12−3+8−12−92+9       =2612−24=212From equation (i), we have      I=412+52+212 = 912

Q.13

Evaluate the definite integrals∫0πxtanxsecx+tanxdx

Ans.

Let  I=∫0πxtanxsecx+tanxdx       =∫0πxsinx1+sinxdx              ...(i)       =∫0π(π−x)sin(π−x)1+sin(π−x)dx     [∵∫0af(x)dx=∫0af(a−x)]       I=∫0π(π−x)sinx1+sinxdx         ...(ii)Adding equation(i) and equation(ii), we get    2I=π∫0π1+sinx−11+sinxdx     =π∫0π1+sinx1+sinxdx−π∫0π(1−sinx)(1+sinx)(1−sinx)dx     =π∫0π1dx−π∫0π(1−sinx)1−sin2xdx    2I=π[x]0π−π∫0π(1−sinx)cos2xdx    2I=π[π−0]−π∫0π1cos2xdx+π∫0πsinxcos2xdx    2I=π2−π∫0πsec2xdx+π∫0πtanxsecx dx    2I=π2−π[tanx+secx]0π    2I=π2−π[tanπ+secπ−tan0−sec0]    2I=π2−π[0−1−tan0−1]      =π2+2π      Ι=π2(π+2)

Q.14

Evaluate the definite integrals∫0π2sin2x tan–1(sinx)dx

Ans.

Let I=∫0π2sin2x tan–1(sinx)dx         =∫0π22sinxcosx tan–1(sinx)dxWhen x=0, t=0 and when x=π2, t=1∴  I=∫012t tan–1(t)dt        =2{tan−1t∫t dt−∫(ddttan−1t∫t dt) dt}01        =2{tan−1t.t22−∫(11+t2.t22) dt}01       =2{t22.tan−1t−12∫(1+t2−11+t2) dt}01Let  t=sinx⇒dtdx=cosx       =2{t22.tan−1t−12∫(1+t21+t2−11+t2) dt}01       =2{t22.tan−1t−12∫(1−11+t2) dt}01       =2{t22.tan−1t−12(t−tan−1t)}01       =2{122.tan−11−12(1−tan−11)}−2{022.tan−10−12(0−tan−10)}       =2{12.π4−12(1−π4)}−0       =π4−1+π4=π2−1

Q.15

Evaluate the definite integrals∫0π4sinx+cosx9+16sin2xdx

Ans.

∫0π4sinx+cosx9+16sin2xdx=∫0π4sinx+cosx9+16(1−1+sin2x)dx                     =∫0π4sinx+cosx9+16−16(sin2x+cos2x−2sinxcosx)dx                     =∫0π4sinx+cosx25−16(sinx−cosx)2dxLet  t=sinx−cosx⇒dtdx=cosx+sinxWhen  x=0, t=−1 and when x=π4, t=0∴∫0π4sinx+cosx9+16sin2xdx=116∫−101(54)2−t2dt                       =116×12(54)[log(54+t54−t)]−10                       =140{log(54+054−0)−log(54−154+1)}                       =140{log(1)−log(19)}                      =140log(9)

Q.16

Evaluate the definite integrals∫01dx1+x+x

Ans.

 ∫01dx1+x+x=∫011+x−x(1+x)2−(x)2dx                 =∫011+x−x1+x−xdx                 =∫011+xdx−∫01xdx                 =∫01(1+x)12dx−∫01x12dx                 =[(1+x)3232−x3232]01                 =23[(1+x)32−x32]01                 =23{(1+1)32−132−(1+0)32+032}                 =23{(2)32−1−1}                 =23×(2)32=2523

Q.17

Evaluate the definite integrals∫π6π3sinx+cosxsin2xdx

Ans.

