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# NCERT Solutions for Class 12 Maths Chapter 7 Integrals

Last updated date: 17th Jul 2024
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## Class 12 Maths Chapter 7 Integrals NCERT Solutions - Free PDF Download

NCERT Solutions for Class 12 Mathematics, Chapter 7 Integrals were created by subject matter experts at Vedantu to assist students with their board exam preparation. Chapter 7 of the NCERT Maths Book for Class 12 covers the idea of integrals. This chapter of the NCERT Solutions for Class 12 Maths teaches students about integral calculus (definite and indefinite), its properties, and much more. The topic is crucial for both the CBSE board test and competitive exams. These NCERT Solutions for Class 12 Maths integrals are quite easy and can help students quickly grasp the problem-solving process. Students may access these NCERT Solutions for Class 12 Maths Chapter 7 and download them for free to practise offline as well. Access the latest Class 12 Maths Syllabus here.

Table of Content
1. Class 12 Maths Chapter 7 Integrals NCERT Solutions - Free PDF Download
2. Glance on Integration for Class 12 Chapter 7 Maths
3. Exercises Under NCERT Class 12 Maths Chapter 7 – Integrals
4. Access NCERT Solutions for Class 12 Mathematics  Chapter 7- Integrals
4.1Exercise 7.1
4.2Exercise 7.2
4.3Exercise 7.3
4.4Exercise 7.4
4.5Exercise 7.5
4.6Exercise 7.6
4.7Exercise 7.7
4.8Exercise 7.8
4.9Exercise 7.9
4.10Exercise 7.10
4.11Miscellaneous Exercise
5. NCERT Solutions for Class 12 Integration Maths
6. Overview of Deleted Syllabus for CBSE Class 12 Maths Chapter 7 Integrals
7. Class 12 Maths Chapter 7: Exercises Breakdown
8. CBSE Class 12 Maths Chapter 7 Other Study Materials
9. Chapter-Specific NCERT Solutions for Class 12 Maths
FAQs

## Glance on Integration for Class 12 Chapter 7 Maths

• Integration helps you find the exact area under that curve.

• This chapter covers various formulas and techniques to solve area problems.

• You can find questions based on the general area (indefinite integrals) and the exact area (definite integrals).

• Special methods are used to solve tricky integral problems like substitution or trigonometric identities.

• This article contains chapter notes and important questions for Chapter 7 Integration.

• There are exercise links provided. It has solutions for each question, from Integration as an Inverse Process of Differentiation, Methods of Integration, Integrals of Some Particular Functions, Definite Integrals, Application of Integrals, and Integration using Trigonometric Identities.

• There are ten exercises and miscellaneous exercise (240 fully solved questions) in class 12 maths Ch 7 Integrals.

 S.No. Current Syllabus Exercises of Class 12 Maths Chapter 7 1 Class 12 Maths NCERT Solutions of Integrals Exercise 7.1 2 3 Class 12 Maths NCERT Solutions of Integrals Exercise 7.3 4 5 6 7 8 9 Class 12 Maths NCERT Solutions of Integrals Exercise 7.9 10 Class 12 Maths NCERT Solutions of Integrals Exercise 7.10 11 Class 12 Maths NCERT Solutions of Integrals Miscellaneous Exercise
Competitive Exams after 12th Science
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Revision notes
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## Exercises Under NCERT Class 12 Maths Chapter 7 – Integrals

### Exercise 7.1

1. Find an antiderivative (or integral) of the following functions by the method of inspection. sin 2x

Ans:  We use the method of inspection as follows:

$\dfrac{d}{dx}\left( \cos 2x \right)=-2\sin 2x\Rightarrow -\dfrac{1}{2}\dfrac{d}{dx}\left( \cos 2x \right)$

$\therefore \sin 2x=\dfrac{d}{dx}\left( -\dfrac{1}{2}\cos 2x \right)$

Thus, the anti-derivative of sin $2x$is $-\dfrac{1}{2}\cos 2x$.

2. Find an antiderivative (or integral) of the following functions by the method of inspection. cos 3x

Ans: We use the method of inspection as follows:

$\dfrac{d}{dx}\left( \sin 3x \right)=3\cos 3x\Rightarrow \dfrac{1}{3}\dfrac{d}{dx}\left( \sin 3x \right)$

$\therefore \cos 3x=\dfrac{d}{dx}\left( \dfrac{1}{3}\sin 3x \right)$

Thus, the anti - derivative of cos $3x$is $\dfrac{1}{3}\sin 3x$.

3. Find an antiderivative (or integral) of the following functions by the method of inspection. $\mathbf{{{e}^{2x}}}$

Ans: We use the method of inspection as follows:

$\dfrac{d}{dx}\left( {{e}^{2x}} \right)\Rightarrow 2{{e}^{2x}}=\dfrac{1}{2}\dfrac{d}{dx}\left( {{e}^{2x}} \right)$

$\therefore {{e}^{2x}}=\dfrac{d}{dx}\left( \dfrac{1}{2}{{e}^{2x}} \right)$

Thus, the anti-derivative of ${{e}^{2x}}$is $\dfrac{1}{2}{{e}^{2x}}$.

4. Find an antiderivative (or integral) of the following functions by the method of inspection.$\mathbf{{{\left( ax+b \right)}^{2}}}$

Ans: We use the method of inspection as follows:

$\dfrac{d}{dx}{{\left( ax+b \right)}^{3}}=3a{{\left( ax+b \right)}^{2}}$

$\Rightarrow {{\left( ax+b \right)}^{2}}=\dfrac{1}{3a}\dfrac{d}{dx}{{\left( ax+b \right)}^{3}}$

$\therefore {{\left( ax+b \right)}^{2}}=\dfrac{d}{dx}\left( \dfrac{1}{3a}{{\left( ax+b \right)}^{3}} \right)$

Thus, the anti-derivative of ${{\left( ax+b \right)}^{2}}$ is $\dfrac{1}{3a}{{\left( ax+b \right)}^{3}}$.

5. Find an antiderivative (or integral) of the following functions by the method of inspection. sin $\mathbf{2x-4{{e}^{3x}}}$

Ans: We use the method of inspection as follows:

$\dfrac{d}{dx}\left( -\dfrac{1}{2}\cos 2x-\dfrac{4}{3}{{e}^{3x}} \right)=\left( \sin 2x-4{{e}^{3x}} \right)$

Thus, the anti-derivative of $\left( \sin 2x-4{{e}^{3x}} \right)$ is $\left( -\dfrac{1}{2}\cos 2x-\dfrac{4}{3}{{e}^{3x}} \right)$.

6. $\int{\left( 4{{e}^{3x}}+1 \right)dx}$

Ans:

$\int{\left( 4{{e}^{3x}}+1 \right)dx}$

$=4\int{{{e}^{3x}}dx}+\int{1}dx$

$=4\left( \dfrac{{{e}^{3x}}}{3} \right)+x+C$

$=\dfrac{4}{3}{{e}^{3x}}+x+C$

7. $\int{{{x}^{2}}\left( 1-\dfrac{1}{{{x}^{2}}} \right)dx}$

Ans: $\int{{{x}^{2}}\left( 1-\dfrac{1}{{{x}^{2}}} \right)dx}$

$=\int{\left( {{x}^{2}}-1 \right)dx}$

$=\dfrac{{{x}^{3}}}{3}-x+C$

8. $\int{\left( a{{x}^{2}}+bx+c \right)dx}$

Ans:$\int{\left( a{{x}^{2}}+bx+c \right)}dx$

$=a\int{{{x}^{2}}dx+b\int{xdx+c\int{1.dx}}}$

$=a\left( \dfrac{{{x}^{3}}}{3} \right)+b\left( \dfrac{{{x}^{2}}}{2} \right)+cx+C$

9. $\int{\left( 2{{x}^{2}}+{{e}^{x}} \right)dx}$

Ans:  $\int{\left( 2{{x}^{2}}+{{e}^{x}} \right)dx}$

$=2\int{{{x}^{2}}dx+\int{{{e}^{x}}dx}}$

$=2\left( \dfrac{{{x}^{3}}}{3} \right)+{{e}^{x}}+C$

$=\dfrac{2}{3}{{x}^{3}}+{{e}^{x}}+C$

10. $\int{{{\left( \sqrt{x}-\dfrac{1}{\sqrt{x}} \right)}^{2}}dx}$

Ans:$\int{{{\left( \sqrt{x}-\dfrac{1}{\sqrt{x}} \right)}^{2}}}dx$

$=\int{\left( x+\dfrac{1}{x}-2 \right)dx}$

$=\int{xdx}+\int{\dfrac{1}{x}dx}-2\int{1.dx}$

$=\dfrac{{{x}^{2}}}{2}+\log \left| x \right|-2x+C$

11. $\int{\dfrac{{{x}^{3}}+5{{x}^{2}}-4}{{{x}^{2}}}dx}$

Ans: $\int{\dfrac{{{x}^{3}}+5{{x}^{2}}-4}{{{x}^{2}}}dx}$

$\int{\left( x+5-4{{x}^{-2}} \right)dx}$

$=\int{xdx}+5\int{1.dx}-4\int{{{x}^{-2}}dx}$

$=\dfrac{{{x}^{2}}}{2}+5x+\dfrac{4}{x}+C$

12. $\int{\dfrac{{{x}^{3}}+3x+4}{\sqrt{x}}dx}$

Ans:$\int{\dfrac{{{x}^{3}}+3x+4}{\sqrt{x}}dx}$

$=\int{\left( {{x}^{\dfrac{5}{2}}}+3{{x}^{\dfrac{1}{2}}}+4{{x}^{-\dfrac{1}{2}}} \right)dx}$

$=\dfrac{{{x}^{\dfrac{7}{2}}}}{\dfrac{7}{2}}+\dfrac{3\left( {{x}^{\dfrac{3}{2}}} \right)}{\dfrac{3}{2}}+\dfrac{4\left( {{x}^{\dfrac{1}{2}}} \right)}{\dfrac{1}{2}}+C$

$=\dfrac{2}{7}{{x}^{\dfrac{7}{2}}}+2{{x}^{\dfrac{3}{2}}}+8\sqrt{x}+C$

13. $\int{\dfrac{{{x}^{3}}-{{x}^{2}}+x-1}{x-1}dx}$

Ans: $\int{\dfrac{{{x}^{3}}-{{x}^{2}}+x-1}{x-1}dx}$

We obtain, on dividing:

$=\int{\left( {{x}^{2}}+1 \right)dx}$

$=\int{{{x}^{2}}dx}+\int{1.dx}$

$=\dfrac{{{x}^{3}}}{3}+x+C$

14. $\int{\left( 1-x \right)}\sqrt{x}dx$

Ans: $\int{\left( 1-x \right)}\sqrt{x}dx$

$=\int{\left( \sqrt{x}-{{x}^{\dfrac{3}{2}}} \right)dx}$

$=\int{{{x}^{\dfrac{1}{2}}}dx-\int{{{x}^{\dfrac{3}{2}}}dx}}$

$=\dfrac{2}{3}{{x}^{\dfrac{3}{2}}}-\dfrac{2}{5}{{x}^{\dfrac{5}{2}}}+C$

15. $\int{\sqrt{x}\left( 3{{x}^{2}}+2x+3 \right)dx}$

Ans: $\int{\sqrt{x}\left( 3{{x}^{2}}+2x+3 \right)dx}$

$=3\int{\left( 2{{x}^{\dfrac{5}{2}}}+2{{x}^{\dfrac{3}{2}}}+3{{x}^{\dfrac{1}{2}}} \right)}$

$=3\int{{{x}^{\dfrac{5}{2}}}dx+2\int{{{x}^{\dfrac{3}{2}}}dx+3\int{{{x}^{\dfrac{1}{2}}}}dx}}$

$=\dfrac{6}{7}{{x}^{\dfrac{7}{2}}}+\dfrac{4}{5}{{x}^{\dfrac{5}{2}}}+2{{x}^{\dfrac{3}{2}}}+C$

16. $\int{\left( 2x-3\cos x+{{e}^{x}} \right)dx}$

Ans: $\int{\left( 2x-3\cos x+{{e}^{x}} \right)dx}$

$=2\int{xdx-3\int{\cos xdx}+\int{{{e}^{x}}dx}}$

$=\dfrac{2{{x}^{2}}}{2}-3\left( \sin x \right)+{{e}^{x}}+C$

$={{x}^{2}}-3\sin x+{{e}^{x}}+C$

17. $\int{\left( 2{{x}^{2}}-3\sin x+5\sqrt{x} \right)}dx$

Ans: $\int{\left( 2{{x}^{2}}-3\sin x+5\sqrt{x} \right)}dx$

=$2\int{{{x}^{2}}dx-3\int{\sin xdx+5\int{{{x}^{\dfrac{1}{2}}}}dx}}$

= $\dfrac{2{{x}^{3}}}{3}-3\left( -\cos x \right)+5\left(\dfrac{{{x}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right)+C$

=$\dfrac{2}{3}{{x}^{3}}+3\cos x+\dfrac{10}{3}{{x}^{\dfrac{3}{2}}}+C$

18. $\int{\sec x\left( \sec x+\tan x \right)dx}$

Ans: $\int{\sec x\left( \sec x+\tan x \right)dx}$

$=\int{\left( {{\sec }^{2}}x+\sec x\tan x \right)dx}$

$=\int{{{\sec }^{2}}xdx+\int{\sec x\tan xdx}}$

$=\tan x+\sec x+C$

19. $\int{\dfrac{{{\sec }^{2}}x}{\cos e{{c}^{2}}x}dx}$

Ans: $\int{\dfrac{{{\sec }^{2}}x}{\cos e{{c}^{2}}x}dx}$

$=\int{\dfrac{\dfrac{1}{{{\cos }^{2}}x}}{\dfrac{1}{{{\sin }^{2}}x}}dx}$

$=\int{\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}dx}$

$=\int{{{\tan }^{2}}xdx}$

$=\int{{{\sec }^{2}}xdx}-\int{1dx}$

$=\tan x-x+C$

20. $\int{\dfrac{2-3\sin x}{{{\cos }^{2}}x}dx}$

Ans: $\int{\dfrac{2-3\sin x}{{{\cos }^{2}}x}dx}$

$=\int{\left( \dfrac{2}{{{\cos }^{2}}x}-\dfrac{3\sin x}{{{\cos }^{2}}x} \right)dx}$

$=\int{2{{\sec }^{2}}xdx}-3\int{\tan x\sec xdx}$

$=2\tan x-3\sec x+C$

21. The anti – derivative of $\left( \sqrt{x}+\dfrac{1}{\sqrt{x}} \right)$ equals

1. $\dfrac{1}{3}{{x}^{\dfrac{1}{3}}}+2{{x}^{\dfrac{1}{2}}}+C$
2. $\dfrac{2}{3}{{x}^{\dfrac{2}{3}}}+\dfrac{1}{2}{{x}^{2}}+C$
3. $\dfrac{2}{3}{{x}^{\dfrac{3}{2}}}+2{{x}^{\dfrac{1}{2}}}+C$
4. $\dfrac{3}{2}{{x}^{\dfrac{3}{2}}}+\dfrac{1}{2}{{x}^{\dfrac{1}{2}}}+C$

Ans:

$\left( \sqrt{x}+\dfrac{1}{\sqrt{x}} \right)$

$=\int{{{x}^{\dfrac{1}{2}}}dx}+\int{{{x}^{-\dfrac{1}{2}}}}dx=\dfrac{{{x}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}+\dfrac{{{x}^{\dfrac{1}{2}}}}{\dfrac{1}{2}}+C$

$=\dfrac{2}{3}{{x}^{\dfrac{3}{2}}}+2{{x}^{\dfrac{1}{2}}}+C$

Thus, the correct answer is C.

22. If $\dfrac{d}{dx}f\left( x \right)=4{{x}^{3}}-\dfrac{3}{{{x}^{4}}}$ such that $f\left( 2 \right)=0$ then $f\left( x \right)$ is

1. ${{x}^{4}}+\dfrac{1}{{{x}^{3}}}-\dfrac{129}{8}$

2. ${{x}^{3}}+\dfrac{1}{{{x}^{4}}}+\dfrac{129}{8}$

3. ${{x}^{4}}+\dfrac{1}{{{x}^{3}}}+\dfrac{129}{8}$

4. ${{x}^{3}}+\dfrac{1}{{{x}^{4}}}-\dfrac{129}{8}$

Ans: Given, $\dfrac{d}{dx}f\left( x \right)=4{{x}^{3}}-\dfrac{3}{{{x}^{4}}}$

Anti-derivative of $4{{x}^{3}}-\dfrac{3}{{{x}^{4}}}=f\left( x \right)$

$\therefore f\left( x \right)=\int{4{{x}^{3}}-\dfrac{3}{{{x}^{4}}}=f\left( x \right)}$

$f\left( x \right)=4\int{{{x}^{3}}dx-3\int{\left( {{x}^{-4}} \right)}dx}$

$f\left( x \right)=4\left( \dfrac{{{x}^{4}}}{4} \right)-3\left( \dfrac{{{x}^{-3}}}{-3} \right)+C$

$f\left( x \right)={{x}^{4}}+\dfrac{1}{{{x}^{3}}}+C$

Also,

$f\left( 2 \right)=0$

$\therefore f\left( 2 \right)={{\left( 2 \right)}^{4}}+\dfrac{1}{{{\left( 2 \right)}^{3}}}+C=0$

$\Rightarrow 16+\dfrac{1}{8}+C=0$

$\Rightarrow C=\dfrac{-129}{8}$

$\therefore f\left( x \right)={{x}^{4}}+\dfrac{1}{{{x}^{3}}}-\dfrac{129}{8}$

Thus, the correct answer is A.

