NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2

The subject of Mathematics is one of the most important in students’ academic careers. For students to succeed in the subject, they must practice meticulously. As Mathematics is a conceptual subject, students should also have a clear understanding of the subject in order to do well in their Class 12 board exams. Chapter 7 of Class 12 Mathematics is Integrals, and it is a topic that students might find challenging. The NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 are a great resource for the students to understand the concepts of Chapter 7 Integrals. The solutions provide students with a consistent learning experience. Consequently, Extramarks allows students to access the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 so that they can achieve high scores in their board examinations.

Class 12 is an extremely significant academic session for students. The academic session not only prepares them for competitive exams but also provides them with the foundation they need to succeed academically. The scores of Class 12 indicate the academic competence of the students. In addition, the scores of this academic session are also evaluated during the admissions to colleges, seeking job opportunities and much more. Therefore, Extramarks provides students with the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 so that they can practice and improve their academic performance.

NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.2) Exercise 7.2

Mathematics is a subject that many students may find challenging. The first and foremost step that students must take for the preparation of their Class 12 Mathematics board examination is to thoroughly go through the NCERT curriculum. However, the NCERT textbooks do not contain the solutions to all the questions included in them. Therefore, Extramarks provides students with the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 to help them prepare well for their examinations. The NCERT solutions provided by the Extramarks website help students review each step of the problem and understand the logic behind it. Practising the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 helps students solve problems accurately and at a better pace. This way, they can perform well in their examinations.

Students can easily download the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 from the Extramarks website. Mathematics is a subject that requires an ample amount of practice so that students can improve their conceptual clarity. The NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 are considered one of the best resources for the preparation of the board examinations. They help students solve problems faster and more efficiently. The NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 are easily accessible on any device as they are available in PDF format through the Extramarks website.

Important Properties and Formulas to Remember

Students can subscribe to the Extramarks website for complete and credible study material. There are various important properties and formulas in Class 12 Chapter 7 Integrals. It is very important for the students to readily remember these properties and formulas in order to perform well in the board examinations. Some of them are-

  1. The methods of Differentiation and Integration are inverses of each other. Students can refer to the Extramarks website for a better understanding of this property.
  2. Two Indefinite Integrals with the same derivative result in the same family of curves, and so, they are equivalent. Students can review the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 to understand how these properties are practically applied to the questions.

There are multiple other formulas and functions that should be well-known to the students to be able to score better in the examinations. Students can refer to the Extramarks website to access the essential properties and formulas of Class 12 Chapter 7 Integrals. Extramarks is an organization that aims at the holistic development of students. It provides students with comprehensive study material so that students can succeed in any examination. NCERT content forms the conceptual academic base of students, therefore Extramarks offers students NCERT Solutions for all the subjects and classes. Students can refer to the Extramarks website for :

NCERT Solutions Class 12

NCERT Solutions Class 11

NCERT Solutions Class 10

NCERT Solutions Class 9

NCERT Solutions Class 8

NCERT Solutions Class 7

NCERT Solutions Class 6

NCERT Solutions Class 5

NCERT Solutions Class 4

NCERT Solutions Class 3

NCERT Solutions Class 2

NCERT Solutions Class 1.

NCERT Solutions for Class 12 Maths Chapter 7 – Exercise 7.2 Questions

The NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 provided by Extramarks are one of the best resources that help students prepare well to score high in board examinations. This is because the NCERT textbook covers all the topics that can appear in board examinations. Moreover, the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 are curated by experienced and certified educationalists. NCERT textbooks are written by professionals who are experts in their subjects. Furthermore, they help students to understand the basic concepts of the subject, and they are ideal for students who find Mathematics difficult. Practising the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 helps students increase their pace of solving the problems of Mathematics, which is very essential for them to score well in the board examinations. Once students go through the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2, they will be able to solve any complicated problem that they may encounter in their examinations. Extramarks recommends students practice Class 12th Maths Chapter 7 Exercise 7.2 thoroughly prior to their examinations.

NCERT Solutions for Class 12 Maths Chapter 7 – Exercise 7.2 Questions

The major concepts involved in Class 12 Chapter 7 Integrals are the Introduction of the Chapter, Integration as an Inverse Process of Differentiation, Geometrical Interpretation of Indefinite Integral, Some Properties of Indefinite Integral, Comparison between Differentiation and Integration, and Methods of Integration. Other included topics are Integrals of Some Particular Functions, Integration by Partial Fractions, Integration by Partial Fractions, Integration by Parts, Definite Integral, Fundamental Theorem of Calculus, Evaluation of Definite Integrals by Substitution, and Some Properties of Definite Integrals. Exercise 7.2 Class 12 Maths Solutions are based on Integration by Substitution, Integration using Partial Fractions and Integration by Parts. Students can practice the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 to structure their answers to Exercise 7.2 Class 12th.

Ex 7.2 Class 12 Maths NCERT Solutions: Introduction

The NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 are based on the various Methods of Integration. Students can also practice the exemplar questions given before Exercise 7.2  to have a better understanding of the concepts of the exercise. Overall, there are eleven exercises in the chapter; therefore, it could be a huge challenge for the students to understand such a wide range of concepts. Therefore, Extramarks provides students with NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 so that they can rigorously practice the NCERT textbook to be able to perform well in their examinations.

NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2: Formalization

Extramarks provides students with the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 so that they have access to authentic solutions without having to look anywhere else. This helps students save time as they do not look for solutions elsewhere on the internet. Along with the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2, Extramarks also provides students with various learning tools, such as K12 study material, in-depth performance reports, and much more. Live doubt-solving sessions allow students one-on-one discussions with the subject-matter experts in order to help them resolve all their doubts. As a result, students are able to concentrate on their goals and succeed in their board examinations. Extramarks also provides students with various revision methodologies to help them remember the formulas and properties of the chapter so that they can complete their examinations efficiently.

Exercise 7.2 Class 12 Maths NCERT Solutions

One of the best ways to improve the mathematical skills of students is to thoroughly review the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2. These solutions help students build strong fundamentals of the curriculum of the subject and also assist them in practising the application of those skills in real-life scenarios. The NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 are designed for students to improve their mathematical skills quickly and effectively. Students can find the solutions complicated, but there are helpful resources to practice Chapter 7 Integrals. By revising with the help of the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 students can practice the concepts and functions of the chapter topics, which are very essential to scoring well in any in-school, board, or competitive examinations. Practising the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 is the primary step that students should take in preparation for their board examinations.

Class 12 Maths Ch Ex 7.2

In Mathematics, an integral can be defined as a numerical identity or a function of which the given function is a derivative. Furthermore, there are two types of Integrals – Definite and In-Definite Integrals. For a better understanding of the Chapter Integrals, students can refer to the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 and subscribe to the Extramarks website.

Why Choose Extramarks?

In addition to the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2, Extramarks also provides students with multiple learning modules so that they can learn comprehensively and enjoy their learning process.

Extramarks provides students with learning modules like-

  • Self-Learning App – Students can learn and excel at their own pace on the Extramarks website. They can practice customised tests and also access their performance.
  • Visual Learning Journey- Extramarks provides students with Graphics and Animations that make the learning process enjoyable, so they do not find their studies monotonous.
  • Practice and Analysis – Extramarks provides students with chapter-specific worksheets, unlimited practice questions, interactive activities, and much more, to gain expertise in all subjects. As a result, students are able to analyse their preparation levels and track their progress.
  • Comprehensive Study Material – The Extramarks website provides students with authentic study material like, the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 and much more, to help them perform better in their examinations.
  • Complete Syllabus Coverage – Sometimes, students are not able to go through their entire curriculum on time. With the guidance of Extramarks, students have access to complete syllabus coverage to ensure that they can do well in any in-school or board examinations.
  • Curriculum Mapping- Extramarks provides students with a curriculum-mapped learning experience so that they do not need to look elsewhere for revision.
  • Learn with Expert Guidance- Extramarks provides students with trained professionals who are experts in their subject and who are highly experienced and qualified to provide students with the most appropriate guidance.

Mathematics is a subject that students may find challenging. Extramarks provides students with NCERT solutions, like the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 to help them structure their answers better in the final examination.

Extramarks is a great resource that allows students to access all the required learning material easily and at their convenience. Mathematics has always been a subject that multiple students find difficult to accomplish, but with the expert guidance of Extramarks, they will find it a lot easier to learn Mathematics. Combined with the practice of the students and the guidance of Extramarks, they will be able to achieve the best results. Extramarks is an organization that provides a variety of educational courses to enable young learners to learn, practice, and excel in their studies. As a paradigm, this e-learning portal proves that technology has its share of positive effects in making learning easier and more efficient. Extramarks as an organization is committed to maintaining the growth and flourishing of this community of young minds in order to promote academic success.

Q1. ∫0π2 sin x − cos x1 + sin x cos x dx

Ans.

Let  I=∫0π2sinx−cosx1+sinxcosxdx             ...(i)          =∫0π2sin(π2−x)−cos(π2−x)1+sin(π2−x)cos(π2−x)dx         =∫0π2cosx−sinx1+cosxsinxdx            ...(ii)Adding equation (i) and equation (ii), we get     2I=∫0π2sinx−cosx+cosx−sinx1+sinxcosx dx        =∫0π201+sinxcosxdx        =0       I=0∴ ∫0π2sinx−cosx1+sinxcosxdx=0

Q2. ∫02πcos5x dx

Ans.

Let  I=∫02πcos5x dxsince, cos5(2π−x)=cos5x∵  ∫02af(x)dx={2∫0af(x)dx, if f(2a−x)=f(x)0,                        if  f(2a−x)=−f(x)∴  ∫02πcos5x dx=2∫0πcos5x dx                                     =0         [∵cos5(π−x)=−cos5x]

Q3. ∫–π2π2sin7x dx

Ans.

Let  I=∫–π2π2sin7x dxHere, f(x)=sin7x  and​  f(−x)=sin7(−x)                                                                     =−sin7x=−f(x)So,​ f(x)​ is​​ an odd function, then ∫−aaf(x)dx=0∴  I=∫–π2π2sin7x dx=0

Q4. ∫0πx1+ sinx dx

Ans.