Let  I=∫π6π3sinx+cosxsin2xdx       =∫π6π3sinx+cosx1−(1−sin2x)dx       =∫π6π3sinx+cosx1−(sin2x+cos2x−2sinxcosx)dx       =∫π6π3sinx+cosx1−(sinx−cosx)2dxLet t=sinx−cosx⇒dtdx=cosx+sinxWhen x=π6,  t=sinπ6−cosπ6=12−32and when x=π3, t=sinπ3−cosπ3=32−12=−(12−32)∴I= ∫−(12−32)(32−12)11−t2dtSince, 11−(−t)2=11−t2, therefore, 11−t2 is an even function.Since, ∫−aaf(x)dx=2∫0af(x)∴I = 2∫0(32−12)11−t2dt    =2[sin−1t]0(32−12)    =2sin−1(3−12)

Q.18

Evaluate the definite integrals∫0π2cos2x.dxcos2x+4sin2x

Ans.

Let  I=∫0π2cos2x.dxcos2x+4sin2x       =∫0π2cos2x.dxcos2x+4(1−cos2x)       =∫0π2cos2x.dx4−3cos2x       =−13∫0π2−3cos2x.dx4−3cos2x       =−13∫0π2(4−3cos2x−4).dx4−3cos2x       =−13∫0π2(4−3cos2x).dx4−3cos2x+43∫0π2dx4−3cos2x       =−13∫0π2dx+43∫0π2sec2xdx4sec2x−3       =−13[x]0π2+43∫0π2sec2xdx4(tan2x+1)−3       =−13[π2−0]+43∫0π2sec2xdx4tan2x+4−3       =−π6+43∫0π2sec2xdx1+(2tanx)2Let  t=2tanx⇒dtdx=2sec2xWhen x=0, t=0 and when x=π2,​ t=∞∴      I=−π6+43∫0∞11+t2.dt2       =−π6+43×12∫0∞dt1+t2      =−π6+23[tan−1t]0∞       =−π6+23[tan−1∞−tan−10]       =−π6+23(π2−0)       =−π6+π3       =π6

Q.19

Evaluate the definite integrals∫0π4sinxcosxcos4x+sin4x dx

Ans.

∫0π4sinxcosxcos4x+sin4x dx=∫0π4sinxcosxcos4x(1+sin4xcos4x) dx                       =∫0π4tanxsec2x(1+tan4x) dx                      =∫0π4tanxsec2x1+(tan2x)2 dxLet  t=tan2x⇒dtdx=2tanxsec2xWhen  x=0, t=0  and  when  x=π4, t=1∴∫0π4sinxcosxcos4x+sin4x dx=∫01tanxsec2x1+t2 . dt2tanxsec2x                         =12∫0111+t2.dt                         =12[tan−1t]01                         =12[tan−11−tan−10]                         =12×π4                         =π8

Q.20

Evaluate the definite integrals∫π2πex(1–sinx1–cosx) dx

Ans.

Let  I=∫π2πex(1−sinx1−cosx) dx       =∫π2πex(1−2sinx2cosx22sin2x2) dx       =∫π2πex(12sin2x2−sinx2cosx2sin2x2) dx       =∫π2πex(12cosec2x2−cotx2) dxLet  f(x)=−cotx2⇒f‘(x)=−(−12cosec2x2)=12cosec2x2∴     I=∫π2πex{f‘(x)+f(x)} dx       =[exf(x)]π2π=−[excotx2]π2π       =−[eπcotπ2−eπ2cot(π4)]       =−[eπ.0−eπ2.1]       =eπ2

Q.21

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

Ans.

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+1x2⇒dtdx=0−2x3⇒dx=−x32dt∴∫x2+1[log(x2+1)−2logx]x4 dx=∫1x31+1x2logt ×−x32dt                                        =−12∫tlogt dt                                        =−12{logt∫t12 dt−∫(ddtlogt∫t12 dt) dt}                                        =−12{logt.t3232−∫1t.t3232 dt}                                        =−12{23logt.t32−23∫t12 dt}                                        =−12{23t32logt−23.t3232}+C                                        =−12{23t32logt−49t32}+C                                        =−13t32(logt−23)+C                                        =−13(1+1x2)32{log(1+1x2)−23}+C

Q.22

Integrate the functionstan–11−x1+x

Ans.