### Exercise 7.2

1. $\dfrac{2x}{1+{{x}^{2}}}$

Ans: Substitute $1+{{x}^{2}}=t$

$\therefore 2xdx=dt$

$\Rightarrow \int{\dfrac{2x}{1+{{x}^{2}}}dx}=\int{\dfrac{1}{t}}dt$

$=\log \left| t \right|+C$

$=\log \left| 1+{{x}^{2}} \right|+C$

$=\log \left( 1+{{x}^{2}} \right)+C$

2. $\dfrac{{{\left( \log x \right)}^{2}}}{x}$

Ans: Substitute $\log \left| x \right|=t$

$\therefore \dfrac{1}{x}dx=dt$

$\Rightarrow \int{\dfrac{{{\left( \log \left| x \right| \right)}^{2}}}{x}}dx=\int{{{t}^{2}}dt}$

$=\dfrac{{{t}^{3}}}{3}+C$

$=\dfrac{{{\left( \log \left| x \right| \right)}^{3}}}{3}+C$

3. ${\dfrac{1}{x+x\log x}}$

Ans: $\dfrac{1}{x+x\log x}=\dfrac{1}{x\left( 1+\log x \right)}$

Substitute $1+\log x=t$

$\therefore \dfrac{1}{x}dx=dt$

$\Rightarrow \int{\dfrac{1}{x\left( 1+\log x \right)}dx}=\int{\dfrac{1}{t}dt}$

$=\log \left| t \right|+C$

$=\log \left| 1+\log x \right|+C$

4. ${Sinx.\sin \left( \cos x \right)}$

Ans: $\operatorname{Sin}x.\sin \left( \cos x \right)$

$Put,\cos x=t$

$\therefore -\sin xdx=dt$

$\Rightarrow \int{\sin x.\sin \left( \cos x \right)dx}=-\int{\sin tdt}$

$=-\left[ -\cos t \right]+C$

$=\cos t+C$

$=\cos \left( \cos x \right)+C$

5. $\operatorname{sin}\left( ax+b \right)\cos \left( ax+b \right)$

Ans: $\operatorname{sin}\left( ax+b \right)\cos \left( ax+b \right)=\dfrac{2\sin \left( ax+b \right)\cos \left( ax+b \right)}{2}=\dfrac{\sin 2\left( ax+b \right)}{2}$

Substitute $2\left( ax+b \right)=t$

$\therefore 2adx=dt$

$\Rightarrow \int{\dfrac{\sin 2\left( ax+b \right)}{2}}dx=\dfrac{1}{2}\int{\dfrac{\sin t}{2a}}dt$

$=\dfrac{1}{4a}\left[ -\cos t \right]+C$

$=\dfrac{-1}{4a}\cos 2\left( ax+b \right)+C$

6. $\sqrt{ax+b}$

Ans: Substitute $ax+b=t$

$\Rightarrow adx=dt$

$\therefore dx=\dfrac{1}{a}dt$

$\Rightarrow \int{{{\left( ax+b \right)}^{\dfrac{1}{2}}}dx}=\dfrac{1}{a}\int{{{t}^{\dfrac{1}{2}}}}dt$

$=\dfrac{1}{a}\left( \dfrac{{{t}^{\dfrac{1}{2}}}}{\dfrac{3}{2}} \right)+C=\dfrac{2}{3a}{{\left( ax+b \right)}^{\dfrac{3}{2}}}+C$

7. $x\sqrt{x+2}$

Ans: Substitute $x+2=t$

$\therefore dx=dt$

$\Rightarrow \int{x\sqrt{x+2}}=\int{\left( t-2 \right)\sqrt{t}}dt$

$=\int{\left( {{t}^{\dfrac{3}{2}}}-2{{t}^{\dfrac{1}{2}}} \right)}dt$

$=\int{{{t}^{\dfrac{3}{2}}}dt-2\int{{{t}^{\dfrac{1}{2}}}}dt}$

$=\dfrac{2}{5}{{t}^{\dfrac{5}{2}}}-\dfrac{4}{3}{{t}^{\dfrac{3}{2}}}+C$

8. $x\sqrt{1+2{{x}^{2}}}$

Ans: Substitute $1+2{{x}^{2}}=t$

$\therefore 4xdx=dt$

$\Rightarrow \int{x\sqrt{1+2{{x}^{2}}}dx}=\int{\dfrac{\sqrt{t}dt}{4}}$

$=\dfrac{1}{4}\int{{{t}^{\dfrac{1}{2}}}}dt$

$=\dfrac{1}{4}\left( \dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right)+C$

$=\dfrac{1}{6}{{\left( 1+2{{x}^{2}} \right)}^{\dfrac{3}{2}}}+C$

9. $\left( 4x+2 \right)\sqrt{{{x}^{2}}+x+1}$

Ans: Substitute ${{x}^{2}}+x+1=t$

$\therefore \left( 2x+1 \right)dx=dt$

$\int{\left( 4x+2 \right)\sqrt{{{x}^{2}}+x+1}}dx$

$=\int{2\sqrt{t}dt}$

$=2\int{\sqrt{t}}dt$

$=2\left( \dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right)+C=\dfrac{4}{3}{{\left( {{x}^{2}}+x+1 \right)}^{\dfrac{3}{2}}}+C$

10. $\dfrac{1}{x-\sqrt{x}}$

Ans: $\dfrac{1}{x-\sqrt{x}}=\dfrac{1}{\sqrt{x}\left( \sqrt{x}-1 \right)}$

Substitute $\left( \sqrt{x-1} \right)=t$

$\Rightarrow \int{\dfrac{1}{\sqrt{x}\left( \sqrt{x}-1 \right)}dx=\int{\dfrac{2}{t}dt}}$

$=2\log \left| t \right|+C$

$=2\log \left| \sqrt{x}-1 \right|+C$

11. $\dfrac{x}{\sqrt{x+4}},x>0$

Ans: Substitute $x+4=t$

$\therefore dx=dt$

$\int{\dfrac{x}{\sqrt{x+4}}dx}=\int{\dfrac{\left( t-4 \right)}{\sqrt{t}}dt=\int{\left( \sqrt{t}-\dfrac{4}{\sqrt{t}} \right)}}dt$

$=\dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}-4\left( \dfrac{{{t}^{\dfrac{1}{2}}}}{\dfrac{1}{2}} \right)+C=\dfrac{2}{3}{{\left( t \right)}^{\dfrac{3}{2}}}-8{{\left( t \right)}^{\dfrac{1}{2}}}+C$

$=\dfrac{2}{3}t.{{t}^{\dfrac{1}{2}}}-8{{t}^{\dfrac{1}{2}}}+C$

$=\dfrac{2}{3}{{t}^{\dfrac{1}{2}}}\left( t-12 \right)+C$

$=\dfrac{2}{3}\sqrt{x+4}\left( x-8 \right)+C$

12. ${{\left( {{x}^{3}}-1 \right)}^{\dfrac{1}{3}}}{{x}^{5}}$

Ans: Substitute ${{x}^{3}}-1=t$

$\therefore 3{{x}^{2}}dx=dt$

$\Rightarrow \int{{{\left( {{x}^{3}}-1 \right)}^{\dfrac{1}{3}}}{{x}^{5}}dx=\int{{{\left( {{x}^{3}}-1 \right)}^{\dfrac{1}{3}}}{{x}^{3}}.{{x}^{2}}dx}}$

$=\int{{{t}^{\dfrac{1}{3}}}\left( t+1 \right)\dfrac{dt}{3}=\dfrac{1}{3}\int{\left( {{t}^{\dfrac{4}{3}}}+{{t}^{\dfrac{1}{3}}} \right)dt}}$

$=\dfrac{1}{3}\left[ \dfrac{3}{7}{{t}^{\dfrac{7}{3}}}+\dfrac{3}{4}{{t}^{\dfrac{4}{3}}} \right]+C$

$=\dfrac{1}{7}{{\left( {{x}^{3}}-1 \right)}^{\dfrac{7}{3}}}+\dfrac{1}{4}{{\left( {{x}^{3}}-1 \right)}^{\dfrac{4}{3}}}+C$

13. $\dfrac{{{x}^{2}}}{{{\left( 2+3{{x}^{3}} \right)}^{3}}}$

Ans: Substitute $2+3{{x}^{3}}=t$

$\therefore 9{{x}^{2}}dx=dt$

$\Rightarrow \int{\dfrac{{{x}^{2}}}{{{\left( 2+3{{x}^{3}} \right)}^{3}}}dx}=\dfrac{1}{9}\int{\dfrac{dt}{{{\left( t \right)}^{3}}}}$

$=\dfrac{1}{9}\left[ \dfrac{{{t}^{-2}}}{-2} \right]+C$

$=\dfrac{-1}{18{{\left( 2+3{{x}^{3}} \right)}^{2}}}+C$

14. $\dfrac{1}{x{{\left( \log x \right)}^{m}}},x>0$

Ans: Substitute $\log x=t$

$\dfrac{1}{x}dx=dt$

$\Rightarrow \int{\dfrac{1}{x{{\left( \log x \right)}^{m}}}dx}=\int{\dfrac{dt}{{{\left( t \right)}^{m}}}=\left( \dfrac{{{t}^{-m-1}}}{1-m} \right)}+C$

$=\dfrac{{{\left( \log x \right)}^{1-m}}}{\left( 1-m \right)}+C$

15. $\dfrac{x}{9-4{{x}^{2}}}$

Ans: Substitute $9-4{{x}^{2}}=t$

$\therefore -8xdx=dt$

$\Rightarrow \int{\dfrac{x}{9-4{{x}^{2}}}dx=\dfrac{-1}{8}\int{\dfrac{1}{t}dt}}$

$=\dfrac{-1}{8}\log \left| t \right|+C$

$=\dfrac{-1}{8}\log \left| 9-4{{x}^{2}} \right|+C$

16. ${{e}^{2x+3}}$

Ans: Substitute $2x+3=t$

$\therefore 2dx=dt$

$\Rightarrow \int{{{e}^{2x+3}}dx}=\dfrac{1}{2}\int{{{e}^{t}}dt}$

$=\dfrac{1}{2}\left( {{e}^{t}} \right)+C$

$=\dfrac{1}{2}{{e}^{\left( 2x+3 \right)}}+C$

17. $\dfrac{x}{{{e}^{{{x}^{2}}}}}$

Ans: Substitute ${{x}^{2}}=t$

$\therefore 2xdx=dt$

$\Rightarrow \int{\dfrac{x}{{{e}^{{{x}^{2}}}}}dx=\dfrac{1}{2}\int{\dfrac{1}{{{e}^{t}}}dt}}$

$=\dfrac{1}{2}\int{{{e}^{-t}}dt}$

$=\dfrac{1}{2}\left( \dfrac{{{e}^{-t}}}{-1} \right)+C$

$=-\dfrac{1}{2}{{e}^{-{{x}^{2}}}}+C$

$=\dfrac{-1}{2{{e}^{{{x}^{2}}}}}+C$

18. $\dfrac{{{e}^{{{\tan }^{-1}}x}}}{1+{{x}^{2}}}$

Ans: Substitute ${{\tan }^{-1}}x=t$

$\therefore \dfrac{1}{1+{{x}^{2}}}dx=dt$

$\Rightarrow \int{\dfrac{{{e}^{{{\tan }^{-1}}x}}}{1+{{x}^{2}}}dx=dt}$

$={{e}^{t}}+C$

$={{e}^{{{\tan }^{-1}}x}}+C$

19. Integrate $\dfrac{{{e}^{2x}}-1}{{{e}^{2x}}+1}$

Ans:

Dividing the given function’s numerator and denominator by  ${{e}^{x}}$ , we obtain,

$\dfrac{\dfrac{\left( {{e}^{2x}}-1 \right)}{{{e}^{x}}}}{\dfrac{\left( {{e}^{2x}}+1 \right)}{{{e}^{x}}}}=\dfrac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}}$

Let ${{e}^{x}}+{{e}^{-x}}=t$

$\left( {{e}^{x}}-{{e}^{-x}} \right)dx=dt$

$\Rightarrow \int{\dfrac{{{e}^{2x}}-1}{{{e}^{2x}}+1}}dx=\int{\dfrac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}}}dx$

$=\int{\dfrac{dt}{t}}$

$=\log |t|+C$

$=\log |{{e}^{x}}-{{e}^{-x}}|+C$

where C is an arbitrary constant.

20. Solve the following: $\dfrac{{{e}^{2x}}-{{e}^{-2x}}}{{{e}^{2x}}+{{e}^{-2x}}}$.

Ans: Given expression $\dfrac{{{e}^{2x}}-{{e}^{-2x}}}{{{e}^{2x}}+{{e}^{-2x}}}$.

Let us substitute ${{e}^{2x}}+{{e}^{-2x}}=t$, we get

$\left( 2{{e}^{2x}}+2{{e}^{-2x}} \right)dx=dt$

$\Rightarrow 2\left( {{e}^{2x}}-{{e}^{-2x}} \right)dx=dt$

Integration of given expression is

$\Rightarrow \int{\dfrac{{{e}^{2x}}-{{e}^{-2x}}}{{{e}^{2x}}+{{e}^{-2x}}}}=\int{\dfrac{dt}{2t}}$

$\Rightarrow \int{\dfrac{{{e}^{2x}}-{{e}^{-2x}}}{{{e}^{2x}}+{{e}^{-2x}}}}=\dfrac{1}{2}\int{\dfrac{1}{t}dt}$

$\Rightarrow \int{\dfrac{{{e}^{2x}}-{{e}^{-2x}}}{{{e}^{2x}}+{{e}^{-2x}}}}=\dfrac{1}{2}\log \left| t \right|+C$

Again substitute $t={{e}^{2x}}+{{e}^{-2x}}$, we get

$\therefore\int{\dfrac{{{e}^{2x}}-{{e}^{-2x}}}{{{e}^{2x}}+{{e}^{-2x}}}}=\dfrac{1}{2}\log \left| {{e}^{2x}}+{{e}^{-2x}} \right|+C$

21.  Solve the following: ${{\tan }^{2}}\left( 2x-3 \right)$.

Ans: Given expression ${{\tan }^{2}}\left( 2x-3 \right)$.

We can apply the identity ${{\tan }^{2}}x={{\sec }^{2}}x-1$, we get

${{\tan }^{2}}\left( 2x-3 \right)={{\sec }^{2}}\left( 2x-3 \right)-1$

Substitute $2x-3=t$, we get

$2dx=dt$

Integration of given expression is

$\Rightarrow \int{{{\tan }^{2}}\left( 2x-3 \right)dx}=\int{{{\sec }^{2}}\left( 2x-3 \right)-1dx}$

$\Rightarrow \int{{{\tan }^{2}}\left( 2x-3 \right)dx}=\dfrac{1}{2}\int{{{\sec }^{2}}tdt-\int{1dx}}$

$\Rightarrow \int{{{\tan }^{2}}\left( 2x-3 \right)dx}=\dfrac{1}{2}\tan t-x+C$

Substitute $2x-3=t$

$\therefore \int{{{\tan }^{2}}\left( 2x-3 \right)dx}=\dfrac{1}{2}\tan \left( 2x-3 \right)-x+C$

22.  Solve the following: ${{\sec }^{2}}\left( 7-4x \right)$.

Ans: Given expression ${{\sec }^{2}}\left( 7-4x \right)$.

Put $7-4x=t$, we get

$\therefore -4dx=dt$

Integration of given expression is

$\Rightarrow \int{{{\sec }^{2}}\left( 7-4x \right)dx=-\dfrac{1}{4}\int{{{\sec }^{2}}tdt}}$

$\Rightarrow \int{{{\sec }^{2}}\left( 7-4x \right)dx=-\dfrac{1}{4}\tan t+C}$

Substitute $7-4x=t$, we get

$\therefore \int{{{\sec }^{2}}\left( 7-4x \right)dx=-\dfrac{1}{4}\tan \left( 7-4x \right)+C}$

23.  Solve the following: $\dfrac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}$.

Ans: Given expression $\dfrac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}$.