Let  I=∫0πx1+ sinx dx           =∫0ππ−x1+ sin(π−x) dx          =∫0ππ−x1+ sinx dx      I=∫0ππ1+ sinx dx−∫0πx1+ sinx dx      I=∫0ππ1+ sinx dx−I   2I=∫0ππ(1− sinx)(1+ sinx)(1− sinx) dx      =∫0ππ(1− sinx)1− sin2x dx    =π∫0π(1− sinx)cos2x dx    =π∫0π(sec2x−secxtanx) dx    =π[tanx−secx]0π    =π{(tanπ−secπ)−(tan0−sec0)}    =π{0−(−1)−0+1}    =2π  Ι=π

Q5. ∫–π2π2sin2x dx

Ans.

Let           I=∫–π2π2sin2x dxHere,  f(x)=sin2xso,       f(−x)=sin2(−x)                          =sin2x                          =f(x)∴            I=2∫0π2sin2x dx         [∫−aaf(x)dx=2∫0af(x)dx,  if f(−x)=f(x)]                I=2∫0π21−cos2x2 dx                 =∫0π2(1−cos2x) dx                 =[x−sin2x2]0π2                =[π2−12sin2(π2)]−[0−12sin2(0)]                 =π2−12sinπ                  =π2

Q6. ∫0π2 (2 log sin x − log sin 2x) dx

Ans.

Let   I=∫0π2(2logsinx–logsin2x) dx            =∫0π2(2logsinx−log2sinxcosx) dx          =∫0π2(2logsinx−log2−logsinx−logcosx) dx         I=∫0π2(logsinx−log2−logcosx) dx          ...(i)         =∫0π2{logsin(π2−x)−log2−logcos(π2−x)} dx                              [P4:∫0af(x) dx=∫0af(a−x) dx]        =∫0π2{logcosx−log2−logsinx} dx           ...(ii)Adding​ equation (i) and equation (ii),​ we get     2I=∫0π2(logsinx−log2−logcosx+logcosx−log2−logsinx) dx     2I=∫0π2{−2log2} dx        =−2log2[x]0π2        =−2log2[π2−0]       I=−(π2)log2=(π2)log12

Q7. ∫02x2−x dx

Ans.

Let   I=∫02x2−x dx         I=∫02(2−x)2−(2−x) dx      [P4:∫0af(x) dx=∫0af(a−x) dx]         I=∫02(2−x)x dx         I=∫02(2x12−x32) dx           =[2.x12+112+1−x32+132+1]02           =[2.23232−25252]−[2.012+112+1−032+132+1]           =[223.232−2.2525]           =2723−2725           =272(13−15)           =272(5−315)           =272(215)           =29215         I=16215

Q8. ∫0π4log(1+tanx) dx

Ans.

Let   I=∫0π4log(1+tanx) dx            =∫0π4log{1+tan(π4−x)} dx      [   P4:∫0af(x) dx=∫0af(a−x) dx]           =∫0π4log{1+tanπ4−tanx1+tanπ4tanx} dx          =∫0π4log{1+1−tanx1+1.tanx} dx         =∫0π4log{1+tanx+1−tanx1+tanx} dx         =∫0π4log(21+tanx) dx      I =∫0π4log2 dx−∫0π4log(1+tanx) dx      I =log2[x]0π4−I     2I=[π4−0]log2        I =π8log2

Q9. ∫01x(1−x)n dx

Ans.

Let  I=∫01x(1−x)n dx           =∫01(1−x){1−(1−x)}n dx      [  P4:∫0af(x) dx=∫0af(a−x) dx]           =∫01(1−x)xn dx          =∫01(xn−xn+1) dx         =[xn+1n+1−xn+2n+2]01         =[1n+1n+1−1n+2n+2]−[0n+1n+1−0n+2n+2]          =n+2−n−1(n+1)(n+2)=1(n+1)(n+2)

Q10. ∫28|x−5| dx

Ans.

Let  I=∫28|x−5| dxSince,​ (x−5)≤0 on [2,5] and (x−5)≥0 on [5,8].∴  I=∫25−(x−5) dx+∫58(x−5) dx         =[−x22+5x]25+[x22−5x]58        =[−522+5(5)−{−222+5(2)}]+[822−5(8)−{522−5(5)}]       =[−252+25+42−10]+[642−40−252+25]       =[252−8]+[−8+252]       =25−16       =9

Q11. ∫–55|x+2| dx

Ans.

Let      I=∫–55|x+2| dxSince,​ (x+2)≤0 on [−5,−2] and (x+2)≥0 on [−2,5].   ∴      I=∫–5−2−(x+2) dx+∫–25(x+2) dx                =[−x22−2x]−5−2+[x22+2x]−25                =−[(−2)22+2(−2)−(−5)22−2(−5)] + [522+2(5)−(−2)22−2(−2)]                =−[2−4−252+10]+[252+10−2+4]                =−8+252+252+12                =29

Q12. ∫0π2cos5xsin5x+ cos5x dx

Ans.