∫tan–11−x1+x dxPutting​ x=cosθ⇒dxdθ=−sinθ∴∫tan–11−x1+x dx=∫tan–11−cosθ1+cosθ ×−sinθ dθ                     =∫tan–12sin2θ22cos2θ2 ×−sinθ dθ                     =∫tan–1tanθ2 ×−sinθ dθ                     =−∫θ2sinθ dθ                     =−12{θ∫sinθ dθ−∫(ddθθ∫sinθ dθ) dθ}                     =−12{θ(−cosθ)−∫1(−cosθ) dθ}                     =−12{θ(−cosθ)+sinθ}+C                     =12θcosθ−121−cos2θ+C                     =12(xcos−1x−1−x2)+C

Q.23

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

Ans.

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)2 dx+∫3(x+2) dx     =−2log|x+1|−1(x+1)+3log|x+2|+C     =−2log|x+1|+3log|x+2|−1(x+1)+C

Q.24

Integrate the functions2+sin2x(1+cos2x)ex

Ans.

 Let        I=∫2+sin2x(1+cos2x)ex dx=∫2+2sinxcosx2cos2xex dx        =∫1+sinxcosxcos2xex dx        =∫(sec2x+tanx) dxLet f(x)=tanx⇒f‘(x)=sec2x∴     I=∫[f(x)+t‘(x)]ex dx           =ex f(x)+C           =extanx+C

Q.25

Integrate the functions 1−x1+x

Ans.

1−x1+xLet  x=cos2t  ⇒  dxdt=−2costsint∫1−x1+x dx=∫1−cos2t1+cos2t ×−2costsint dt                            =∫1−cost1+cost ×−2costsint dt                           =∫2sin2(t2)2cos2(t2) ×−2costsint dt                          =∫sin(t2)cos(t2) ×−2cost{2sin(t2)cos(t2)} dt                         =−4∫sin(t2)cos(t2) ×cost{sin(t2)cos(t2)} dt                        =−4∫sin2(t2)cost dt                       =−4∫sin2(t2)(2cos2(t2)−1 )dt                      =−4∫{2sin2(t2)cos2(t2)−sin2(t2)}dt                     =−2∫4sin2(t2)cos2(t2) dt+4∫sin2(t2) dt                    =−2∫sin2t dt+2∫(1−cost) dt                   =−∫(1−cos2t) dt+2∫(1−cost) dt                   =−(t−sin2t2)+2(t−sint)+C                  =sin2t2−t+2t−2sint+C                  =t+2sintcost2−2sint+C                  =t+sintcost−2sint+C                  =t+1−cos2tcost−21−cos2t+C                  =cos−1(x)+1−xx−21−x+C                  =cos−1(x)+x−x2−21−x+C                  =−21−x+x−x2+cos−1(x)+C

Q.26

Integrate the functionssin-1x − cos–1xsin-1x + cos–1x,    x∈[0, 1]

Ans.

sin–1x−cos–1xsin–1x+cos–1x,    x∈[0,1]                      =sin–1x−cos–1xπ2        [∵sin–1x+cos–1x=π2]                     =2π(π2−cos–1x−cos–1x)                     =2π(π2−2cos–1x)∴∫sin–1x−cos–1xsin–1x+cos–1x dx                     =2π∫(π2−2cos–1x) dx                     =∫1 dx−4π∫cos–1x dx                     =x−4π∫cos–1t .2tdt       [Let  t=x⇒dtdx=12x=12t]                     =x−8π∫tcos–1t dt                     =x−8π[cos–1t∫t dt−∫{ddtcos–1t∫t dt} dt]                     =x−8π[cos–1t.t22−∫{−11−t2.t22} dt]                     =x−8π[cos–1t.t22+12∫{t21−t2} dt]                     =x−8π[cos–1t.t22−12∫{1−t2−11−t2} dt]                     =x−8π[12t2cos–1t−12∫{1−t21−t2−11−t2} dt]                     =x−8π[12t2cos–1t−12∫{1−t2−11−t2} dt]                     =x−8π.12t2cos–1t+8π.12(t21−t2+12sin−1t             −sin−1t)+C                     =x−4πxcos–1x+4π(x21−x−12sin−1x)+C                     =x−4πx(π2−sin−1x)+2πx−x2−2πsin−1x+C                     =x−2x+2πx−x2+4πxsin−1x−2πsin−1x+C                     =−x+2πx−x2+2π(2x−1)sin−1x+C

Q.27

∫1sin3xsin(x+α) dx

Ans.