Put ${{\sin }^{-1}}x=t$, we get

$\Rightarrow \dfrac{1}{\sqrt{1-{{x}^{2}}}}dx=dt$

Integration of given expression is

$\Rightarrow \int{\dfrac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}dx=\int{tdt}}$

$\Rightarrow \int{\dfrac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}dx=\dfrac{{{t}^{2}}}{2}+C}$

Substitute ${{\sin }^{-1}}x=t$, we get

$\therefore \int{\dfrac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}dx=\dfrac{{{\left( {{\sin }^{-1}}x \right)}^{2}}}{2}+C}$

24.  Solve the following: $\dfrac{2\cos x-3\sin x}{6\cos x+4\sin x}$.

Ans: Given expression is $\dfrac{2\cos x-3\sin x}{6\cos x+4\sin x}$.

Given expression can be written as

$\dfrac{2\cos x-3\sin x}{6\cos x+4\sin x}=\dfrac{2\cos x-3\sin x}{2\left( 3\cos x+2\sin x \right)}$

Let $3\cos x+2\sin x=t$, we get

$\left( -3\sin x+2\cos x \right)dx=dt$

Integration of given expression is

$\Rightarrow \int{\dfrac{2\cos x-3\sin x}{6\cos x+4\sin x}}dx=\int{\dfrac{2\cos x-3\sin x}{2\left( 3\cos x+2\sin x \right)}}dx$

$\Rightarrow \int{\dfrac{2\cos x-3\sin x}{6\cos x+4\sin x}}dx=\int{\dfrac{dt}{2t}}$

$\Rightarrow \int{\dfrac{2\cos x-3\sin x}{6\cos x+4\sin x}}dx=\dfrac{1}{2}\int{\dfrac{1}{t}dt}$

$\Rightarrow \int{\dfrac{2\cos x-3\sin x}{6\cos x+4\sin x}}dx=\dfrac{1}{2}\log \left| t \right|+C$

Substitute $3\cos x+2\sin x=t$

$\therefore \int{\dfrac{2\cos x-3\sin x}{6\cos x+4\sin x}}dx=\dfrac{1}{2}\log \left| 2\sin x+3\cos x \right|+C$

25. Solve the following: $\dfrac{1}{{{\cos }^{2}}x{{\left( 1-\tan x \right)}^{2}}}$.

Ans: Given expression $\dfrac{1}{{{\cos }^{2}}x{{\left( 1-\tan x \right)}^{2}}}$.

Given expression can be written as

$\dfrac{1}{{{\cos }^{2}}x{{\left( 1-\tan x \right)}^{2}}}=\dfrac{{{\sec }^{2}}x}{{{\left( 1-\tan x \right)}^{2}}}$

Let $\left( 1-\tan x \right)=t$, we get

$-{{\sec }^{2}}xdx=dt$

Integration of given expression is

$\Rightarrow \int{\dfrac{1}{{{\cos }^{2}}x{{\left( 1-\tan x \right)}^{2}}}}dx=\int{\dfrac{{{\sec }^{2}}x}{{{\left( 1-\tan x \right)}^{2}}}}dx$

$\Rightarrow \int{\dfrac{1}{{{\cos }^{2}}x{{\left( 1-\tan x \right)}^{2}}}}dx=\int{\dfrac{-dt}{{{t}^{2}}}}$

$\Rightarrow \int{\dfrac{1}{{{\cos }^{2}}x{{\left( 1-\tan x \right)}^{2}}}}dx=-\int{{{t}^{-2}}dt}$

$\Rightarrow \int{\dfrac{1}{{{\cos }^{2}}x{{\left( 1-\tan x \right)}^{2}}}}dx=\dfrac{1}{t}+C$

Substitute $\left( 1-\tan x \right)=t$,

$\therefore \int{\dfrac{1}{{{\cos }^{2}}x{{\left( 1-\tan x \right)}^{2}}}}dx=\dfrac{1}{\left( 1-\tan x \right)}+C$

26. Solve the following: $\dfrac{\cos \sqrt{x}}{\sqrt{x}}$.

Ans: Given expression is $\dfrac{\cos \sqrt{x}}{\sqrt{x}}$.

Let $\sqrt{x}=t$, we get

$\dfrac{1}{2\sqrt{x}}dx=dt$

Integration of given expression is

$\Rightarrow \int{\dfrac{\cos \sqrt{x}}{\sqrt{x}}dx}=2\int{\cos tdt}$

$\Rightarrow \int{\dfrac{\cos \sqrt{x}}{\sqrt{x}}dx}=2\sin t+C$

Substitute $\sqrt{x}=t$

$\therefore \int{\dfrac{\cos \sqrt{x}}{\sqrt{x}}dx}=2\sin \sqrt{x}+C$

27. Solve the following: $\sqrt{\sin 2x}\cos 2x$.

Ans: Given expression is $\sqrt{\sin 2x}\cos 2x$.

Let $\sin 2x=t$, we get

$2\cos 2xdx=dt$

Integration of given expression is

$\Rightarrow \int{\sqrt{\sin 2x}\cos 2xdx=\dfrac{1}{2}\int{\sqrt{t}dt}}$

$\Rightarrow \int{\sqrt{\sin 2x}\cos 2xdx=\dfrac{1}{2}\left( \dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right)+C}$

$\Rightarrow \int{\sqrt{\sin 2x}\cos 2xdx=\dfrac{1}{3}{{t}^{\dfrac{3}{2}}}+C}$

Substitute $\sin 2x=t$

$\therefore \int{\sqrt{\sin 2x}\cos 2xdx=\dfrac{1}{3}{{\left( \sin 2x \right)}^{\dfrac{3}{2}}}+C}$

28. Solve the following: $\dfrac{\cos x}{\sqrt{1+\sin x}}$.

Ans: Given expression $\dfrac{\cos x}{\sqrt{1+\sin x}}$.

Let $1+\sin x=t$

$\therefore \cos xdx=dt$

Integration of given expression is

$\Rightarrow \int{\dfrac{\cos x}{\sqrt{1+\sin x}}dx}=\int{\dfrac{dt}{\sqrt{t}}}$

$\Rightarrow \int{\dfrac{\cos x}{\sqrt{1+\sin x}}dx}=\dfrac{{{t}^{\dfrac{1}{2}}}}{\dfrac{1}{2}}+C$

$\Rightarrow \int{\dfrac{\cos x}{\sqrt{1+\sin x}}dx}=2\sqrt{t}+C$

Substitute $1+\sin x=t$,

$\therefore \int{\dfrac{\cos x}{\sqrt{1+\sin x}}dx}=2\sqrt{1+\sin x}+C$

29.  Solve the following: $\cot x\log \sin x$.

Ans: Given expression $\cot x\log \sin x$.

Let $\log \sin x=t$, we get

$\dfrac{1}{\sin x}\cos xdx=dt$

$\Rightarrow \cot xdx=dt$

Integration of given expression is

$\int{\cot x\log \sin xdx=\int{tdt}}$

$\Rightarrow \int{\cot x\log \sin xdx=\dfrac{{{t}^{2}}}{2}+C}$

Substitute $\log \sin x=t$,

$\therefore \int{\cot x\log \sin xdx=\dfrac{1}{2}{{\left( \log \sin x \right)}^{2}}+C}$

30.  Solve the following: $\dfrac{\sin x}{1+\cos x}$.

Ans: Given expression $\dfrac{\sin x}{1+\cos x}$.

Let $1+\cos x=t$

$\therefore -\sin xdx=dt$

Integration of given expression is

$\Rightarrow \int{\dfrac{\sin x}{1+\cos x}dx=\int{-\dfrac{dt}{t}}}$

$\Rightarrow \int{\dfrac{\sin x}{1+\cos x}dx=-\log \left| t \right|+C}$

Substitute $1+\cos x=t$,

$\therefore \int{\dfrac{\sin x}{1+\cos x}dx=-\log \left| 1+\cos x \right|+C}$

31.  Solve the following: $\dfrac{\sin x}{{{\left( 1+\cos x \right)}^{2}}}$.

Ans: Given expression $\dfrac{\sin x}{{{\left( 1+\cos x \right)}^{2}}}$.

Let $1+\cos x=t$

$\therefore -\sin xdx=dt$

Integration of given expression is

$\Rightarrow \int{\dfrac{\sin x}{{{\left( 1+\cos x \right)}^{2}}}dx=\int{-\dfrac{dt}{{{t}^{2}}}}}$

$\Rightarrow \int{\dfrac{\sin x}{{{\left( 1+\cos x \right)}^{2}}}dx=-\int{{{t}^{-2}}dt}}$

$\Rightarrow \int{\dfrac{\sin x}{{{\left( 1+\cos x \right)}^{2}}}dx=\dfrac{1}{t}+C}$

Substitute $1+\cos x=t$,

$\therefore \int{\dfrac{\sin x}{{{\left( 1+\cos x \right)}^{2}}}dx=\dfrac{1}{1+\cos x}+C}$

.

32.Solve the following: $\dfrac{1}{1+\cot x}$.

Ans: Given expression $\dfrac{1}{1+\cot x}$.

Let  $I=\int{\dfrac{1}{1+\cot x}}dx$

Integration of given expression is

$\Rightarrow I=\int{\dfrac{1}{1+\cot x}}dx$

$\Rightarrow I=\int{\dfrac{1}{1+\dfrac{\cos x}{\sin x}}}dx$

$\Rightarrow I=\int{\dfrac{\sin x}{\sin x+\cos x}}dx$

$\Rightarrow I=\dfrac{1}{2}\int{\dfrac{2\sin x}{\sin x+\cos x}}dx$

$\Rightarrow I=\dfrac{1}{2}\int{\dfrac{\left( \sin x+\cos x \right)+\left( \sin x-\cos x \right)}{\sin x+\cos x}}dx$

$\Rightarrow I=\dfrac{1}{2}\int{\dfrac{\left( \sin x+\cos x \right)}{\sin x+\cos x}+\dfrac{\left( \sin x-\cos x \right)}{\sin x+\cos x}}dx$

$\Rightarrow I=\dfrac{1}{2}\int{1dx+\dfrac{1}{2}\int{\dfrac{\left( \sin x-\cos x \right)}{\sin x+\cos x}}}dx$

Let $\sin x+\cos x=t$

$\therefore \left( \cos x-\sin x \right)dx=dt$

Substitute in above obtained equation, we get

$\Rightarrow I=\dfrac{1}{2}x+\dfrac{1}{2}\int{-\dfrac{dt}{t}}$

$\Rightarrow I=\dfrac{x}{2}-\dfrac{1}{2}\log \left| t \right|+C$

Substitute $\sin x+\cos x=t$,

$\therefore I=\dfrac{x}{2}-\dfrac{1}{2}\log \left| \sin x+\cos x \right|+C$

33.  Solve the following: $\dfrac{1}{1-\tan x}$.

Ans: Given expression $\dfrac{1}{1-\tan x}$.

Let  $I=\int{\dfrac{1}{1-\tan x}}dx$

Integration of given expression is

$\Rightarrow I=\int{\dfrac{1}{1-\tan x}}dx$

$\Rightarrow I=\int{\dfrac{1}{1-\dfrac{\sin x}{\cos x}}}dx$

$\Rightarrow I=\int{\dfrac{\cos x}{\cos x-\sin x}}dx$

$\Rightarrow I=\dfrac{1}{2}\int{\dfrac{2\cos x}{\cos x-\sin x}}dx$

$\Rightarrow I=\dfrac{1}{2}\int{\dfrac{\left( \cos x-\sin x \right)+\left( \cos x+\sin x \right)}{\cos x-\sin x}}dx$

$\Rightarrow I=\dfrac{1}{2}\int{\dfrac{\left( \cos x-\sin x \right)}{\cos x-\sin x}+\dfrac{\left( \cos x+\sin x \right)}{\cos x-\sin x}}dx$

$\Rightarrow I=\dfrac{1}{2}\int{1dx+\dfrac{1}{2}\int{\dfrac{\left( \cos x+\sin x \right)}{\cos x-\sin x}}}dx$

Let $\cos x-\sin x=t$

$\therefore \left( -\sin x-\cos x \right)dx=dt$

Substitute in above obtained equation, we get

$\Rightarrow I=\dfrac{1}{2}x+\dfrac{1}{2}\int{-\dfrac{dt}{t}}$

$\Rightarrow I=\dfrac{x}{2}-\dfrac{1}{2}\log \left| t \right|+C$

Substitute $\cos x-\sin x=t$,

$\therefore I=\dfrac{x}{2}-\dfrac{1}{2}\log \left| \cos x-\sin x \right|+C$

34. Solve the following: $\dfrac{\sqrt{\tan x}}{\sin x\cos x}$.

Ans: Given expression $\dfrac{\sqrt{\tan x}}{\sin x\cos x}$.

Let $I=\int{\dfrac{\sqrt{\tan x}}{\sin x\cos x}dx}$

Multiply and divide by $\cos x$, we get

$\Rightarrow I=\int{\dfrac{\sqrt{\tan x}\times \cos x}{\sin x\cos x\times \cos x}dx}$

$\Rightarrow I=\int{\dfrac{\sqrt{\tan x}\times \cos x}{\tan x\times {{\cos }^{2}}x}dx}$

$\Rightarrow I=\int{\dfrac{{{\sec }^{2}}x}{\sqrt{\tan x}}dx}$

Let $\tan x=t$

$\therefore {{\sec }^{2}}xdx=dt$

Substitute in above obtained equation, we get

$\Rightarrow I=\int{\dfrac{dt}{\sqrt{t}}dx}$

$\Rightarrow I=2\sqrt{t}+C$

Substitute $\tan x=t$,

$\therefore I=2\sqrt{\tan x}+C$

35.  Solve the following: $\dfrac{{{\left( 1+\log x \right)}^{2}}}{x}$.

Ans: Given expression $\dfrac{{{\left( 1+\log x \right)}^{2}}}{x}$.

Let $1+\log x=t$

$\therefore \dfrac{1}{x}dx=dt$

Integration of given expression is

$\int{\dfrac{{{\left( 1+\log x \right)}^{2}}}{x}dx=\int{{{t}^{2}}dt}}$

$\Rightarrow \int{\dfrac{{{\left( 1+\log x \right)}^{2}}}{x}dx=\dfrac{{{t}^{3}}}{3}+C}$

Substitute $1+\log x=t$

$\therefore \int{\dfrac{{{\left( 1+\log x \right)}^{2}}}{x}dx=\dfrac{{{\left( 1+\log x \right)}^{3}}}{3}+C}$

36. Solve the following: $\dfrac{\left( x+1 \right){{\left( x+\log x \right)}^{2}}}{x}$.

Ans: Given expression $\dfrac{\left( x+1 \right){{\left( x+\log x \right)}^{2}}}{x}$.

Given expression can be written as

$\dfrac{\left( x+1 \right){{\left( x+\log x \right)}^{2}}}{x}=\left( \dfrac{x+1}{x} \right){{\left( x+\log x \right)}^{2}}$

$\Rightarrow \dfrac{\left( x+1 \right){{\left( x+\log x \right)}^{2}}}{x}=\left( 1+\dfrac{1}{x} \right){{\left( x+\log x \right)}^{2}}$

Let $x+\log x=t$

$\therefore \left( 1+\dfrac{1}{x} \right)dx=dt$

Integration of given expression is

$\Rightarrow \int{\dfrac{\left( x+1 \right){{\left( x+\log x \right)}^{2}}}{x}dx=\int{{{t}^{2}}dt}}$

$\Rightarrow \int{\dfrac{\left( x+1 \right){{\left( x+\log x \right)}^{2}}}{x}dx=\dfrac{{{t}^{3}}}{3}+C}$

Substitute $x+\log x=t$

$\therefore \int{\dfrac{\left( x+1 \right){{\left( x+\log x \right)}^{2}}}{x}dx=\dfrac{1}{3}{{\left( x+\log x \right)}^{3}}+C}$

37. Solve the following: $\dfrac{{{x}^{3}}\sin \left( {{\tan }^{-1}}{{x}^{4}} \right)}{1+{{x}^{8}}}$.

Ans: Given expression $\dfrac{{{x}^{3}}\sin \left( {{\tan }^{-1}}{{x}^{4}} \right)}{1+{{x}^{8}}}$.