Let         I=∫0π2cos5xsin5x+ cos5x dx        ...(i)         =∫0π2cos5(π2−x)sin5(π2−x)+ cos5(π2−x) dx     ∵P4:∫0afxdx=∫0afa−xdx      I=∫0π2sin5xcos5x+ sin5x dx       ...(ii)Adding  equation (i) and equation (ii), we get    2 I=∫0π2cos5xsin5x+ cos5x dx + ∫0π2sin5xcos5x+ sin5x dx         =∫0π2cos5x+sin5xsin5x+ cos5x dx         =∫0π21. dx         =[x]0π2         =π2−0         =π2     I =π4

Q13. ∫0π2sin32xsin32x + cos32x dx

Ans.

Let   I=∫0π2sin32xsin32x+cos32x dx       ...(i)           =∫0π2sin32(π2−x)sin32(π2−x)+cos32(π2−x) dx                         [∵P4:∫0af(x) dx=∫0af(a−x) dx]        I=∫0π2cos32xcos32x+sin32x dx      ...(ii)Adding  equation (i) and equation (ii), we get    2 I=∫0π2sin32xsin32x+cos32x dx +∫0π2cos32xcos32x+sin32x dx         =∫0π2sin32x+cos32xcos32x+sin32x dx         =∫0π21. dx         =[x]0π2         =π2−0         =π2

Q14.∫0π2 sin xsin x + cos x  dx

Ans.

Let   I=∫0π2sinxsinx+cosx dx         ...(i)           =∫0π2sin(π2−x)sin(π2−x)+cos(π2−x) dx                         [∵  P4:∫0af(x) dx=∫0af(a−x) dx]        I=∫0π2cosxcosx+sinx dx        ...(ii)Adding  equation (i) and equation (ii), we get    2 I=∫0π2sinxsinx+cosx dx +∫0π2cosxcosx+sinx dx          =∫0π2sinx+cosxsinx+cosx dx          =∫0π21. dx         =[x]0π2         =π2−0         =π2        I =π4

Q15. ∫0π2cos2x dx

Ans.

Let    I=∫0π2cos2x dx  ...   (i)           =∫0π2cos2(π2−x) dx    [∵  P4:∫0af(x) dx=∫0af(a−x) dx]      I =∫0π2sin2x dx   ...(ii)Adding equation (i)​ and  (ii),​ we get       2I=∫0π2(cos2x+sin2x) dx       2I=∫0π21 dx       [∵ cos2x+sin2x=1]           =[x]0π2           =π2−0           =π2        I =π4

Q16.

Choose  the  correct  answerIf fx=∫0xt sint dt, then f‘x isA cosx+x sinxB x sinxC x cosxD sinx+x cosx

Ans.

We have  f(x)= ∫0xt sint dt               =[t∫sint dt−∫ddtt∫sint dt dt]0x               =[t(−cost)−∫1.(−cost) dt]0x               =[t(−cost)+∫cost dt]0x               =[t(−cost)+sint]0x               =x(−cosx)+sinx−0(−cos0)−sin0               =−xcosx+sinxDifferentiating​ w.r.t. x, we get                  f’(x)=−(xddxcosx+cosxddxx)+ddxsinx               =−(x×−sinx+cosx×1)+cosx               =xsinx−cosx+cosx               =xsinxHence, the correct option is B.

Q17.

Choose  the  correct  answerThe value of the integral ∫−131x−x313x4dx isA 6B 0C 3D 4

Ans.

∫–131(x−x3)13x4 dxLet  x=sinθ⇒dxdθ=cosθWhen x=−13, θ=sin−1(−13) and when x=1, θ=π2∫–131(x−x3)13x4 dx=∫sin–1(−13)π2(sinθ−sin3θ)13sin4θ .cosθ dθ                =∫sin–1(−13)π2(sinθ)13(1−sin2θ)13sin4θ .cosθ dθ                =∫sin–1(−13)π2(sinθ)13(cos2θ)13sin4θ .cosθ dθ               =∫sin–1(−13)π2(sinθ)13(cosθ)23sin2θ.sin2θ .cosθ dθ               =∫sin–1(−13)π2(cosθ)53sin53θ .cosec2θ dθ               =∫sin–1(−13)π2(cotθ)53 .cosec2θ dθLet  t=cotθ⇒dtdθ=−cosec2θWhen θ=sin−1(−13), t=22 and when θ=π2,t=0               =−∫220(t)53 .dt=−[t83(83)]220               =−38[0−(22)83]               =−38(−232)83=               =38×24               =38×16=6Thus, the correct option is A.

Q18. Evaluate the integrals ∫12 (1x−12x2) e2x dx

Ans.

Let  I=∫12(1x−12x2) e2x dx         =∫12(22x−2(2x)2) e2x dxAgain let​ t=2x⇒dtdx=2When x=1, t=2 and when x=2, t=4       I=∫24(2t−2t2) et dt2       =∫24(1t−1t2) etdtLet F(t)=1t,  Then F‘(t)=−1t2⇒∫24(1t−1t2) etdt=∫24(F(t)−F‘(t)) etdt                    =[etF(t)]24                    =[et.1t]24                    =e4.24−e2.12                   =14e2(e2−2)Thus,∫12(1x−12x2) e2x dx=14e2(e2−2)

Q19.Evaluate the integrals ∫–11 dxx2+2x+5 dx

Ans.