∫1sin3xsin(x+α) dx=∫1sin3x(sinxcosα+cosxsinα) dx                         =∫1sin4x(cosα+cotxsinα) dx                        =∫cosec2x(cosα+cotxsinα) dx[Let  t=cosα+cotxsinα⇒dtdx=−cosec2x.sinα]                       =∫cosec2xt dt−cosec2x.sinα                       =−1sinα∫t−12 dt                       =−1sinα[t1212]+C                       =−2sinα[cosα+cotxsinα]+C                       =−2sinα[cosα+cosxsinxsinα]+C                       =−2sinα[sinxcosα+cosxsinαsinx]+C                       =−2sinα[sin(x+α)sinx]+C

Q.28

Integrate the functions f‘ax+b fax+bn

Ans.

∫f’(ax + b)[f(ax + b)]n dxLet t=f(ax + b)⇒dtdx=af‘(ax + b)∴∫f’(ax + b)[f(ax + b)]n dx=∫f’(ax + b)tn dtaf‘(ax + b)                                                                       =1a∫tn dx                                                                      =1atn+1n+1+C                                                                     =1a[f(ax + b)]n+1n+1+C

Q.29

Integrate the functions e3logxx4+1−1

Ans.

     e3logx(x4+1)−1=elogx3(x4+1)−1                                            =x3x4+1          [∵elogx=x]∴∫e3logx(x4+1)−1 dx=∫x3x4+1 dx   =∫x3t dt4x3      [∵Let  t=x4+1⇒dtdx=4x3]     =∫1t dt    =logt+C         =log(x4+1)+C

Q.30

Integrate the functions cos3x elogsinx

Ans.

       cos3x elogsinx=cos3xsinx      [∵  elogx=x]∴∫cos3x elogsinx dx=∫cos3xsinx dx                                                 =−∫t3sinx dtsinx     [Let  t=cosx⇒dtdx=−sinx]                                                =−∫t3 dt                                               =−t44+C                                               =−∫t3 dt                                              =−cos4x4+C

Q.31

Integrate the functions 1(x2+1)(x2+4)

Ans.

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=0 and 4B+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=13∫1(x2+1) dx−13∫1(x2+4) dx                                                    =13tan−1(x)−13×2tan−1(x2)+C                                                    =13tan−1(x)−16tan−1(x2)+C

Q.32

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

Ans.

Given,  ∫ex(1+ex)(2+ex) dxLet  t=ex⇒dtdx=ex⇒dx=dtex∴∫ex(1+ex)(2+ex) dx=∫ex(1+t)(2+t) dtex                                                =∫(12+t−11+t) dt                                               =∫12+t dt−∫11+t dt                                               =log|2+t|−log|1+t|+C                                               =log|2+t1+t|+C                                              =log|2+ex1+ex|+C

Q.33

Integrate the functions x31−x8

                                          =log|2+ex1+ex|+C

Ans.

Since, x31−x8=x31−(x4)2Let  t=x4⇒dtdx=4x3⇒dx=dt4x3∴∫x31−x8 dx=∫x31−t2 dt4x3                                  =14∫11−t2 dt                                  =14sin−1t+C                                  =14sin−1x4+C

                                      =log|2+ex1+ex|+C

Q.34

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

                                             =log|2+ex1+ex|+C

Ans.

Since, x31−x8=x31−(x4)2Let  t=x4⇒dtdx=4x3⇒dx=dt4x3∴∫x31−x8 dx=∫x31−t2 dt4x3                                  =14∫11−t2 dt                                  =14sin−1t+C                                  =14sin−1x4+C

Sr34 (1701534)

Instruction

Question

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

Marks 1
Answer

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

Q.35

Integrate the functionssin8x−cos8x1−2sin2xcos2x

Ans.