Let ${{x}^{4}}=t$,

$\therefore 4{{x}^{3}}dx=dt$

Integration of given expression is

$\Rightarrow \int{\dfrac{{{x}^{3}}\sin \left( {{\tan }^{-1}}{{x}^{4}} \right)}{1+{{x}^{8}}}dx=\dfrac{1}{4}\int{\dfrac{\sin \left( {{\tan }^{-1}}t \right)}{1+{{t}^{2}}}dt}}$ ………(1)

Let ${{\tan }^{-1}}t=u$

$\therefore \dfrac{1}{1+{{t}^{2}}}dt=du$

Substitute in eq. (1), we get

$\Rightarrow \int{\dfrac{{{x}^{3}}\sin \left( {{\tan }^{-1}}{{x}^{4}} \right)}{1+{{x}^{8}}}dx=\dfrac{1}{4}\int{\sin udu}}$

$\Rightarrow \int{\dfrac{{{x}^{3}}\sin \left( {{\tan }^{-1}}{{x}^{4}} \right)}{1+{{x}^{8}}}dx=\dfrac{1}{4}\left( -\cos u \right)+C}$

Substitute ${{\tan }^{-1}}t=u$,

$\Rightarrow \int{\dfrac{{{x}^{3}}\sin \left( {{\tan }^{-1}}{{x}^{4}} \right)}{1+{{x}^{8}}}dx=-\dfrac{1}{4}\cos \left( {{\tan }^{-1}}t \right)+C}$

Substitute ${{x}^{4}}=t$,

$\therefore \int{\dfrac{{{x}^{3}}\sin \left( {{\tan }^{-1}}{{x}^{4}} \right)}{1+{{x}^{8}}}dx=-\dfrac{1}{4}\cos \left( {{\tan }^{-1}}{{x}^{4}} \right)+C}$

38. Choose the correct answer in Question 38 and 39.

38. $\int{\dfrac{10{{x}^{9}}+{{10}^{x}}{{\log }_{e}}10}{{{x}^{10}}+{{10}^{x}}}}dx$ equals

1. ${{10}^{x}}-{{x}^{10}}+C$

2. ${{10}^{x}}+{{x}^{10}}+C$

3. ${{\left( {{10}^{x}}-{{x}^{10}} \right)}^{-1}}+C$

4. $\log \left( {{10}^{x}}+{{x}^{10}} \right)+C$

Ans: Given expression $\int{\dfrac{10{{x}^{9}}+{{10}^{x}}{{\log }_{e}}10}{{{x}^{10}}+{{10}^{x}}}dx}$.

Let ${{x}^{10}}+{{10}^{x}}=t$,

$\therefore \left( 10{{x}^{9}}+{{10}^{x}}{{\log }_{e}}10 \right)dx=dt$

Integration of given expression is

$\Rightarrow \int{\dfrac{10{{x}^{9}}+{{10}^{x}}{{\log }_{e}}10}{{{x}^{10}}+{{10}^{x}}}dx}=\int{\dfrac{dt}{t}}$

$\Rightarrow \int{\dfrac{10{{x}^{9}}+{{10}^{x}}{{\log }_{e}}10}{{{x}^{10}}+{{10}^{x}}}dx}=\log t+C$

Substitute ${{x}^{10}}+{{10}^{x}}=t$,

$\therefore \int{\dfrac{10{{x}^{9}}+{{10}^{x}}{{\log }_{e}}10}{{{x}^{10}}+{{10}^{x}}}dx}=\log \left( {{10}^{x}}+{{x}^{10}} \right)+C$

Therefore, option D is the correct answer.

39. $\int{\dfrac{dx}{{{\sin }^{2}}{{\cos }^{2}}x}}$ equals

1. $\tan x+\cot x+C$

2. $\tan x-\cot x+C$

3. $\tan x\cot x+C$

4. $\tan x-\cot 2x+C$

Ans: Given expression $\int{\dfrac{dx}{{{\sin }^{2}}x{{\cos }^{2}}x}}$.

Let  $I=\int{\dfrac{dx}{{{\sin }^{2}}x{{\cos }^{2}}x}}$

$I=\int{\dfrac{1}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}$

We know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, we get

$\Rightarrow I=\int{\dfrac{{{\sin }^{2}}x+{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}$

$\Rightarrow I=\int{\dfrac{{{\sin }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}+\int{\dfrac{{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}$

$\Rightarrow I=\int{\dfrac{1}{{{\cos }^{2}}x}dx}+\int{\dfrac{1}{{{\sin }^{2}}x}dx}$

$\Rightarrow I=\int{{{\sec }^{2}}xdx}+\int{cose{{c}^{2}}xdx}$

$\Rightarrow I=\tan x-\cot x+C$

Therefore, option B is the correct answer.

### Exercise 7.3

1.  Solve the following: ${{\sin }^{2}}\left( 2x+5 \right)$.

Ans: Given expression ${{\sin }^{2}}\left( 2x+5 \right)$.

Given expression can be written as

${{\sin }^{2}}\left( 2x+5 \right)=\dfrac{1-\cos 2\left( 2x+5 \right)}{2}$

$\Rightarrow {{\sin }^{2}}\left( 2x+5 \right)=\dfrac{1-\cos \left( 4x+10 \right)}{2}$

Integration of given expression is

$\Rightarrow \int{{{\sin }^{2}}\left( 2x+5 \right)dx}=\int{\dfrac{1-\cos \left( 4x+10 \right)}{2}dx}$

$\Rightarrow \int{{{\sin }^{2}}\left( 2x+5 \right)dx}=\dfrac{1}{2}\int{1dx-\dfrac{1}{2}\int{\cos \left( 4x+10 \right)dx}}$

$\Rightarrow \int{{{\sin }^{2}}\left( 2x+5 \right)dx}=\dfrac{1}{2}x-\dfrac{1}{2}\left( \dfrac{\sin \left( 4x+10 \right)}{4} \right)+C$

$\therefore \int{{{\sin }^{2}}\left( 2x+5 \right)dx}=\dfrac{1}{2}x-\dfrac{1}{8}\sin \left( 4x+10 \right)+C$

2. Solve the following: $\sin 3x\cos 4x$.

Ans: Given expression $\sin 3x\cos 4x$.

Using the identity $\sin A\cos B=\dfrac{1}{2}\left\{ \sin \left( A+B \right)+\sin \left( A-B \right) \right\}$ given expression can be written as

$\sin 3x\cos 4x=\dfrac{1}{2}\left\{ \sin \left( 3x+4x \right)+\sin \left( 3x-4x \right) \right\}$

Integration of above expression is

$\Rightarrow \int{\sin 3x\cos 4x}dx=\int{\dfrac{1}{2}\left\{ \sin 7x+\sin \left( -x \right) \right\}dx}$

$\Rightarrow \int{\sin 3x\cos 4x}dx=\dfrac{1}{2}\int{\left\{ \sin 7x+\sin x \right\}dx}$

$\Rightarrow \int{\sin 3x\cos 4x}dx=\dfrac{1}{2}\int{\sin 7xdx+\dfrac{1}{2}\int{\sin x}dx}$

$\Rightarrow \int{\sin 3x\cos 4x}dx=\dfrac{1}{2}\left( \dfrac{-\cos 7x}{7} \right)-\dfrac{1}{2}\left( -\cos x \right)+C$

$\therefore \int{\sin 3x\cos 4x}dx=\dfrac{-\cos 7x}{14}+\dfrac{\cos x}{2}+C$

3.  Solve the following: $\cos 2x\cos 4x\cos 6x$.

Ans: Given expression $\cos 2x\cos 4x\cos 6x$.

Using the identity $\cos A\cos B=\dfrac{1}{2}\left\{ \cos \left( A+B \right)+\cos \left( A-B \right) \right\}$ given expression can be written as

$\cos 2x\left( \cos 4x\cos 6x \right)=\cos 2x\dfrac{1}{2}\left\{ \cos \left( 4x+6x \right)+\cos \left( 4x-6x \right) \right\}$

Integration of the above expression is

$\Rightarrow \int{\cos 2x\left( \cos 4x\cos 6x \right)}=\int{\cos 2x}\left[ \dfrac{1}{2}\left\{ \cos 10x+\cos \left( -2x \right) \right\} \right]dx$

$\Rightarrow \int{\cos 2x\left( \cos 4x\cos 6x \right)}=\int{\left[ \dfrac{1}{2}\left\{ \cos 2x\cos 10x+\cos 2x\cos \left( -2x \right) \right\} \right]}dx$

$\Rightarrow \int{\cos 2x\left( \cos 4x\cos 6x \right)}=\dfrac{1}{2}\int{\left[ \left\{ \cos 2x\cos 10x+{{\cos }^{2}}2x \right\} \right]}dx$

Again applying the identity $\cos A\cos B=\dfrac{1}{2}\left\{ \cos \left( A+B \right)+\cos \left( A-B \right) \right\}$, we get