∫–11dxx2+2x+5 dx=∫–11dx(x+1)2+22 dxLet t=x+1⇒dtdx=1When x=−1, t=0​  and when x=1, t=2∴∫–11dx(x+1)2+22 dx=∫02dxt2+22 dt                                                   ={12tan−1(t2)}02                                                   =12tan−1(22)−12tan−1(02)                                                   =12tan−1(1)−12tan−1(0)                                                   =12×π4−12×0=π8Thus,   ∫–11dxx2+2x+5 dx=π8

Q20. Evaluate the integrals ∫02 dxx+4−x2 

Ans.

Let   I=∫02dxx+4−x2          =∫02dx−(x2−x−4)          =∫02dx−(x2−x+14−14−4)         =−∫02dx(x−12)2−174         =−∫02dx(x−12)2−(174)2Let t=x−12⇒dtdx=1When x=0, t=−12 and when x=2, t=32 I=−∫−1232dt(174)2−t2  =12(174)log{(174)+t(174)−t}−1232  =12(174)[log{(174)+32(174)−32}−log{(174)−12(174)+12}]  =12(174)[log{17+317−3}−log{17−117+1}]  =12(174)log{17+317−3×17+117−1}  =12(174)log{17+17+317+317−17−317+3}  =217log{20+41720−417}  =217log{5+175−17}  =217log{5+175−17×5+175+17}  =217log{25+17+101725−17}  =217log{42+10178}  =217log{21+5174}

Q21.Evaluate the integrals ∫0π2sinx1+cos2x dx

Ans.

Let   I=∫0π2sinx1+cos2x dxAgain, let t=cosx⇒dtdx=−sinxWhen x=0, t=1  and  when  x=π2, t=0∴     I=∫10sinx1+t2 dt−sinx          =−∫1011+t2 dt         =−[tan−1t]10         =−(tan−10−tan−11)         =−(0−π4)=π4

Q22. Evaluate the integrals ∫02xx+2  dx

Ans.

I=∫02xx+2  dxLet​ t2=x+2⇒dtdx=2tWhen x=0,  t=2  and when x=2,  t=2∴     I=∫22(t2−2)t2 2t dt          =2∫22(t4−2t2) dt         =2[t55−2×t33]22       =2[255−2×233]− 2[(2)55−2×(2)33]         =2[255−243]− 2[425−423]        =25[25−13]− 232[15−13]         =25(6−515)− 232(3−515)         =2515−232(−215)         =24(215+215)         =162(2+115)

Q23. Evaluate the integrals ∫01sin–1(2x1+x2) dx

Ans.

Let  I=∫01sin–1(2x1+x2) dxAlso, let x=tanθ⇒dxdθ=sec2θWhen  x=0,  θ=0 and when x=1,  θ=π4∴     I=∫0π4sin–1(2tanθ1+tan2θ) sec2θ dθ=∫0π4sin–1(sin2θ) sec2θdθ=∫0π42θ sec2θ dθNow,∫2θ sec2θ dθ=2[θ∫sec2θ dθ−∫(ddθθ∫sec2θ dθ)  dθ]=2[θ tanθ−∫(1. tanθ) dθ]=2[θ tan θ+log cos θ]∴     I=2[θ tan θ+log cos θ]0π4=2[(π4)tan π4+log cos π4]−[2(0)tan 0+log cos 0]=2[π4×1+log 12]−[0+log 1]=[π2−2 × 12 log2]−[0+0]=(π2−log 2)

Q24.Evaluate the integrals ∫0π2sinϕcos5ϕ dϕ

Ans.

Let  I=∫0π2sinϕcos5ϕ  dϕ          =∫0π2sinϕcos4ϕ cosϕ dϕ          =∫0π2sinϕ(cos2ϕ)2 cosϕ  dϕ          =∫0π2sinϕ(1−sin2ϕ)2 cosϕ dϕ          =∫0π2sinϕ(1−2sin2ϕ+sin4ϕ) cosϕ  dϕLet t=sinϕ⇒dtdϕ=cosϕWhen ϕ=0, t=0 and when ϕ=π2, t=1∴     I=∫01t(1−2t2+t4) dt=∫01(t12−2t52+t92) dt=[t3232−2t7272+t112112]01=[13232−217272+1112112]−[03232−207272+0112112]=[23−2.27+211]=23−47+211=154−132+42231=64231

Q25. Evaluate the integrals ∫01xx2+1 dx

Ans.

∫01xx2+1 dxLet  t=x2+1⇒dtdx=2xWhen x=0, t=1 and when x=1,  t=2∴∫01xx2+1 dx=12∫121t dt       =12[log|t|]12       =12[log2−log1]       =12log2

Q26.

Choose the correct answer∫023dx4+9x2 equalsA π6B π12C π24D π4

Ans.

Let               I=∫023dx4+9x2∫14+9x2 dx=∫122+(3x)2 dx                   =12tan−1(3x2)×13     =16tan−1(3x2)=F(x)Therefore, by the second fundamental theorem, we have               I=F(23)−F(0)     =16tan−1(32×23)−16tan−1(32×0)     =16tan−1(1)−16tan−1(0)     =16×π4=π24Hence, the correct option is C.