Here,  sin8x−cos8x1−2sin2xcos2x=(sin4x)2−(cos4x)21−2sin2xcos2x                                =(sin4x+cos4x)(sin4x−cos4x)(sin2x+cos2x)2−2sin2xcos2x                               =(sin4x+cos4x)(sin2x−cos2x)(sin2x+cos2x)sin4x+cos4x+2sin2xcos2x−2sin2xcos2x                               =(sin4x+cos4x)(sin2x−cos2x)(sin2x+cos2x)(sin4x+cos4x)                               =(sin2x−cos2x)                               =−cos2x∴∫sin8x−cos8x1−2sin2xcos2x dx                            =−∫cos2x dx                            =−(sin2x2)+C

Q.36

Integrate the functions cosx4−sin2x

Ans.

Given,  ∫cosx4−sin2x dxLet     t=sinx⇒dtdx=cosx∴∫cosx4−sin2x dx=∫cosx22−t2 dtcosx                                   =sin−1(t2)+C                                   =sin−1(sinx2)+C

Q.37

Integrate the functions e5logx − e4logxe3logx − e2logx

Ans.

Here,  e5logx−e4logxe3logx−e2logx=e4logx(elogx−1)e2logx(elogx−1)                                                =e2logx                                                =elogx2∴  ∫e5logx−e4logxe3logx−e2logx dx=∫elogx2 dx                                                  =∫x2 dx                                                  =x33+C

Q.38

Integrate the functions sinxsin(x–a)

Ans.

We have,  ∫sinxsin(x−a) dxLet  t=x−a⇒dtdx=1∴ ∫sinxsin(x−a) dx=∫sin(t+a)sint dt    =∫sintcosa+sinacostsintdt    =∫(cosa+sinacott) dt    =cosa(t)+sina(logsint)+C    =cosa(x−a)+sina{logsin(x−a)}+C    =xcosa−acosa+sinalogsin(x−a)+C    =xcosa+sinalogsin(x−a)+C−acosa    =xcosa+sinalogsin(x−a)+C1                [Let  C1=C−acosa]

Q.39

Integrate the functions 5x(x+1)(x2+9)

Ans.

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=−12∫1x+1 dx+12∫xx2+9 dx+92∫1x2+33 dx                                        =−12log|x+1|+12×12log|x2+9|+92×13tan−1(x3)+C                                        =−12log|x+1|+14log(x2+9)+32tan−1(x3)+C

Q.40

Integrate the functions 1x12 + x13

Ans.

We have,∫1x12+x13dx=∫1x13(x12−13+1)dx                           =∫1x13(1+x16)dxPutting x=t6⇒dxdt=6t5⇒dx=6t5dt  =∫1t2(1+t)6t5dt  =6∫t3(1+t) dt =6∫{(t2−t+1)−1t+1} dx =6[t33−t22+t−log(t+1)]+C   =6[(x16)33−(x16)22+(x16)−log{(x16)+1}]+C =6[x3−x32+(x16)−log{(x16)+1}]+C =6x−3x3+6x16−6log(1+x16)+C

Q.41

Integrate the functions 1x2(x4+1)34

Ans.

We have, ∫1x2(x4+1)34 dx=∫1x2{x4(1+1x4)}34 dx                                     =∫1x2.x3(1+1x4)34 dx                                     =∫1x5t34 x5dt−4 Let  t=1+1x4⇒dtdx=−4x5⇒dx=x5dt−4                                     =−14∫t−34 dt                                    =−14[t1414] +C                                    =−(1+1x4)14+C

Q.42

Integrate the functions 1xax−x2

Ans.

We have, ∫1xax−x2 dx=∫1xx2(ax−1) dx                                   =∫1x2(ax−1) dx                                  =∫1x2t ×−x2adt  [Let t=ax−1⇒dtdx=−ax2⇒dx=−x2adt]                                 =−1a∫t−12 dt=−1a[t1212]+C                                =−2a(ax−1)+C                                =−2a(a−xx)+C

Q.43

Integrate the functions 1x+a+x+b

Ans.