$\Rightarrow \int{\cos 2x\left( \cos 4x\cos 6x \right)}=\dfrac{1}{2}\int{\left[ \left\{ \dfrac{1}{2}\cos \left( 2x+10x \right)+\cos \left( 2x-10x \right)+\left( \dfrac{1+\cos 4x}{2} \right) \right\} \right]}dx$$\Rightarrow \int{\cos 2x\left( \cos 4x\cos 6x \right)}=\dfrac{1}{4}\int{\left[ \cos 12x+\cos 8x+\cos 4x \right]}dx \therefore \int{\cos 2x\cos 4x\cos 6x}=\dfrac{1}{4}\left[ \dfrac{\sin 12x}{12}+\dfrac{\sin 8x}{8}+\dfrac{\sin 4x}{4} \right]+C 4. Solve the following: {{\sin }^{3}}\left( 2x+1 \right). Ans: Given expression {{\sin }^{3}}\left( 2x+1 \right). Let I=\int{{{\sin }^{3}}\left( 2x+1 \right)dx} \Rightarrow I=\int{{{\sin }^{2}}\left( 2x+1 \right)\sin \left( 2x+1 \right)dx} \Rightarrow I=\int{\left( 1-{{\cos }^{2}}\left( 2x+1 \right) \right)\sin \left( 2x+1 \right)dx} Let \cos \left( 2x+1 \right)=t \therefore -2\sin \left( 2x+1 \right)dx=dt Integration becomes \Rightarrow I=-\dfrac{1}{2}\int{\left( 1-{{t}^{2}} \right)dt} \Rightarrow I=-\dfrac{1}{2}\left( t-\dfrac{{{t}^{3}}}{3} \right)+C Substitute \cos \left( 2x+1 \right)=t, \Rightarrow I=-\dfrac{1}{2}\left( \cos \left( 2x+1 \right)-\dfrac{{{\cos }^{3}}\left( 2x+1 \right)}{3} \right)+C \therefore \int{{{\sin }^{3}}\left( 2x+1 \right)dx}=\dfrac{-\cos \left( 2x+1 \right)}{2}+\dfrac{{{\cos }^{3}}\left( 2x+1 \right)}{3}+C 5. Solve the following: {{\sin }^{3}}x{{\cos }^{3}}x. Ans: Given expression {{\sin }^{3}}x{{\cos }^{3}}x. Let I=\int{{{\sin }^{3}}x{{\cos }^{3}}x}dx \Rightarrow I=\int{{{\sin }^{2}}x\sin x{{\cos }^{3}}x}dx \Rightarrow I=\int{{{\cos }^{3}}x\left( 1-{{\cos }^{2}}x \right)\sin x}dx Let \cos x=t \therefore -\sin dx=dt Integration becomes \Rightarrow I=-\int{{{t}^{3}}\left( 1-{{t}^{2}} \right)}dt \Rightarrow I=-\int{\left( {{t}^{3}}-{{t}^{5}} \right)}dt \Rightarrow I=-\left[ \dfrac{{{t}^{4}}}{4}-\dfrac{{{t}^{6}}}{6} \right]+C Substitute \cos x=t, \Rightarrow I=-\left[ \dfrac{{{\cos }^{4}}x}{4}-\dfrac{{{\cos }^{6}}x}{6} \right]+C \therefore \int{{{\sin }^{3}}x{{\cos }^{3}}x}dx=\dfrac{{{\cos }^{6}}x}{6}-\dfrac{{{\cos }^{4}}x}{4}+C 6. Solve the following: \sin x\sin 2x\sin 3x. Ans: Given expression \sin x\sin 2x\sin 3x. Using the identity \sin A\sin B=\dfrac{1}{2}\cos \left( A-B \right)-\cos \left( A+B \right), given expression can be written as \Rightarrow \sin x\sin 2x\sin 3x=\sin x.\dfrac{1}{2}\cos \left( 2x-3x \right)-\cos \left( 2x+3x \right) Integration of given expression is \Rightarrow \int{\sin x\sin 2x\sin 3x}dx=\dfrac{1}{2}\int{\left( \sin x\cos \left( -x \right)-\sin x\cos \left( 5x \right) \right)dx} \Rightarrow \int{\sin x\sin 2x\sin 3x}dx=\dfrac{1}{2}\int{\left( \sin x\cos x-\sin x\cos 5x \right)dx} \Rightarrow \int{\sin x\sin 2x\sin 3x}dx=\dfrac{1}{2}\int{\dfrac{\sin 2x}{2}dx-\dfrac{1}{2}\int{\left( \sin x\cos 5x \right)}dx} \Rightarrow \int{\sin x\sin 2x\sin 3x}dx=\dfrac{1}{4}\left( \dfrac{-\cos 2x}{2} \right)-\dfrac{1}{2}\int{\left\{ \dfrac{1}{2}\left( \sin \left( x+5x \right)+\sin \left( x-5x \right) \right) \right\}dx} \Rightarrow \int{\sin x\sin 2x\sin 3x}dx=\dfrac{-\cos 2x}{8}-\dfrac{1}{4}\int{\left\{ \left( \sin 6x+\sin \left( -4x \right) \right) \right\}dx} \Rightarrow \int{\sin x\sin 2x\sin 3x}dx=\dfrac{-\cos 2x}{8}-\dfrac{1}{4}\int{\left\{ \left( \sin 6x+\sin 4x \right) \right\}dx} \Rightarrow \int{\sin x\sin 2x\sin 3x}dx=\dfrac{-\cos 2x}{8}-\dfrac{1}{8}\left[ \dfrac{-\cos 6x}{3}+\dfrac{\cos 4x}{4} \right]+C \therefore \int{\sin x\sin 2x\sin 3x}dx=\dfrac{1}{8}\left[ \dfrac{\cos 6x}{3}-\dfrac{\cos 4x}{4}-\cos 2x \right]+C 7. Solve the following: \sin 4x\sin 8x. Ans: Given expression \sin 4x\sin 8x. Using the identity \sin A\sin B=\dfrac{1}{2}\cos \left( A-B \right)-\cos \left( A+B \right), given expression can be written as \Rightarrow \sin 4x\sin 8x=\dfrac{1}{2}\cos \left( 4x-8x \right)-\cos \left( 4x+8x \right) Integration of given expression is \Rightarrow \int{\sin 4x\sin 8x}dx=\dfrac{1}{2}\int{\left( \cos \left( -4x \right)-\cos \left( 12x \right) \right)dx} \Rightarrow \int{\sin 4x\sin 8x}dx=\dfrac{1}{2}\int{\left( \cos 4x-\cos 12x \right)dx} \therefore \int{\sin 4x\sin 8x}dx=\dfrac{1}{2}\left[ \dfrac{\sin 4x}{4}-\dfrac{\sin 12x}{12} \right]+C 8. Solve the following: \dfrac{1-\cos x}{1+\cos x}. Ans: Given expression \dfrac{1-\cos x}{1+\cos x}. Using the identities 2{{\sin }^{2}}\dfrac{x}{2}=1-\cos x and \cos x=2{{\cos }^{2}}\dfrac{x}{2}-1 given expression can be written as \Rightarrow \dfrac{1-\cos x}{1+\cos x}=\dfrac{2{{\sin }^{2}}\dfrac{x}{2}}{2{{\cos }^{2}}\dfrac{x}{2}} \Rightarrow \dfrac{1-\cos x}{1+\cos x}={{\tan }^{2}}\dfrac{x}{2} Integration of given expression is \Rightarrow \int{\dfrac{1-\cos x}{1+\cos x}dx}=\int{\left[ {{\tan }^{2}}\dfrac{x}{2} \right]dx} \Rightarrow \int{\dfrac{1-\cos x}{1+\cos x}dx}=\int{\left[ {{\sec }^{2}}\dfrac{x}{2}-1 \right]dx} \Rightarrow \int{\dfrac{1-\cos x}{1+\cos x}dx}=\left[ \dfrac{\tan \dfrac{x}{2}}{\dfrac{1}{2}}-x \right]+C \therefore \int{\dfrac{1-\cos x}{1+\cos x}dx}=2\tan \dfrac{x}{2}-x+C 9. Solve the following: \dfrac{\cos x}{1+\cos x}. Ans: Given expression \dfrac{\cos x}{1+\cos x}. Using the identity \cos x={{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2} and \cos x=2{{\cos }^{2}}\dfrac{x}{2}-1 given expression can be written as \Rightarrow \dfrac{\cos x}{1+\cos x}=\dfrac{{{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2}}{2{{\cos }^{2}}\dfrac{x}{2}} \Rightarrow \dfrac{\cos x}{1+\cos x}=\dfrac{1}{2}\left[ 1-\dfrac{{{\sin }^{2}}\dfrac{x}{2}}{{{\cos }^{2}}\dfrac{x}{2}} \right] \Rightarrow \dfrac{\cos x}{1+\cos x}=\dfrac{1}{2}\left[ 1-{{\tan }^{2}}\dfrac{x}{2} \right] Integration of given expression is \Rightarrow \int{\dfrac{\cos x}{1+\cos x}dx}=\dfrac{1}{2}\int{\left[ 1-{{\tan }^{2}}\dfrac{x}{2} \right]dx} \Rightarrow \int{\dfrac{\cos x}{1+\cos x}dx}=\dfrac{1}{2}\int{\left[ 1-{{\sec }^{2}}\dfrac{x}{2}+1 \right]dx} \Rightarrow \int{\dfrac{\cos x}{1+\cos x}dx}=\dfrac{1}{2}\int{\left[ 2-{{\sec }^{2}}\dfrac{x}{2} \right]dx} \Rightarrow \int{\dfrac{\cos x}{1+\cos x}dx}=\dfrac{1}{2}\left[ 2x-\dfrac{\tan \dfrac{x}{2}}{\dfrac{1}{2}} \right]+C \therefore \int{\dfrac{\cos x}{1+\cos x}dx}=x-\tan \dfrac{x}{2}+C 10. Solve the following: {{\sin }^{4}}x. Ans: Given expression {{\sin }^{4}}x. Given expression can be written as {{\sin }^{4}}x={{\sin }^{2}}x{{\sin }^{2}}x \Rightarrow {{\sin }^{4}}x=\left( \dfrac{1-\cos 2x}{2} \right)\left( \dfrac{1-\cos 2x}{2} \right) \Rightarrow {{\sin }^{4}}x=\dfrac{1}{4}{{\left( 1-\cos 2x \right)}^{2}} \Rightarrow {{\sin }^{4}}x=\dfrac{1}{4}\left( 1+{{\cos }^{2}}2x-2\cos 2x \right) \Rightarrow {{\sin }^{4}}x=\dfrac{1}{4}\left( 1+\dfrac{1+\cos 4x}{2}-2\cos 2x \right) \Rightarrow {{\sin }^{4}}x=\dfrac{1}{4}\left( 1+\dfrac{1}{2}+\dfrac{\cos 4x}{2}-2\cos 2x \right) \Rightarrow {{\sin }^{4}}x=\dfrac{1}{4}\left( \dfrac{3}{2}+\dfrac{1}{2}\cos 4x-2\cos 2x \right) Integration of given expression is \Rightarrow \int{{{\sin }^{4}}x}dx=\dfrac{1}{4}\int{\left( \dfrac{3}{2}+\dfrac{1}{2}\cos 4x-2\cos 2x \right)}dx \Rightarrow \int{{{\sin }^{4}}x}dx=\dfrac{1}{4}\left[ \dfrac{3}{2}x+\dfrac{\sin 4x}{8}-\sin 2x \right]+C \therefore \int{{{\sin }^{4}}x}dx=\dfrac{3x}{8}+\dfrac{\sin 4x}{32}-\dfrac{1}{4}\sin 2x+C 11. Solve the following: {{\cos }^{4}}2x. Ans: Given expression {{\cos }^{4}}2x. Given expression can be written as {{\cos }^{4}}2x={{\left( {{\cos }^{2}}2x \right)}^{2}} \Rightarrow {{\cos }^{4}}2x={{\left( \dfrac{1+\cos 4x}{2} \right)}^{2}} \Rightarrow {{\cos }^{4}}2x=\dfrac{1}{4}\left( 1+{{\cos }^{2}}4x+2\cos 4x \right) \Rightarrow {{\cos }^{4}}2x=\dfrac{1}{4}\left( 1+\dfrac{1+\cos 8x}{2}+2\cos 4x \right) \Rightarrow {{\cos }^{4}}2x=\dfrac{1}{4}\left( 1+\dfrac{1}{2}+\dfrac{\cos 8x}{2}+2\cos 4x \right) \Rightarrow {{\cos }^{4}}2x=\dfrac{1}{4}\left( \dfrac{3}{2}+\dfrac{\cos 8x}{2}+2\cos 4x \right) Integration of given expression is \Rightarrow \int{{{\cos }^{4}}2xdx}=\int{\left( \dfrac{3}{8}+\dfrac{\cos 8x}{8}+\dfrac{\cos 4x}{2} \right)dx} \therefore \int{{{\cos }^{4}}2xdx}=\dfrac{3}{8}x+\dfrac{\sin 8x}{64}+\dfrac{\sin 4x}{8}+C 12. Solve the following: \dfrac{{{\sin }^{2}}x}{1+\cos x}. Ans: Given expression \dfrac{{{\sin }^{2}}x}{1+\cos x}. By applying the identity \sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2} and \cos x=2{{\cos }^{2}}\dfrac{x}{2}-1, given expression can be written as \Rightarrow \dfrac{{{\sin }^{2}}x}{1+\cos x}=\dfrac{{{\left( 2\sin \dfrac{x}{2}\cos \dfrac{x}{2} \right)}^{2}}}{2{{\cos }^{2}}\dfrac{x}{2}} \Rightarrow \dfrac{{{\sin }^{2}}x}{1+\cos x}=\dfrac{4{{\sin }^{2}}\dfrac{x}{2}{{\cos }^{2}}\dfrac{x}{2}}{2{{\cos }^{2}}\dfrac{x}{2}} \Rightarrow \dfrac{{{\sin }^{2}}x}{1+\cos x}=2{{\sin }^{2}}\dfrac{x}{2} \Rightarrow \dfrac{{{\sin }^{2}}x}{1+\cos x}=1-\cos x Integration of given expression is \Rightarrow \int{\dfrac{{{\sin }^{2}}x}{1+\cos x}dx}=\int{1dx-\int{\cos xdx}} \therefore \int{\dfrac{{{\sin }^{2}}x}{1+\cos x}dx}=x-\sin x+C 13. Solve the following: \dfrac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }. Ans: Given expression \dfrac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }. We can apply the identity \cos C-\cos D=-2\sin \dfrac{C+D}{2}\sin \dfrac{C-D}{2} , we get \Rightarrow \dfrac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }=\dfrac{-2\sin \dfrac{2x+2\alpha }{2}\sin \dfrac{2x-2\alpha }{2}}{-2\sin \dfrac{x+\alpha }{2}\sin \dfrac{x-\alpha }{2}} \Rightarrow \dfrac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }=\dfrac{\sin \dfrac{2\left( x+\alpha \right)}{2}\sin \dfrac{2\left( x-\alpha \right)}{2}}{\sin \dfrac{x+\alpha }{2}\sin \dfrac{x-\alpha }{2}} \Rightarrow \dfrac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }=\dfrac{\sin \left( x+\alpha \right)\sin \left( x-\alpha \right)}{\sin \dfrac{x+\alpha }{2}\sin \dfrac{x-\alpha }{2}} We can apply the identity \sin 2x=2\sin x\cos x, we get \Rightarrow \dfrac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }=\dfrac{\left[ 2\sin \dfrac{x+\alpha }{2}\cos \dfrac{x+\alpha }{2} \right]\left[ 2\sin \dfrac{x-\alpha }{2}\cos \dfrac{x-\alpha }{2} \right]}{\sin \dfrac{x+\alpha }{2}\sin \dfrac{x-\alpha }{2}} \Rightarrow \dfrac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }=4\cos \dfrac{x+\alpha }{2}\cos \dfrac{x-\alpha }{2} \Rightarrow \dfrac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }=2\left[ \cos \dfrac{x+\alpha }{2}+\dfrac{x-\alpha }{2}+\cos \dfrac{x+\alpha }{2}-\dfrac{x-\alpha }{2} \right] \Rightarrow \dfrac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }=2\left[ \cos x+\cos \alpha \right] Integration of given expression is \Rightarrow \int{\dfrac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }dx}=2\int{\left[ \cos x+\cos \alpha \right]}dx \therefore \int{\dfrac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }dx}=2\left[ \sin x+x\cos \alpha \right]+C 14. Solve the following: \dfrac{\cos x-\sin x}{1+\sin 2x}. Ans: Given expression \dfrac{\cos x-\sin x}{1+\sin 2x}. We know that {{\sin }^{2}}x+{{\cos }^{2}}x=1. Given expression can be written as \Rightarrow \dfrac{\cos x-\sin x}{1+\sin 2x}=\dfrac{\cos x-\sin x}{{{\sin }^{2}}x+{{\cos }^{2}}x+\sin 2x} . We can apply the identity \sin 2x=2\sin x\cos x, we get \Rightarrow \dfrac{\cos x-\sin x}{1+\sin 2x}=\dfrac{\cos x-\sin x}{{{\sin }^{2}}x+{{\cos }^{2}}x+2\sin x\cos x} \Rightarrow \dfrac{\cos x-\sin x}{1+\sin 2x}=\dfrac{\cos x-\sin x}{{{\left( \sin x+\cos x \right)}^{2}}} Let \sin x+\cos x=t \therefore \left( \cos x-\sin x \right)dx=dt Integration of given expression is \Rightarrow \int{\dfrac{\cos x-\sin x}{1+\sin 2x}dx=\int{\dfrac{\cos x-\sin x}{{{\left( \sin x+\cos x \right)}^{2}}}dx}} \Rightarrow \int{\dfrac{\cos x-\sin x}{1+\sin 2x}dx=\int{\dfrac{dt}{{{t}^{2}}}}} \Rightarrow \int{\dfrac{\cos x-\sin x}{1+\sin 2x}dx=\int{{{t}^{-2}}dt}} \Rightarrow \int{\dfrac{\cos x-\sin x}{1+\sin 2x}dx=-{{t}^{-1}}+C} \Rightarrow \int{\dfrac{\cos x-\sin x}{1+\sin 2x}dx=-\dfrac{1}{t}+C} Substitute \sin x+\cos x=t, \therefore \int{\dfrac{\cos x-\sin x}{1+\sin 2x}dx=-\dfrac{1}{\sin x+\cos x}+C} 15. Solve the following: {{\tan }^{3}}2x\sec 2x. Ans: Given expression {{\tan }^{3}}2x\sec 2x. Given expression can be written as {{\tan }^{3}}2x\sec 