Q27.

Choose the correct answer∫13dx1+x2 equalsA π3B 2π3C π6D π12

Ans.

Let            I=∫13dx1+x2∫11+x2 dx=tan−1xTherefore, by the second fundamental theorem, we have                I=F(3)−F(1)                 =tan−1(3)−tan−1(1)                 =π3−π4                 =4π−3π12                 =π12Hence, the correct option is D.

Q28. ∫01(x ex+sinπx4)dx

Ans.

Let   I=∫01(x ex+sinπx4)dx∫(x ex+sinπx4) dx=∫x ex dx+∫sinπx4 dx                                      =x∫ex dx−∫(ddxx∫ex dx) dx−1(π4)cosπx4+C                                      =xex−∫ex dx−4πcosπx4                                      =xex−ex−4πcosπx4                                      =ex(x−1)−4πcosπx4=F(x)Therefore, by the second fundamental theorem, we have        I=F(1)−F(0)         ={e1(1−1)−4πcosπ(1)4}−{e0(0−1)−4πcosπ(0)4}         ={−4πcosπ4}−{−1−4π×1}         ={−4π×12}−{−1−4π}         =−42π+1+4π         =1+4π−22π

Q29. ∫026x+3x2+4 dx

Ans.

            Let  I=∫026x+3x2+4dx∫6x+3x2+4 dx=∫6xx2+4 dx+∫3x2+4 dx                         =3∫2xx2+4 dx+3∫1x2+22 dx                         =3[log(x2+4)]+32[tan−1x2]=F(x)Therefore, by the second fundamental theorem, we have                 I=F(2)−F(0)                   =3[log(22+4)−32tan−122]−[log(02+4)+32tan−102]                  =3[log8−log4]−32[tan−1(1)−tan−1(0)]                  =3log84−32[π4−0]∫026x+3x2+4dx=3log2−3π8

Q30. ∫0π(sin2x2−cos2x2)dx

Ans.

Let                    I=∫0π(sin2x2−cos2x2)dx∫(sin2x2−cos2x2) dx=−∫(cos2x2−sin2x2) dx                           =−∫cosx dx      [∵cos2θ−sin2θ=cos2θ]                          =−sinx=F(x)Therefore, by the second fundamental theorem, we have                         I=F(π)−F(0)             =−sinπ−(−sin0)  =0

Q31. ∫0π4(2sec2 x+x3+2) dx

Ans.

Let  I=∫0π4(2sec2 x+x3+2) dx∫(2sec2 x+x3+2) dx      =2∫sec2 x dx+∫x3 dx+∫2 dx      =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[tanπ4−tan0]+14[(π4)4−0]+2[π4−0]       =2[1−0]+14[(π4)4]+2[π4]       =2+π41024+π2

Q32. ∫125x2x2+4x+3  dx

Ans.

Let                 I=∫125x2x2+4x+3 dx∫5x2x2+4x+3 dx=∫(5−20x+15x2+4x+3) dx                                 =∫5 dx−∫20x+15x2+4x+3 dx  ...(i)Let     20x+15=Addx(x2+4x+3)+B                               =A(2x+4)+B                               =2Ax+4A+BEquating the coefficients of x and constant term, we get⇒ 2A=20 and 4A+B=15⇒ A=10  and  B=−25∴        20x+15=10(2x+4)−25From equation(i), we have  ∫5x2x2+4x+3 dx=∫5 dx−∫10(2x+4)−25x2+4x+3 dx                                    =∫5 dx−10∫(2x+4)x2+4x+3 dx+25∫1x2+4x+3 dx                                    =5x−10log(x2+4x+3)+25∫1(x+2)2−1 dx                                    =5x−10log(x2+4x+3)+25×12log|x+2−1x+2+1|+C                                    =5x−10log(x2+4x+3)+252log|x+1x+3|∫125x2x2+4x+3 dx=[5x−10log(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+3 dx=[5x−10log(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|]                               =(10−10log15+252log35)−(5−10log8+252log24)                            =5−10(log15−log8)+252(log35−log24)                                  =5−10(log5+log3−log4−log2)                                     +252(log3−log5−log2+log4)                                  =5−(10+252)log5+(10+252)log4+(−10+252)log3                                   +(10−252)log2                               =5−452log5+452log4+52log3−52log2∫125x2x2+4x+3 dx =5−452log54+52log32  =5−52(9log54−log32)

Q33. ∫01x ex2dx

Ans.

Let       I=∫01x ex2dx∵∫x ex2 dx=∫x et dt2x        [Let​ t=x2⇒dtdx=2x⇒dx=dt2x]                      =12et                      =12ex2=F(x)Therefore, by the second fundamental theorem, we have             I=F(1)−F(0)            =12e12−12e0 =12(e−1)

Q34. ∫012x+3(5x2+1)dx

Ans.