We have 1x+a+x+b=1x+a+x+b×x+a−x+bx+a−x+b                                                 =x+a−x+b(x+a)−(x+b)                                                 =x+a−x+b(a−b)       ∫1x+a+x+b dx=1(a−b)∫(x+a−x+b) dx                                                   =1(a−b)∫{(x+a)12−(x+b)12} dx                                                    =1(a−b){(x+a)3232−(x+b)3232}+C∴           ∫1x+a+x+b dx=23(a−b){(x+a)32−(x+b)32}+C

Q.44

Integrate the functions 1x−x3

Ans.

We  have1x−x3=1x(1−x)(1+x)             =Ax+B(1−x)+C(1+x)         1=A(1−x)(1+x)+Bx(1+x)+Cx(1−x)   ...(i)Substituting x = 0, 1 and –1 respectively in equation (i), we getA=1,   B=12 and C=−12⇒   1x−x3=1x+12(1−x)−12(1+x)∴∫1x−x3 dx=∫1x dx+12∫11−x dx−12∫11+x dx      =log|x|−12log|1−x|−12log|1+x|+D      =12(2log|x|−log|(1−x)(1+x)|)+D      =12log|x21−x2|+D

Q.45

Choose  the  correct  answerThe value of∫0π2log4 + 3sinx4 + 3cosxdx isA 2B 34C 0D −2

Ans.

Let  I=∫0π2log(4+3sinx4+3cosx) dx   ...(i)          =∫0π2log{4+3sin(π2−x)4+3cos(π2−x)} 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.

Q.46

Choose  the  correct  answerThe value of ∫−π2π2x3+ xcosx+tan5x+1dx isA 0B 2C πD 1

Ans.

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

Q.47

Show that∫0af(x)g(x) dx=2∫0af(x)dx, if f and g are defined as f(x)=f(a−x) and g(x)+g(a−x)=4

Ans.

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

Q.48

∫04|x−1| dx

Ans.

Let  I=∫04|x−1| dxSince,​ (x−1)≤0 on [0,2] and (x−1)≥0 on [1,4].∴  I=∫01−(x−1) dx+∫14(x−1) dx         =[−x22+x]01+[x22−x]14         =[−122+1+0−0]+[422−4−122+1]         =−12+1+8−4−12+1         =5∴  ∫04|x−1| dx=5

Q.49

∫0axx+x−a dx

Ans.

Let  I=∫0axx+a−x dx      ...(i)           =∫0aa−xa−x+a−(a−x) dx           =∫0aa−xa−x+x dx      ...(ii)Adding equation (i) and equation (ii), we get   2I=∫0ax+a−xx+a−x dx       =∫0a1 dx       =[x]0a     I=a2a

Q.50

∫0πlog (1 + cos x) dx

Ans.

Let  I=∫0πlog(1+cosx) dx          ...(i)         =∫0πlog{1+cos(π−x)} dx       =∫0πlog(1−cosx) dx         ...(ii)Adding equation (i)​  and equation (ii),​ we get  2I=∫0πlog(1+cosx) dx+∫0πlog(1−cosx) dx        =∫0πlog{(1+cosx)(1−cosx)} dx       =∫0πlog(1−cos2x) dx       =∫0πlogsin2x dx  2I=2∫0πlogsinx dx    I=∫0πlogsinx dx       ...(iii)    I=2∫0π2logsinx dx      ...(iv)           [∵sin(π−x)=sinx]    I=2∫0π2logsin(π2−x) dx     =2∫0π2logcosx dx      ...(v)Adding equation (iv)​  2I=2∫0π2(logsinx+logcosx) dx   I=∫0π2(logsinx+logcosx) dx   =∫0π2log(2sinxcosx2) dx  =∫0π2logsin2x dx−∫0π2log2 dxLet t=2x⇒dtdx=2When x=0, t=0 and when x=π2,​ t=π. Ι=∫0πlogsint dt2−log2[x]0π2 Ι=12∫0πlogsinx dx−log2[π2−0][∴∫abf(x)dx=∫abf(t)dt] I=12I−π2log2 I−12I=−π2log2    12I=−π2log2I=−πlog2

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