2x={{\tan }^{2}}2x\tan 2x\sec 2x \Rightarrow {{\tan }^{3}}2x\sec 2x=\left( {{\sec }^{2}}2x-1 \right)\tan 2x\sec 2x \Rightarrow {{\tan }^{3}}2x\sec 2x={{\sec }^{2}}2x\tan 2x\sec 2x-\tan 2x\sec 2x Integration of given expression is \Rightarrow \int{{{\tan }^{3}}2x\sec 2xdx}=\int{{{\sec }^{2}}2x\tan 2x\sec 2xdx}-\int{\tan 2x\sec 2xdx} \Rightarrow \int{{{\tan }^{3}}2x\sec 2xdx}=\int{{{\sec }^{2}}2x\tan 2x\sec 2xdx}-\dfrac{\sec 2x}{2}+C Let \sec 2x=t \therefore 2\sec 2x\tan 2xdx=dt Above integral becomes \Rightarrow \int{{{\tan }^{3}}2x\sec 2xdx}=\dfrac{1}{2}\int{{{t}^{2}}dt}-\dfrac{\sec 2x}{2}+C \Rightarrow \int{{{\tan }^{3}}2x\sec 2xdx}=\dfrac{{{t}^{3}}}{6}-\dfrac{\sec 2x}{2}+C Substitute \sec 2x=t, \therefore \int{{{\tan }^{3}}2x\sec 2xdx}=\dfrac{{{\left( \sec 2x \right)}^{3}}}{6}-\dfrac{\sec 2x}{2}+C 16. Solve the following: {{\tan }^{4}}x. Ans: Given expression {{\tan }^{4}}x. Given expression can be written as \Rightarrow {{\tan }^{4}}x={{\tan }^{2}}x{{\tan }^{2}}x \Rightarrow {{\tan }^{4}}x=\left( {{\sec }^{2}}x-1 \right){{\tan }^{2}}x \Rightarrow {{\tan }^{4}}x={{\sec }^{2}}x{{\tan }^{2}}x-{{\tan }^{2}}x \Rightarrow {{\tan }^{4}}x={{\sec }^{2}}x{{\tan }^{2}}x-\left( {{\sec }^{2}}x-1 \right) \Rightarrow {{\tan }^{4}}x={{\sec }^{2}}x{{\tan }^{2}}x-{{\sec }^{2}}x+1 Integration of given expression is \Rightarrow \int{{{\tan }^{4}}xdx}=\int{\left( {{\sec }^{2}}x{{\tan }^{2}}x-{{\sec }^{2}}x+1 \right)}dx \Rightarrow \int{{{\tan }^{4}}xdx}=\int{\left( {{\sec }^{2}}x{{\tan }^{2}}x \right)}dx-\int{{{\sec }^{2}}xdx+\int{1dx}} \Rightarrow \int{{{\tan }^{4}}xdx}=\int{{{\sec }^{2}}x{{\tan }^{2}}x}dx-\tan x+x+C Let \tan x=t \therefore {{\sec }^{2}}xdx=dt \Rightarrow \int{{{\tan }^{4}}xdx}=\int{{{t}^{2}}}dt-\tan x+x+C \Rightarrow \int{{{\tan }^{4}}xdx}=\dfrac{{{t}^{3}}}{3}-\tan x+x+C Substitute \tan x=t, \therefore \int{{{\tan }^{4}}xdx}=\dfrac{1}{3}{{\tan }^{3}}x-\tan x+x+C 17. Solve the following: \dfrac{{{\sin }^{3}}x+{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}. Ans: Given expression \dfrac{{{\sin }^{3}}x+{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}. Given expression can be written as \Rightarrow \dfrac{{{\sin }^{3}}x+{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}=\dfrac{{{\sin }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}+\dfrac{{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x} \Rightarrow \dfrac{{{\sin }^{3}}x+{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}=\dfrac{\sin x}{{{\cos }^{2}}x}+\dfrac{\cos x}{{{\sin }^{2}}x} \Rightarrow \dfrac{{{\sin }^{3}}x+{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}=\tan x\sec x+\cot xcosecx Integration of given expression is \Rightarrow \int{\dfrac{{{\sin }^{3}}x+{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}=\int{\tan x\sec xdx}+\int{\cot x\operatorname{cosecx}dx} \therefore \int{\dfrac{{{\sin }^{3}}x+{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}=\sec x-cosecx+C 18. Solve the following: \dfrac{\cos 2x+2{{\sin }^{2}}x}{{{\cos }^{2}}x}. Ans: Given expression \dfrac{\cos 2x+2{{\sin }^{2}}x}{{{\cos }^{2}}x}. By applying the identity \cos 2x=1-2{{\sin }^{2}}x, we get \Rightarrow \dfrac{\cos 2x+2{{\sin }^{2}}x}{{{\cos }^{2}}x}=\dfrac{\cos 2x+1-\cos 2x}{{{\cos }^{2}}x} \Rightarrow \dfrac{\cos 2x+2{{\sin }^{2}}x}{{{\cos }^{2}}x}=\dfrac{1}{{{\cos }^{2}}x} \Rightarrow \dfrac{\cos 2x+2{{\sin }^{2}}x}{{{\cos }^{2}}x}={{\sec }^{2}}x Integration of given expression is \Rightarrow \int{\dfrac{\cos 2x+2{{\sin }^{2}}x}{{{\cos }^{2}}x}}dx=\int{{{\sec }^{2}}xdx} \therefore \int{\dfrac{\cos 2x+2{{\sin }^{2}}x}{{{\cos }^{2}}x}}dx=\tan x+C 19. Solve the following: \dfrac{1}{\sin x{{\cos }^{3}}x}. Ans: Given expression \dfrac{1}{\sin x{{\cos }^{3}}x}. We can apply the identity {{\sin }^{2}}x+{{\cos }^{2}}x=1, we get \Rightarrow \dfrac{1}{\sin x{{\cos }^{3}}x}=\dfrac{{{\sin }^{2}}x+{{\cos }^{2}}x}{\sin x{{\cos }^{3}}x} \Rightarrow \dfrac{1}{\sin x{{\cos }^{3}}x}=\dfrac{{{\sin }^{2}}x}{\sin x{{\cos }^{3}}x}+\dfrac{{{\cos }^{2}}x}{\sin x{{\cos }^{3}}x} \Rightarrow \dfrac{1}{\sin x{{\cos }^{3}}x}=\dfrac{\sin x}{{{\cos }^{3}}x}+\dfrac{1}{\sin x\cos x} \Rightarrow \dfrac{1}{\sin x{{\cos }^{3}}x}=\tan x{{\sec }^{2}}x+\dfrac{{{\cos }^{2}}x}{\dfrac{\sin x\cos x}{{{\cos }^{2}}x}} \Rightarrow \dfrac{1}{\sin x{{\cos }^{3}}x}=\tan x{{\sec }^{2}}x+\dfrac{{{\sec }^{2}}x}{\tan x} Integration of given expression is \Rightarrow \int{\dfrac{1}{\sin x{{\cos }^{3}}x}}dx=\int{\tan x{{\sec }^{2}}x}dx+\int{\dfrac{{{\sec }^{2}}x}{\tan x}}dx Let \tan x=t \therefore {{\sec }^{2}}xdx=dt \Rightarrow \int{\dfrac{1}{\sin x{{\cos }^{3}}x}}dx=\int{t}dt+\int{\dfrac{1}{\operatorname{t}}}dt \Rightarrow \int{\dfrac{1}{\sin x{{\cos }^{3}}x}}dx=\dfrac{{{t}^{2}}}{2}+\log \left| t \right|+C Substitute \tan x=t, \therefore \int{\dfrac{1}{\sin x{{\cos }^{3}}x}}dx=\dfrac{1}{2}{{\tan }^{2}}x+\log \left| \tan x \right|+C 20. Solve the following: \dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}. Ans: Given expression \dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}. Given expression can be written as \Rightarrow \dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}=\dfrac{\cos 2x}{{{\cos }^{2}}x+{{\sin }^{2}}x+2\sin x\cos x} We know that {{\sin }^{2}}x+{{\cos }^{2}}x=1 and 2\sin x\cos x=\sin 2x, we get \Rightarrow \dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}=\dfrac{\cos 2x}{1+\sin 2x} Integration of given expression is \Rightarrow \int{\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=\int{\dfrac{\cos 2x}{1+\sin 2x}dx} Let 1+\sin 2x=t \therefore 2\cos 2xdx=dt Integration becomes \Rightarrow \int{\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=\dfrac{1}{2}\int{\dfrac{1}{t}dt} \Rightarrow \int{\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=\dfrac{1}{2}\log \left| t \right|+C Substitute 1+\sin 2x=t \Rightarrow \int{\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=\dfrac{1}{2}\log \left| 1+\sin 2x \right|+C \Rightarrow \int{\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=\dfrac{1}{2}\log \left| {{\left( \cos x+\sin x \right)}^{2}} \right|+C \therefore \int{\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=\log \left| \left( \cos x+\sin x \right) \right|+C 21. Solve the following: {{\sin }^{-1}}\left( \cos x \right). Ans: Given expression {{\sin }^{-1}}\left( \cos x \right). Let \cos x=t \therefore \sin x=\sqrt{1-{{t}^{2}}} \Rightarrow -\sin xdx=dt \Rightarrow dx=-\dfrac{dt}{\sin x} \Rightarrow dx=-\dfrac{dt}{\sqrt{1-{{t}^{2}}}} Integration of given expression is \Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=\int{{{\sin }^{-1}}t\left( \dfrac{-dt}{\sqrt{1-{{t}^{2}}}} \right)} \Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=-\int{\left( \dfrac{{{\sin }^{-1}}t}{\sqrt{1-{{t}^{2}}}} \right)dt} Let {{\sin }^{-1}}t=u \Rightarrow \dfrac{1}{\sqrt{1-{{t}^{2}}}}dt=du Integration becomes \Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=\int{4du} \Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=-\dfrac{{{u}^{2}}}{2}+C Substitute {{\sin }^{-1}}t=u \Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=-\dfrac{{{\left( {{\sin }^{-1}}t \right)}^{2}}}{2}+C Substitute \cos x=t \Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=-\dfrac{{{\left[ {{\sin }^{-1}}\left( \cos x \right) \right]}^{2}}}{2}+C ……..(1) We know that {{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2} \therefore {{\sin }^{-1}}\left( \cos x \right)=\dfrac{\pi }{2}-{{\cos }^{-1}}\left( \cos x \right)=\left( \dfrac{\pi }{2}-x \right) Substitute in eq. (1), we get \Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=\dfrac{-{{\left( \dfrac{\pi }{2}-x \right)}^{2}}}{2}+C \Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=\dfrac{1}{2}\left( \dfrac{{{\pi }^{2}}}{2}+{{x}^{2}}-\pi x \right)+C \Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=-\dfrac{{{\pi }^{2}}}{4}-\dfrac{{{x}^{2}}}{2}+\dfrac{1}{2}\pi x+C \Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=\dfrac{\pi x}{2}-\dfrac{{{x}^{2}}}{2}+\left( C-\dfrac{{{\pi }^{2}}}{4} \right) \therefore \int{{{\sin }^{-1}}\left( \cos x \right)dx}=\dfrac{\pi x}{2}-\dfrac{{{x}^{2}}}{2}+{{C}_{1}} 22. Solve the following: \dfrac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}. Ans: Given expression \dfrac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}. Given expression can be written as \Rightarrow \dfrac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}=\dfrac{1}{\sin \left( a-b \right)}\left[ \dfrac{\sin \left( a-b \right)}{\cos \left( x-a \right)\cos \left( x-b \right)} \right] \Rightarrow \dfrac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}=\dfrac{1}{\sin \left( a-b \right)}\left[ \dfrac{\sin \left[ \left( x-b \right)-\left( x-a \right) \right]}{\cos \left( x-a \right)\cos \left( x-b \right)} \right] \Rightarrow \dfrac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}=\dfrac{1}{\sin \left( a-b \right)}\left[ \dfrac{\sin \left( x-b \right)\cos \left( x-a \right)-\cos \left( x-b \right)\sin \left( x-a \right)}{\cos \left( x-a \right)\cos \left( x-b \right)} \right] \Rightarrow \dfrac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}=\dfrac{1}{\sin \left( a-b \right)}\left[ \tan \left( x-b \right)-\tan \left( x-b \right) \right] Integration of given expression is \Rightarrow \int{\dfrac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}dx}=\int{\dfrac{1}{\sin \left( a-b \right)}\left[ \tan \left( x-b \right)-\tan \left( x-b \right) \right]dx} \Rightarrow \int{\dfrac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}dx}=\int{\dfrac{1}{\sin \left( a-b \right)}\left[ -\log \left| \cos \left( x-b \right) \right|+\log \cos \left( x-a \right) \right]} \Rightarrow \int{\dfrac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}dx}=\int{\dfrac{1}{\sin \left( a-b \right)}\left[ \log \left| \dfrac{\cos \left( x-a \right)}{\cos \left( x-b \right)} \right| \right]}+C 23. Solve the following: \int{\dfrac{{{\sin }^{2}}x-{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx} is equal to 1. \tan x+\cot x+C 2. \tan x+cosecx+C 3. -\tan x+\cot x+C 4. \tan x+\sec x+C Ans: Given expression \int{\dfrac{{{\sin }^{2}}x-{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}. Given expression can be written as \Rightarrow \int{\dfrac{{{\sin }^{2}}x-{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}=\int{\dfrac{{{\sin }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}-\int{\dfrac{{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx} \Rightarrow \int{\dfrac{{{\sin }^{2}}x-{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}=\int{{{\sec }^{2}}xdx}-\int{cose{{c}^{2}}xdx} \therefore \int{\dfrac{{{\sin }^{2}}x-{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}=\tan x+\cot x+C Therefore, option A is the correct answer. 