Let            I=∫012x+3(5x2+1)dx                     =15∫0110x(5x2+1)dx+∫013(5x2+1)dx∫2x+3(5x2+1) dx=15∫10x(5x2+1) dx+3∫1(5x2+1) dx                                =15log|5x2+1|+3∫1(5x)2+1 dx                                =15log|5x2+1|+3tan−15x.15                                =15log|5x2+1|+35tan−1(5x)=F(x)Therefore, by the second fundamental theorem, we have                I=F(1)−F(0)                 =15log|5(1)2+1|+35tan−1{5(1)}−[15log|5(0)2+1|    +35tan−1{5(0)}]                 =15log(6)+35tan−1(5)−15log|1|−35tan−1(0)    =15log(6)+35tan−1(5)

Q35. ∫23x dx(x2+1)

Ans.

Let  I=∫23x dx(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|

Q36. ∫0π2cos2x dx

Ans.

Let  I=∫0π2cos2x dx=∫0π21+cos2x2 dx∵∫1+cos2x2 dx=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

Q37. ∫23dx(x2−1)

Ans.

Let  I=∫23dx(x2−1)∵∫1x2−1 dx=12log|x−1x+1|=F(x)Therefore, by the second fundamental theorem, we have      I=F(3)−F(2)       =12log|3−13+1|−12log|2−12+1|       =12log|24|−12log|13|       =12log|24×31|       =12log|32|

Q38. ∫011(1+x2)dx

Ans.

Let  I=∫0111+x2dx∵∫11+x2 dx=tan−1x=F(x)Therefore, by the second fundamental theorem, we have      I=F(1)−F(0)       =tan−1(1)−tan−1(0)       =π4−0=π4

Q39. ∫0111−x2 dx

Ans.

Let  I=∫0111−x2dx∵∫11−x2 dx=sin−1x=F(x)Therefore, by the second fundamental theorem, we have      I=F(1)−F(0)       =sin−1(1)−sin−1(0)       =π2−0=π2

Q40. ∫π6π4cosec x dx

Ans.

Let  I=∫π6π4cosec x dx∵∫cosec x dx=log|cosecx−cotx|=F(x)Therefore, by the second fundamental theorem, we have      I=F(π4)−F(π6)     =log|cosecπ4−cotπ4|−log|cosecπ6−cotπ6|     =log|2−1|−log|2−3|     =log(2−12−3)

Q41. ∫0π4 tan x dx

Ans.

Let  I=∫0π4tan x dx∵∫tan x dx=−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

Q42. ∫45ex dx

Ans.

Let  I=∫45ex dx∵∫ex dx=ex=F(x)Therefore, by the second fundamental theorem, we have      I=F(5)−F(4)       =e5−e4       =e4(e−1)

Q43. ∫0π2cos2x dx

Ans.

Let  I=∫0π2cos2x dx∵∫cos2x dx=12sin2x=F(x)Therefore, by the second fundamental theorem, we have      I=F(π2)−F(0)       =−12sin2(π2)−(−12sin0)       =−12sinπ+12sin0       =0+0=0

Q44. ∫0π4sin2x dx

Ans.

Let  I=∫0π4sin2x dx∵∫sin2x dx=−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

Q45. ∫12(4x3−5x2+6x+9)dx

Ans.

Let  I=∫12(4x3−5x2+6x+9)dx∵∫(4x3−5x2+6x+9) dx      =∫4x3 dx−∫5x2 dx+∫6x dx+∫9 dx      =x4−53x3+3x2+9x=F(x)Therefore, by the second fundamental theorem, we have      I=F(2)−F(1)       ={24−53(2)3+3(2)2+9(2)}−{(1)4−53(1)3+3(1)2+9(1)}      =(16−403+12+18)−(1−53+3+9)      =(46−403)−(13−53)      =46−403−13+53      =33−353      =643

Q46. ∫231xdx

Ans.

Let  I=∫231xdx∵ ∫ 1x dx=logx=F(x)Therefore, by the second fundamental theorem, we have      I=F(3)−F(2)       =log3−log2=log32

Q47. ∫–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

Q48. Evaluate the following definite integrals as limit of sums.∫04(x+e2x) dx

Ans.

We have, ∫04(x+e2x) dx=∫04x dx+∫04e2x dx                           =I1+I2   (Let)∫abf(x) dx=(b−a)limn→∞1n[f(a)+f(a+h)+...+f(a+(n−1)h)]where,   h=b−anHere,  a=0,  b=4,  f(x)=x,  h=4−0n=4n        I1=∫04x dx∴∫04x dx=(4−0)limn→∞1n[f(0)+f(0+h)+...+f(0+(n−1)h)]             =4limn→∞1n[0+h+...+(n−1)h]             =4limn→∞1n[h+2h+...+(n−1)h]             =4limn→∞1n[{1+2+...+(n−1)}h]             =4limn→∞1n[(n−1)n2h][∵∑n=n(n+1)2]             =4limn→∞[(n−1)2(4n)]             =4limn→∞[42(1−1n)]             =4[2(1−0)]             =4(2)=8        I2=∫04e2x dx             =(4−0)limn→∞1n[f(0)+f(0+h)+...+f(0+(n−1)h)]             =4limn→∞1n[e0+e2h+...+e2(n−1)h]             =4limn→∞1n[1+e2h+...+e2(n−1)h]Using the sum to n terms of a G.P., where a=1, r=e2h, we have ∴∫04e2x dx=4limn→∞1n[1.(e2nh−1)(e2h−1)]           =4limn→∞1n[(e2n.4n−1)(e8n−1)]           =4limn→∞1n[(e8−1)(e8n−1)]           =4(e8−1)limn→∞[e8n−18n].8           =4(e8−1)8         [∵limx→0eh−1h=1]           =e8−12So,  ∫04(x+e2x) dx          =I1+I2           =8+e8−12          =15+e82

Q49. Evaluate the following definite integrals as limit of sums.∫–11ex dx

Ans.