24. Solve the following: \int{\dfrac{{{e}^{x}}\left( 1+x \right)}{{{\cos }^{2}}\left( {{e}^{x}}x \right)}dx} equals 1. -\cot \left( e{{x}^{x}} \right)+C 2. \tan \left( x{{e}^{x}} \right)+C 3. \tan \left( {{e}^{x}} \right)+C 4. \cot \left( {{e}^{x}} \right)+C Ans: Given expression \int{\dfrac{{{e}^{x}}\left( 1+x \right)}{{{\cos }^{2}}\left( {{e}^{x}}x \right)}dx}. Let {{e}^{x}}x=t \therefore \left( {{e}^{x}}.x+{{e}^{x}}.1 \right)dx=dt \Rightarrow {{e}^{x}}\left( x+1 \right)dx=dt Integration of given expression is \Rightarrow \int{\dfrac{{{e}^{x}}\left( 1+x \right)}{{{\cos }^{2}}\left( {{e}^{x}}x \right)}dx}=\int{\dfrac{dt}{{{\cos }^{2}}t}} \Rightarrow \int{\dfrac{{{e}^{x}}\left( 1+x \right)}{{{\cos }^{2}}\left( {{e}^{x}}x \right)}dx}=\int{{{\sec }^{2}}t}dt \Rightarrow \int{\dfrac{{{e}^{x}}\left( 1+x \right)}{{{\cos }^{2}}\left( {{e}^{x}}x \right)}dx}=\tan t+C Substitute {{e}^{x}}x=t, \therefore \int{\dfrac{{{e}^{x}}\left( 1+x \right)}{{{\cos }^{2}}\left( {{e}^{x}}x \right)}dx}=\tan \left( {{e}^{x}}x \right)+C Therefore, option B is the correct answer. ### Exercise 7.4 1. Solve the following: \dfrac{3{{x}^{2}}}{{{x}^{6}}+1}. Ans: Given expression \dfrac{3{{x}^{2}}}{{{x}^{6}}+1}. Let {{x}^{3}}=t \therefore 3{{x}^{2}}dx=dt Integration of given expression is \Rightarrow \int{\dfrac{3{{x}^{2}}}{{{x}^{6}}+1}dx=\int{\dfrac{dt}{{{t}^{2}}+1}}} We know that \int{\dfrac{1}{1+{{x}^{2}}}={{\tan }^{-1}}x} \Rightarrow \int{\dfrac{3{{x}^{2}}}{{{x}^{6}}+1}dx={{\tan }^{-1}}t+C} Substitute {{x}^{3}}=t, \therefore \int{\dfrac{3{{x}^{2}}}{{{x}^{6}}+1}dx={{\tan }^{-1}}\left( {{x}^{3}} \right)+C} 2. Solve the following: \dfrac{1}{\sqrt{1+4{{x}^{2}}}}. Ans: Given expression \dfrac{1}{\sqrt{1+4{{x}^{2}}}}. Let 2x=t \therefore 2dx=dt Integration of given expression is \Rightarrow \int{\dfrac{1}{\sqrt{1+4{{x}^{2}}}}dx=\dfrac{1}{2}\int{\dfrac{dt}{\sqrt{1+{{t}^{2}}}}}} We know that \int{\dfrac{1}{\sqrt{{{x}^{2}}+{{a}^{2}}}}}dt=\log \left| x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right| \Rightarrow \int{\dfrac{1}{\sqrt{1+4{{x}^{2}}}}dx=\dfrac{1}{2}\log \left| t+\sqrt{{{t}^{2}}+1} \right|}+C Substitute 2x=t, \therefore \int{\dfrac{1}{\sqrt{1+4{{x}^{2}}}}dx=\dfrac{1}{2}\log \left| 2x+\sqrt{4{{x}^{2}}+1} \right|}+C 3. Solve the following: \dfrac{1}{\sqrt{{{\left( 2-x \right)}^{2}}+1}}. Ans: Given expression \dfrac{1}{\sqrt{{{\left( 2-x \right)}^{2}}+1}}. Let 2-x=t \therefore -dx=dt Integration of given expression is \Rightarrow \int{\dfrac{1}{\sqrt{{{\left( 2-x \right)}^{2}}+1}}dx=-\int{\dfrac{1}{\sqrt{{{t}^{2}}+1}}dt}} We know that \int{\dfrac{1}{\sqrt{{{x}^{2}}+{{a}^{2}}}}}dt=\log \left| x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right| \Rightarrow \int{\dfrac{1}{\sqrt{{{\left( 2-x \right)}^{2}}+1}}dx=-\log \left| t+\sqrt{{{t}^{2}}+1} \right|+C} Substitute 2-x=t, \Rightarrow \int{\dfrac{1}{\sqrt{{{\left( 2-x \right)}^{2}}+1}}dx=-\log \left| \left( 2-x \right)+\sqrt{{{\left( 2-x \right)}^{2}}+1} \right|+C} \therefore \int{\dfrac{1}{\sqrt{{{\left( 2-x \right)}^{2}}+1}}dx=\log \left| \dfrac{1}{\left( 2-x \right)+\sqrt{{{x}^{2}}-4x+5}} \right|+C} 4. Solve the following: \dfrac{1}{\sqrt{9-25{{x}^{2}}}}. Ans: Given expression \dfrac{1}{\sqrt{9-25{{x}^{2}}}}. Let 5x=t \therefore 5dx=dt Integration of given expression is \Rightarrow \int{\dfrac{1}{\sqrt{9-25{{x}^{2}}}}dx=\dfrac{1}{5}\int{\dfrac{1}{\sqrt{9-{{t}^{2}}}}dt}} \Rightarrow \int{\dfrac{1}{\sqrt{9-25{{x}^{2}}}}dx=\dfrac{1}{5}\int{\dfrac{1}{\sqrt{{{3}^{2}}-{{t}^{2}}}}dt}} \Rightarrow \int{\dfrac{1}{\sqrt{9-25{{x}^{2}}}}dx=\dfrac{1}{5}{{\sin }^{-1}}\left( \dfrac{t}{3} \right)+C} Substitute 5x=t, \therefore \int{\dfrac{1}{\sqrt{9-25{{x}^{2}}}}dx=\dfrac{1}{5}{{\sin }^{-1}}\left( \dfrac{5x}{3} \right)+C} 5. Solve the following: \dfrac{3x}{1+2{{x}^{4}}}. Ans: Given expression \dfrac{3x}{1+2{{x}^{4}}}. Let \sqrt{2}{{x}^{2}}=t \therefore 2\sqrt{2}dx=dt Integration of given expression is \Rightarrow \int{\dfrac{3x}{1+2{{x}^{4}}}dx}=\dfrac{3}{2\sqrt{2}}\int{\dfrac{dt}{1+{{t}^{2}}}} \Rightarrow \int{\dfrac{3x}{1+2{{x}^{4}}}dx}=\dfrac{3}{2\sqrt{2}}{{\tan }^{-1}}t+C Substitute \sqrt{2}{{x}^{2}}=t, \therefore \int{\dfrac{3x}{1+2{{x}^{4}}}dx}=\dfrac{3}{2\sqrt{2}}{{\tan }^{-1}}\left( \sqrt{2}{{x}^{2}} \right)+C 6. Solve the following: \dfrac{{{x}^{2}}}{1-{{x}^{6}}}. Ans: Given expression \dfrac{{{x}^{2}}}{1-{{x}^{6}}}. Let {{x}^{3}}=t \therefore 3{{x}^{2}}dx=dt Integration of given expression is \Rightarrow \int{\dfrac{{{x}^{2}}}{1-{{x}^{6}}}dx=\dfrac{1}{3}\int{\dfrac{dt}{1-{{t}^{2}}}}} \Rightarrow \int{\dfrac{{{x}^{2}}}{1-{{x}^{6}}}dx=\dfrac{1}{3}\left[ \dfrac{1}{2}\log \left| \dfrac{1+t}{1-t} \right| \right]+C} Substitute {{x}^{3}}=t, \therefore \int{\dfrac{{{x}^{2}}}{1-{{x}^{6}}}dx=\dfrac{1}{3}\left[ \dfrac{1}{2}\log \left| \dfrac{1+{{x}^{3}}}{1-{{x}^{3}}} \right| \right]+C} 7. Solve the following: \dfrac{x-1}{\sqrt{{{x}^{2}}-1}}. Ans: Given expression \dfrac{x-1}{\sqrt{{{x}^{2}}-1}}. Given expression can be written as \Rightarrow \dfrac{x-1}{\sqrt{{{x}^{2}}-1}}=\dfrac{x}{\sqrt{{{x}^{2}}-1}}-\dfrac{1}{\sqrt{{{x}^{2}}-1}} Integration of given expression is \Rightarrow \int{\dfrac{x-1}{\sqrt{{{x}^{2}}-1}}dx}=\int{\dfrac{x}{\sqrt{{{x}^{2}}-1}}dx}-\int{\dfrac{1}{\sqrt{{{x}^{2}}-1}}dx} \Rightarrow \int{\dfrac{x-1}{\sqrt{{{x}^{2}}-1}}dx}=\int{\dfrac{x}{\sqrt{{{x}^{2}}-1}}dx}-\log \left| x+\sqrt{{{x}^{2}}-1} \right|+C Let {{x}^{2}}-1=t \therefore 2xdx=dt Integration becomes \Rightarrow \int{\dfrac{x-1}{\sqrt{{{x}^{2}}-1}}dx}=\dfrac{1}{2}\int{\dfrac{dt}{\sqrt{t}}}-\log \left| x+\sqrt{{{x}^{2}}-1} \right|+C \Rightarrow \int{\dfrac{x-1}{\sqrt{{{x}^{2}}-1}}dx}=\dfrac{1}{2}\int{{{t}^{\dfrac{1}{2}}}dt}-\log \left| x+\sqrt{{{x}^{2}}-1} \right|+C \Rightarrow \int{\dfrac{x-1}{\sqrt{{{x}^{2}}-1}}dx}=\dfrac{1}{2}\left( 2{{t}^{\dfrac{1}{2}}} \right)-\log \left| x+\sqrt{{{x}^{2}}-1} \right|+C \Rightarrow \int{\dfrac{x-1}{\sqrt{{{x}^{2}}-1}}dx}=\sqrt{t}-\log \left| x+\sqrt{{{x}^{2}}-1} \right|+C Substitute {{x}^{2}}-1=t \Rightarrow \int{\dfrac{x-1}{\sqrt{{{x}^{2}}-1}}dx}=\sqrt{{{x}^{2}}-1}-\log \left| x+\sqrt{{{x}^{2}}-1} \right|+C 8. Solve the following: \dfrac{{{x}^{2}}}{\sqrt{{{x}^{6}}+{{a}^{6}}}}. Ans: Given expression \dfrac{{{x}^{2}}}{\sqrt{{{x}^{6}}+{{a}^{6}}}}. Let {{x}^{3}}=t \therefore 3{{x}^{2}}dx=dt Integration of given expression is \Rightarrow \int{\dfrac{{{x}^{2}}}{\sqrt{{{x}^{6}}+{{a}^{6}}}}dx=\dfrac{1}{3}\int{\dfrac{dt}{\sqrt{{{t}^{2}}+{{\left( {{a}^{3}} \right)}^{2}}}}}} \Rightarrow \int{\dfrac{{{x}^{2}}}{\sqrt{{{x}^{6}}+{{a}^{6}}}}dx=\dfrac{1}{3}\log \left| t+\sqrt{{{t}^{2}}+{{a}^{6}}} \right|+C} Substitute {{x}^{3}}=t, \therefore \int{\dfrac{{{x}^{2}}}{\sqrt{{{x}^{6}}+{{a}^{6}}}}dx=\dfrac{1}{3}\log \left| {{x}^{3}}+\sqrt{{{x}^{6}}+{{a}^{6}}} \right|+C} 9. Solve the following: \dfrac{{{\sec }^{2}}x}{\sqrt{{{\tan }^{2}}x+4}}. Ans: Given expression \dfrac{{{\sec }^{2}}x}{\sqrt{{{\tan }^{2}}x+4}}. Let \tan x=t \therefore {{\sec }^{2}}xdx=dt Integration of given expression is \Rightarrow \int{\dfrac{{{\sec }^{2}}x}{\sqrt{{{\tan }^{2}}x+4}}dx=\int{\dfrac{dt}{\sqrt{{{t}^{2}}+{{2}^{2}}}}}} \Rightarrow \int{\dfrac{{{\sec }^{2}}x}{\sqrt{{{\tan }^{2}}x+4}}dx=\log \left| t+\sqrt{{{\operatorname{t}}^{2}}+4} \right|}+C Substitute \tan x=t, \therefore \int{\dfrac{{{\sec }^{2}}x}{\sqrt{{{\tan }^{2}}x+4}}dx=\log \left| \tan x+\sqrt{{{\tan }^{2}}x+4} \right|}+C 10. Solve the following: \dfrac{1}{\sqrt{{{x}^{2}}+2x+2}}. Ans: Given expression \dfrac{1}{\sqrt{{{x}^{2}}+2x+2}}. Given expression can be written as \dfrac{1}{\sqrt{{{x}^{2}}+2x+2}}=\dfrac{1}{\sqrt{{{\left( x+1 \right)}^{2}}+{{\left( 1 \right)}^{2}}}} Let x+1=t \therefore dx=dt Integration of given expression is \Rightarrow \int{\dfrac{1}{\sqrt{{{x}^{2}}+2x+2}}dx}=\int{\dfrac{1}{\sqrt{{{t}^{2}}+1}}dt} \Rightarrow \int{\dfrac{1}{\sqrt{{{x}^{2}}+2x+2}}dx}=\log \left| t+\sqrt{{{t}^{2}}+1} \right|+C Substitute x+1=t, \Rightarrow \int{\dfrac{1}{\sqrt{{{x}^{2}}+2x+2}}dx}=\log \left| \left( x+1 \right)+\sqrt{{{\left( x+1 \right)}^{2}}+1} \right|+C \therefore \int{\dfrac{1}{\sqrt{{{x}^{2}}+2x+2}}dx}=\log \left| \left( x+1 \right)+\sqrt{{{x}^{2}}+2x+2} \right|+C 11. Solve the following: \dfrac{1}{\sqrt{9{{x}^{2}}+6x+5}}. Ans: Given expression \dfrac{1}{\sqrt{9{{x}^{2}}+6x+5}}. Given expression can be written as \dfrac{1}{\sqrt{9{{x}^{2}}+6x+5}}=\dfrac{1}{\sqrt{{{\left( 3x+1 \right)}^{2}}+{{\left( 2 \right)}^{2}}}} Let 3x+1=t \therefore 3dx=dt Integration of given expression is \Rightarrow \int{\dfrac{1}{\sqrt{9{{x}^{2}}+6x+5}}dx=\dfrac{1}{3}\int{\dfrac{1}{\sqrt{{{t}^{2}}+{{2}^{2}}}}dt}} \Rightarrow \int{\dfrac{1}{\sqrt{9{{x}^{2}}+6x+5}}dx=\dfrac{1}{3}\left[ \dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{t}{2} \right) \right]+C} Substitute 3x+1=t, \therefore \int{\dfrac{1}{\sqrt{9{{x}^{2}}+6x+5}}dx=\dfrac{1}{3}\left[ \dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{3x+1}{2} \right) \right]+C} 12. Solve the following: \dfrac{1}{\sqrt{7-6x-{{x}^{2}}}}. Ans: Given expression \dfrac{1}{\sqrt{7-6x-{{x}^{2}}}}. Given expression can be written as \Rightarrow \dfrac{1}{\sqrt{7-6x-{{x}^{2}}}}=\dfrac{1}{\sqrt{7-\left( {{x}^{2}}+6x+9-9 \right)}} \Rightarrow \dfrac{1}{\sqrt{7-6x-{{x}^{2}}}}=\dfrac{1}{\sqrt{16-\left( {{x}^{2}}+6x+9 \right)}} \Rightarrow \dfrac{1}{\sqrt{7-6x-{{x}^{2}}}}=\dfrac{1}{\sqrt{16-{{\left( x+3 \right)}^{2}}}} \Rightarrow \dfrac{1}{\sqrt{7-6x-{{x}^{2}}}}=\dfrac{1}{\sqrt{{{4}^{2}}-{{\left( x+3 \right)}^{2}}}} Let x+3=t \therefore dx=dt Integration of given expression is \Rightarrow \int{\dfrac{1}{\sqrt{7-6x-{{x}^{2}}}}dx}=\int{\dfrac{1}{\sqrt{{{4}^{2}}-{{\left( t \right)}^{2}}}}dt} \Rightarrow \int{\dfrac{1}{\sqrt{7-6x-{{x}^{2}}}}dx}={{\sin }^{-1}}\left( \dfrac{t}{4} \right)+C Substitute x+3=t, \therefore \int{\dfrac{1}{\sqrt{7-6x-{{x}^{2}}}}dx}={{\sin }^{-1}}\left( \dfrac{x+3}{4} \right)+C 13. Solve the following: \dfrac{1}{\sqrt{\left( x-1 \right)\left( x-2 \right)}}. Ans: Given expression \dfrac{1}{\sqrt{\left( x-1 \right)\left( x-2 \right)}}. Given expression can be written as \Rightarrow \dfrac{1}{\sqrt{\left( x-1 \right)\left( x-2 \right)}}=\dfrac{1}{\sqrt{{{x}^{2}}-3x+2}} \Rightarrow \dfrac{1}{\sqrt{\left( x-1 \right)\left( x-2 \right)}}=\dfrac{1}{\sqrt{{{x}^{2}}-3x+\dfrac{9}{4}-\dfrac{9}{4}+2}} \Rightarrow \dfrac{1}{\sqrt{\left( x-1 \right)\left( x-2 \right)}}=\dfrac{1}{\sqrt{{{\left( x-\dfrac{3}{2} \right)}^{2}}-{{\left( \dfrac{1}{2} \right)}^{2}}}} Let x-\dfrac{3}{2}=t \therefore dx=dt \Rightarrow \int{\dfrac{1}{\sqrt{\left( x-1 \right)\left( x-2 \right)}}dx}=\int{\dfrac{1}{\sqrt{{{t}^{2}}-{{\left( \dfrac{1}{2} \right)}^{2}}}}dx} \Rightarrow \int{\dfrac{1}{\sqrt{\left( x-1 \right)\left( x-2 \right)}}dx}=\log \left| t+\sqrt{{{t}^{2}}-{{\left( \dfrac{1}{2} \right)}^{2}}} \right|+C Substitute x-\dfrac{3}{2}=t, \therefore \int{\dfrac{1}{\sqrt{\left( x-1 \right)\left( x-2 \right)}}dx}=\log \left| \left( x-\dfrac{3}{2} \right)+\sqrt{{{x}^{2}}-3x+2} \right|+C 14. Solve the following: \dfrac{1}{\sqrt{8+3x-{{x}^{2}}}}. Ans: Given expression \dfrac{1}{\sqrt{8+3x-{{x}^{2}}}}. Given expression can be written as \Rightarrow \dfrac{1}{\sqrt{8+3x-{{x}^{2}}}}=\dfrac{1}{\sqrt{8-\left( {{x}^{2}}-3x+\dfrac{9}{4}-\dfrac{9}{4} \right)}} \Rightarrow \dfrac{1}{\sqrt{8+3x-{{x}^{2}}}}=\dfrac{1}{\sqrt{8-\left( {{x}^{2}}-3x+\dfrac{9}{4}-\dfrac{9}{4} \right)}} \Rightarrow \dfrac{1}{\sqrt{8+3x-{{x}^{2}}}}=\dfrac{1}{\sqrt{\dfrac{41}{4}-{{\left( x-\dfrac{3}{2} \right)}^{2}}}} Let x-\dfrac{3}{2}=t \therefore dx=dt \Rightarrow \int{\dfrac{1}{\sqrt{8+3x-{{x}^{2}}}}dx}=\int{\dfrac{1}{\sqrt{{{\left( \dfrac{\sqrt{41}}{2} \right)}^{2}}-{{\left( t \right)}^{2}}}}dt} \Rightarrow \int{\dfrac{1}{\sqrt{8+3x-{{x}^{2}}}}dx}={{\sin }^{-1}}\left( \dfrac{t}{\dfrac{\sqrt{41}}{2}} \right)+C Substitute x-\dfrac{3}{2}=t \Rightarrow \int{\dfrac{1}{\sqrt{8+3x-{{x}^{2}}}}dx}={{\sin }^{-1}}\left( \dfrac{x-\dfrac{3}{2}}{\dfrac{\sqrt{41}}{2}} \right)+C \therefore \int{\dfrac{1}{\sqrt{8+3x-{{x}^{2}}}}dx}={{\sin }^{-1}}\left( \dfrac{2x-3}{\sqrt{41}} \right)+C 15. Solve the following: \dfrac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}. Ans: Given expression \dfrac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}. Given expression can be written as \Rightarrow \dfrac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}=\dfrac{1}{\sqrt{{{x}^{2}}-\left( a+b \right)x+ab}} \Rightarrow \dfrac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}=\dfrac{1}{\sqrt{{{x}^{2}}-\left( a+b \right)x+\dfrac{{{\left( a+b \right)}^{2}}}{4}-\dfrac{{{\left( a+b \right)}^{2}}}{4}+ab}} \Rightarrow \dfrac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}=\dfrac{1}{\sqrt{{{\left[ x-\dfrac{\left( a+b \right)}{4} \right]}^{2}}-\dfrac{{{\left( a+b \right)}^{2}}}{4}}} \Rightarrow \dfrac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}=\dfrac{1}{\sqrt{{{\left[ x-\dfrac{\left( a+b \right)}{4} \right]}^{2}}-{{\left( \dfrac{a+b}{2} \right)}^{2}}}} Integration of given expression is \Rightarrow \int{\dfrac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}dx}=\int{\dfrac{1}{\sqrt{{{\left[ x-\dfrac{\left( a+b \right)}{4} \right]}^{2}}-{{\left( \dfrac{a+b}{2} \right)}^{2}}}}dx} Let x-\left( \dfrac{a+b}{2} \right)=t \therefore dx=dt \Rightarrow \int{\dfrac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}dx}=\int{\dfrac{1}{\sqrt{{{t}^{2}}-{{\left( \dfrac{a+b}{2} \right)}^{2}}}}dx} \Rightarrow \int{\dfrac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}dx}=\log \left| t+\sqrt{{{t}^{2}}-{{\left( \dfrac{a+b}{2} \right)}^{2}}} \right|+C Substitute x-\left( \dfrac{a+b}{2} \right)=t, \therefore \int{\dfrac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}dx}=\log \left| x-\left( \dfrac{a+b}{2} \right)+\sqrt{\left( x-a \right)\left( x-b \right)} \right|+C 16. \dfrac{4x+1}{\sqrt{2{{x}^{2}}+x-3}} Ans:Consider 4x+1=A\dfrac{d}{dx}\left( 2{{x}^{2}}+x-3 \right)+B Simplifying, \Rightarrow 4x+1=A(4x+1)+B \Rightarrow 4x+1=4Ax+A+B We obtain the below values by equating the coefficients of \text{x} and the constant term on both sides. 