We  have,  ∫–11ex dx∵∫abf(x) dx=(b−a)limn→∞1n[f(a)+f(a+h)+...+f(a+(n−1)h)],where,   h=b−anHere,  a=−1,  b=1,  f(x)=ex,  h=1−(−1)n=2n∴∫–11ex dx=(1+1)limn→∞1n[f(−1)+f(−1+h)+...+f(−1+(n−1)h)]           =2limn→∞1n[e−1+e−1+h+...+e−1+(n−1)h]Using the sum to n terms of a G.P., where a=e−1, r=eh, we have∴∫–11ex dx=2limn→∞1n[e−1(enh−1)(eh−1)]           =2limn→∞1n[e−1(en.2n−1)(e2n−1)]           =2limn→∞1n[e−1(e2−1)(e2n−1)]           =2e−1(e2−1)limn→∞[e2n−12n].2           =2e−1(e2−1)2           =e−1e           ∵limx→0eh−1h=1

Q50. Evaluate the following definite integrals as limit of sums.∫14(x2−x) dx

Ans.

We have, ∫14(x2−x) dx=∫14x2dx−∫14x dx                                                     =I1−I2    (Let)∫abf(x) dx=(b−a)limn→∞1n[f(a)+f(a+h)+...+f(a+(n−1)h)]where,   h=b−anHere,  a=1,  b=4,  f(x)=x2,  h=4−1n=3n∴      I1=∫14x2dx              =(4−1)limn→∞1n[f(1)+f(1+h)+...+f(1+(n−1)h)]                =3limn→∞1n[(1)2+(1+h)2+...+{1+(n−1)h}2]                =3limn→∞1n[1+(1+2h+h2)+...+{1+2(n−1)h+(n−1)2h2}]                =3limn→∞1n[n+{1+2+...+(n−1)}2h+{1+22+32+...+(n−1)2}h2]                =3limn→∞1n[n+(n−1)n22h+(n−1)n(2n−2+1)6h2]                =3limn→∞1n[n+(n−1)n.3n+(n−1)n(2n−2+1)6.9n2]                                      ∵∑n=nn+12,∑n2=nn+12n+16                =3limn→∞[1+(1−1n)×3+(1−1n)(2−1n)6×9]                 =3[1+3(1−0)+3(1−0)(2−0)2]                =3(1+3+3)            =21For​   I2=∫14x dx∴∫14x dx=(4−1)limn→∞1n[f(1)+f(1+h)+...+f(1+(n−1)h)]                       =3limn→∞1n[1+1+h+...+1+(n−1)h]                       =3limn→∞1n[n+h+2h+...+(n−1)h]                       =3limn→∞1n[n+{1+2+...+(n−1)}h]                       =3limn→∞1n[n+(n−1)n2h][∵∑n=n(n+1)2]                       =3limn→∞[1+(n−1)2(3n)]                      =3limn→∞[1+32(1−1n)]                       =3[1+32(1−0)]                      =3(52)=152So,  ∫14(x2−x) dx                     =I1−I2 =21−152 =272

Please register to view this section

FAQs (Frequently Asked Questions)

No, with regular practice on the part of students, and the proper guidance of Extramarks, students can easily understand the NCERT Solutions and excel at the necessary concepts to score well in the board examinations.

Yes, it is beneficial for students to practice all the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 as they help students build a strong conceptual core of the syllabus. Also, practising them increases the pace and efficiency of the students. Every question of the exercise contains a different concept, and rehearsing these helps students avoid small errors that occur while solving problems in Mathematics.

Students can refer to the NCERT Solutions provided by Extramarks to clarify their doubts related to the exercise. Furthermore, they can subscribe to the learning application of Extramarks to have access to live doubt-solving sessions and guidance from expert teachers for their further studies.

Extramarks provides students with NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2. Along with this, Extramarks ensures that students are delivered well-curated K12 study material for their revision schedules and receive live classes with the experts, a self-assessment centre, and much more, to help them succeed in their board examinations.

Yes, for example, according to the changes in the admission pattern of Delhi University, the entrance examinations being conducted are entirely based on the curriculum of NCERT. Also, there are other universities that follow the same pattern of admissions. Moreover, students should have a good understanding of the concepts of Chapter 7 Integrals for JEE. So yes, the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 will help students in the preparation for multiple entrance examinations.

Mathematics, as a subject, requires meticulous practice. Reviewing the examples given in the NCERT book and the NCERT Exemplar will definitely help students revise the concepts of Exercise 7.2. The more questions students practice, the better they can get at writing a good paper and achieving a high score.