4~\text{A}=4\Rightarrow \text{A}=1 \text{A}+\text{B}=1\Rightarrow \text{B}=0 Consider 2{{x}^{2}}+x-3=t \therefore (4x+1)dx=dt \Rightarrow \int{\dfrac{4x+1}{\sqrt{2{{x}^{2}}+x-3}}}dx=\int{\dfrac{1}{\sqrt{t}}}dt Using the power rule of integration, =2\sqrt{t}+C Substitute the value of t, =2\sqrt{2{{x}^{2}}+x-3}+C 17. \dfrac{x+2}{\sqrt{{{x}^{2}}-1}} Ans:Consider \text{x}+2=A\dfrac{d}{dx}\left( {{x}^{2}}-1 \right)+B \Rightarrow x+2=A(2x)+B......\left( 1 \right) We obtain the below values by equating the coefficients of x and the constant term on both sides. 2A=1\Rightarrow A=\dfrac{1}{2} B=2 From (1), we get (x+2)=\dfrac{1}{2}(2x)+2 \int{\dfrac{x+2}{\sqrt{{{x}^{2}}-1}}}dx=\int{\dfrac{\dfrac{1}{2}(2x)+2}{\sqrt{{{x}^{2}}-1}}}dx =\dfrac{1}{2}\int{\dfrac{2x}{\sqrt{{{x}^{2}}-1}}}dx+\int{\dfrac{2}{\sqrt{{{x}^{2}}-1}}}dx \text{In }\dfrac{1}{2}\int{\dfrac{2x}{\sqrt{{{x}^{2}}-1}}}dx\text{ let }{{x}^{2}}-1=t\Rightarrow 2xdx=dt \dfrac{1}{2}\int{\dfrac{2x}{\sqrt{{{x}^{2}}-1}}}dx=\dfrac{1}{2}\int{\dfrac{dt}{\sqrt{t}}}\text{ } Integrating using the power rule =\dfrac{1}{2}[2\sqrt{t}] Simplifying, =\sqrt{t}\text{ } Substitute the value of t, =\sqrt{{{x}^{2}}-1} Then, \int{\dfrac{2}{\sqrt{{{x}^{2}}-1}}}dx=2\int{\dfrac{1}{\sqrt{{{x}^{2}}-1}}}dx=2\log \left| x+\sqrt{{{x}^{2}}-1} \right| From equation (2), we get \int{\dfrac{x+2}{\sqrt{{{x}^{2}}-1}}}dx=\sqrt{{{x}^{2}}-1}+2\log \left| x+\sqrt{{{x}^{2}}-1} \right|+C 18. \dfrac{5x-2}{1+2x+3{{x}^{2}}} Ans: Let 5x-2=A\dfrac{d}{dx}\left( 1+2x+3{{x}^{2}} \right)+B\text{ } \Rightarrow \text{ }5\text{ }x-2=A\left( 2+6\text{ }x \right)+B......\left( 1 \right) We obtain the below values by equating the coefficients of x and the constant term on both sides. 5=6A\Rightarrow A=\dfrac{5}{6} 2A+B=-2\Rightarrow B=-\dfrac{11}{3} Substitute the above values in (1) \therefore 5x-2=\dfrac{5}{6}(2+6x)+\left( -\dfrac{11}{3} \right) \Rightarrow\int{\dfrac{5x-2}{1+2x+3{{x}^{2}}}}dx=\int{\dfrac{\dfrac{5}{6}(2+6x)-\dfrac{11}{3}}{1+2x+3{{x}^{2}}}}dx =\dfrac{5}{6}\int{\dfrac{2+6x}{1+2x+3{{x}^{2}}}}dx-\dfrac{11}{3}\int{\dfrac{1}{1+2x+3{{x}^{2}}}}dx Consider {{I}_{1}}=\int{\dfrac{2+6x}{1+2x+3{{x}^{2}}}}dx and {{I}_{2}}=\int{\dfrac{1}{1+2x+3{{x}^{2}}}}dx \therefore\int{\dfrac{5x-2}{1+2x+3{{x}^{2}}}}dx=\dfrac{5}{6}{{I}_{1}}-\dfrac{11}{3}{{I}_{2}}...\left( 1 \right) {{I}_{1}}=\int{\dfrac{2+6x}{1+2x+3{{x}^{2}}}}dx Put 1+2x+3{{x}^{2}}=t \Rightarrow (2+6x)dx=dt \therefore {{I}_{1}}=\int{\dfrac{dt}{t}} Using the logarithm formula of integration, {{I}_{1}}=\log |t|\text{ } Substitute the value of t, {{I}_{1}}=\log \left| 1+2x+3{{x}^{2}} \right|...\left( 2 \right) Then, {{I}_{2}}=\int{\dfrac{1}{1+2x+3{{x}^{2}}}}dx 1+2x+3{{x}^{2}} can be rewritten as 1+3\left( {{x}^{2}}+\dfrac{2}{3}x \right) Thus, 1+3\left( {{x}^{2}}+\dfrac{2}{3}x \right) By completing square method, =1+3\left( {{x}^{2}}+\dfrac{2}{3}x+\dfrac{1}{9}-\dfrac{1}{9} \right) =1+3{{\left( x+\dfrac{1}{3} \right)}^{2}}-\dfrac{1}{3} Simplifying, =\dfrac{2}{3}+3{{\left( x+\dfrac{1}{3} \right)}^{2}} =3\left[ {{\left( x+\dfrac{1}{3} \right)}^{2}}+\dfrac{2}{9} \right] =3\left[ {{\left( x+\dfrac{1}{3} \right)}^{2}}+{{\left( \dfrac{\sqrt{2}}{3} \right)}^{2}} \right] Therefore {{I}_{2}}can be rewritten as , {{I}_{2}}=\dfrac{1}{3}\int{\dfrac{1}{\left. {{\left[ x+\dfrac{1}{3} \right)}^{2}}+{{\left( \dfrac{\sqrt{2}}{3} \right)}^{2}} \right]}}dx =\dfrac{1}{3}\left[ \dfrac{3}{\sqrt{2}}{{\tan }^{-1}}\left( \dfrac{3x+1}{\sqrt{2}} \right) \right] Simplifying, =\dfrac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \dfrac{3x+1}{\sqrt{2}} \right)...\left( 3 \right) We obtain the below values by substituting equations (2) and (3) in equation (1) \int{\dfrac{5x-2}{1+2x+3{{x}^{2}}}}dx=\dfrac{5}{6}\left[ \log \left| 1+2x+3{{x}^{2}} \right| \right]-\dfrac{11}{3}\left[ \dfrac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \dfrac{3x+1}{\sqrt{2}} \right) \right]+C Simplifying, =\dfrac{5}{6}\log \left| 1+2x+3{{x}^{2}} \right|-\dfrac{11}{3\sqrt{2}}{{\tan }^{-1}}\left( \dfrac{3x+1}{\sqrt{2}} \right)+C 19. \dfrac{6x+7}{\sqrt{\left( x-5 \right)\left( x-4 \right)}} Ans: Consider 6x+7=A\dfrac{d}{dx}\left( {{x}^{2}}-9x+20 \right)+B Differentiating, \Rightarrow 6\text{ }x+7=A\left( 2\text{ }x-9 \right)+B We obtain the below values by equating the coefficients of x and the constant term on both sides. 2~\text{A}=6\Rightarrow \text{A}=3 -9~\text{A}+\text{B}=7\Rightarrow \text{B}=34 \therefore 6\text{x}+7=3(2\text{x}-9)+34 \int{\dfrac{6x+7}{\sqrt{{{x}^{2}}-9x+20}}}=\int{\dfrac{3(2x-9)+34}{\sqrt{{{x}^{2}}-9x+20}}}dx =3\int{\dfrac{2x-9}{\sqrt{{{x}^{2}}-9x+20}}}dx+34\int{\dfrac{1}{\sqrt{{{x}^{2}}-9x+20}}}dx Consider {{I}_{1}}=\int{\dfrac{2x-9}{\sqrt{{{x}^{2}}-9x+20}}}dx\,\,\,and\,\,{{I}_{2}}=\int{\dfrac{1}{\sqrt{{{x}^{2}}-9x+20}}}dx \therefore\int{\dfrac{6x+7}{\sqrt{{{x}^{2}}-9x+20}}}=3{{I}_{1}}+34{{I}_{2}}\,\,\,\,\,\,\,\,....(1) {{I}_{1}}=\int{\dfrac{2x-9}{\sqrt{{{x}^{2}}-9x+20}}}dx Put {{x}^{2}}-9x+20=t \Rightarrow (2x-9)dx=dt \Rightarrow {{I}_{1}}=\int{\dfrac{dt}{\sqrt{t}}} Integrating using the power rule {{I}_{1}}=2\sqrt{t} Substitute the value of t, {{I}_{1}}=2\sqrt{{{x}^{2}}-9x+20}\,\,\,.....(2) {{I}_{2}}=\int{\dfrac{1}{\sqrt{{{x}^{2}}-9x+20}}}dx Consider {{x}^{2}}-9x+20 By completing square methods, ={{x}^{2}}-9x+20+\dfrac{81}{4}-\dfrac{81}{4} ={{\left( x-\dfrac{9}{2} \right)}^{2}}-\dfrac{1}{4} ={{\left( x-\dfrac{9}{2} \right)}^{2}}-{{\left( \dfrac{1}{2} \right)}^{2}} \Rightarrow {{I}_{2}}=\int{\dfrac{1}{{{\left( x-\dfrac{9}{2} \right)}^{2}}-{{\left( \dfrac{1}{2} \right)}^{2}}}}\,\,dx {{I}_{2}}=\log \left| \left( x-\dfrac{9}{2} \right)+\sqrt{{{x}^{2}}-9x+20} \right|.....\left( 3 \right) We obtain the below values by substituting equations (2) and (3) in (1), \int{\dfrac{6x+7}{\sqrt{{{x}^{2}}-9x+20}}}dx=3\left[ 2\sqrt{{{x}^{2}}-9x+20} \right]+34\log \left[ \left( x-\dfrac{9}{2} \right)+\sqrt{{{x}^{2}}-9x+20} \right]+\text{C} Simplifying, =6\sqrt{{{x}^{2}}-9x+20}+34\log \left[ \left( x-\dfrac{9}{2} \right)+\sqrt{{{x}^{2}}-9x+20} \right]+C 20. \dfrac{x+2}{\sqrt{4x-{{x}^{2}}}} Ans:Consider, \text{x}+2=A\dfrac{d}{dx}\left( 4x-{{x}^{2}} \right)+B \Rightarrow x+2=A(4-2x)+B We obtain the below values by equating the coefficients of x and the constant term on both sides. -2A=1\Rightarrow A=-\dfrac{1}{2} 4A+B=2\Rightarrow B=4 \Rightarrow (x+2)=-\dfrac{1}{2}(4-2x)+4 \therefore\int{\dfrac{x+2}{\sqrt{4x-{{x}^{2}}}}}dx=\int{\dfrac{-\dfrac{1}{2}(4-2x)+4}{\sqrt{4x-{{x}^{2}}}}}dx =-\dfrac{1}{2}\int{\dfrac{4-2x}{\sqrt{4x-{{x}^{2}}}}}dx+4\int{\dfrac{1}{\sqrt{4x-{{x}^{2}}}}}dx Let {{I}_{1}}=\int{\dfrac{4-2x}{\sqrt{4x-{{x}^{2}}}}}dx and {{I}_{2}}\int{\dfrac{1}{\sqrt{4x-{{x}^{2}}}}}dx \therefore \int{\dfrac{x+2}{\sqrt{4x-{{x}^{2}}}}}dx=-\dfrac{1}{2}{{I}_{1}}\text{ and }+4{{I}_{2}}\,\,\,\,\,\,\,....\left( 1 \right) Then, {{I}_{1}}=\int{\dfrac{4-2x}{\sqrt{4x-{{x}^{2}}}}}dx Let 4x-{{x}^{2}}=t \Rightarrow (4-2x)dx=dt \Rightarrow{{I}_{1}}=\int{\dfrac{dt}{\sqrt{t}}}=2\sqrt{t}=2\sqrt{4x-{{x}^{2}}}...\left( 2 \right) (Using the logarithm formula of integration,) {{I}_{2}}=\int{\dfrac{1}{\sqrt{4x-{{x}^{2}}}}}dx Integrating using the power rule, \Rightarrow 4x-{{x}^{2}}=-\left( -4x+{{x}^{2}} \right) By completing square methods, =\left( -4x+{{x}^{2}}+4-4 \right) =4-{{(x-2)}^{2}} ={{(2)}^{2}}-{{(x-2)}^{2}} \therefore {{I}_{2}}=\int{\dfrac{1}{\sqrt{{{(2)}^{2}}-{{(x-2)}^{2}}}}}dx={{\sin }^{-1}}\left( \dfrac{x-2}{2} \right)...\left( 3 \right) Using equations (2) and (3) in (1), to obtain \int{\dfrac{x+2}{\sqrt{4x-{{x}^{2}}}}}dx=-\dfrac{1}{2}\left( 2\sqrt{4x-{{x}^{2}}} \right)+4{{\sin }^{-1}}\left( \dfrac{x-2}{2} \right)+C =-\sqrt{4x-{{x}^{2}}}+4{{\sin }^{-1}}\left( \dfrac{x-2}{2} \right)+C 21.\dfrac{x+2}{\sqrt{{{x}^{2}}+2x+3}} Ans:\int{\dfrac{x+2}{\sqrt{{{x}^{2}}+2x+3}}}dx=\dfrac{1}{2}\int{\dfrac{2(x+2)}{\sqrt{{{x}^{2}}+2x+3}}}dx Simplifying, =\dfrac{1}{2}\int{\dfrac{2x+4}{\sqrt{{{x}^{2}}+2x+3}}}dx =\dfrac{1}{2}\int{\dfrac{2x+2}{\sqrt{{{x}^{2}}+2x+3}}}dx+\dfrac{1}{2}\int{\dfrac{2}{\sqrt{{{x}^{2}}+2x+3}}}dx =\dfrac{1}{2}\int{\dfrac{2x+2}{\sqrt{{{x}^{2}}+2x+3}}}dx+\int{\dfrac{1}{\sqrt{{{x}^{2}}+2x+3}}}dx Let {{I}_{1}}=\int{\dfrac{2x+2}{\sqrt{{{x}^{2}}+2x+3}}}dx\,\,\,and\,\,{{I}_{2}}=\int{\dfrac{1}{\sqrt{{{x}^{2}}+2x+3}}}dx \therefore \int{\dfrac{x+2}{\sqrt{{{x}^{2}}+2x+3}}}dx=\dfrac{1}{2}{{I}_{1}}+{{I}_{2}}\,\,\,\,\,\,\,\,....\left( 1 \right) Then, {{I}_{1}}=\int{\dfrac{2x+2}{\sqrt{{{x}^{2}}+2x+3}}}dx Put, {{x}^{2}}+2x+3=t Integrating using the power rule, \Rightarrow (2\text{x}+2)\text{dx}=\text{dt}\,\,\,\,\,\, \,{{I}_{1}}=\int{\dfrac{dt}{\sqrt{t}}}=2\sqrt{t}=2\sqrt{{{x}^{2}}+2x+3}..\left( 2 \right) {{I}_{2}}=\int{\dfrac{1}{\sqrt{{{x}^{2}}+2x+3}}}dx. By completing square methods, \Rightarrow {{x}^{2}}+2x+3={{x}^{2}}+2x+1+2={{(x+1)}^{2}}+{{(\sqrt{2})}^{2}} \therefore {{I}_{2}}=\int{\dfrac{1}{\sqrt{{{(x+1)}^{2}}+{{(\sqrt{2})}^{2}}}}}dx=\log \left| (x+1)+\sqrt{{{x}^{2}}+2x+3} \right|...\left( 3 \right) Using equations (2) and (3) in (1), to obtain \therefore\int{\dfrac{x+2}{\sqrt{{{x}^{2}}+2x+3}}}dx=\sqrt{{{x}^{2}}+2x+3}+\log \left| (x+1)+\sqrt{{{x}^{2}}+2x+3} \right|\,+C 22. \dfrac{x+3}{{{x}^{2}}-2x-5} Ans: Consider (x+3)=A\dfrac{d}{dx}\left( {{x}^{2}}-2x-5 \right)+B =\dfrac{{{x}^{2}}}{2}{{(\log x)}^{2}}-\int{x}\log xdx We obtain the below values by equating the coefficients of x and the constant term on both sides. 2A=1\Rightarrow A=\dfrac{1}{2} -2A+B=3\Rightarrow B=4 \therefore (x+3)=\dfrac{1}{2}(2x-2)+4 \Rightarrow \int{\dfrac{x+3}{{{x}^{2}}-2x-5}}dx=\int{\dfrac{\dfrac{1}{2}(2x-2)+4}{{{x}^{2}}-2x-5}}dx =\dfrac{1}{2}\int{\dfrac{2x-2}{{{x}^{2}}-2x-5}}dx+4\int{\dfrac{1}{{{x}^{2}}-2x-5}}~\text{d}x Consider{{I}_{1}}=\int{\dfrac{2x-2}{{{x}^{2}}-2x-5}}dx and {{I}_{2}}=\int{\dfrac{1}{{{x}^{2}}-2x-5}}dx \therefore \int{\dfrac{x+3}{{{x}^{2}}-2x-5}}dx=\dfrac{1}{2}{{I}_{1}}+4{{I}_{2}}...\left( 1 \right) Then, {{I}_{1}}=\int{\dfrac{2x-2}{{{x}^{2}}-2x-5}}dx Put {{x}^{2}}-2x-5=\text{t} \Rightarrow (2x-2)dx=dt Using the logarithm formula of integration, \Rightarrow {{I}_{1}}=\int{\dfrac{dt}{t}}=\log |t|=\log \left| {{x}^{2}}-2x-5 \right|\,\,\,....\left( 2 \right) {{I}_{2}}=\int{\dfrac{1}{{{x}^{2}}-2x-5}}dx =\int{\dfrac{1}{\left( {{x}^{2}}-2x+1 \right)-6}}dx =\int{\dfrac{1}{{{(x-1)}^{2}}+{{(\sqrt{6})}^{2}}}}dx =\dfrac{1}{2\sqrt{6}}\log \left( \dfrac{x-1-\sqrt{6}}{x-1+\sqrt{6}} \right)....\left( 3 \right) We obtain the below values by substituting (2) and (3) in (1), \int{\dfrac{x+3}{{{x}^{2}}-2x-5}}dx=\dfrac{1}{2}\log \left| {{x}^{2}}-2x-5 \right|+\dfrac{4}{2\sqrt{6}}\log \left| \dfrac{x-1-\sqrt{6}}{x-1+\sqrt{6}} \right|+C =\dfrac{1}{2}\log \left| {{x}^{2}}-2x-5 \right|+\dfrac{2}{\sqrt{6}}\log \left| \dfrac{x-1-\sqrt{6}}{x-1+\sqrt{6}} \right|+C 23. \dfrac{5x+3}{\sqrt{{{x}^{2}}+4x+10}} Ans: 5x+3=A\dfrac{a}{dx}\left( {{x}^{2}}+4x+10 \right)+B \Rightarrow 5x+3=A(2x+4)+B Equating the coefficients of \text{x}and constant term, we get 2A=5\Rightarrow A=\dfrac{5}{2} 4A+B=3\Rightarrow B=-7 \therefore 5x+3=\dfrac{5}{2}(2x+4)-7 \Rightarrow \int{\dfrac{5x+3}{\sqrt{{{x}^{2}}+4x+10}}}dx=\int{\dfrac{\dfrac{5}{2}(2x+4)-7}{\sqrt{{{x}^{2}}+4x+10}}}dx =\dfrac{5}{2}\int{\dfrac{2x+4}{\sqrt{{{x}^{2}}+4x+10}}}dx-7\int{\dfrac{1}{\sqrt{{{x}^{2}}+4x+10}}}dx Let {{I}_{1}}=\int{\dfrac{2x+4}{\sqrt{{{x}^{2}}+4x+10}}}dx and {{I}_{2}}=\int{\dfrac{1}{\sqrt{{{x}^{2}}+4x+10}}}dx \therefore \int{\dfrac{5x+3}{\sqrt{{{x}^{2}}+4x+10}}}dx=\dfrac{5}{2}{{I}_{1}}-7{{I}_{2}}\,\,\,\,\,\,\,\,\,....\left( 1 \right) Then, {{I}_{1}}=\int{\dfrac{2x+4}{\sqrt{{{x}^{2}}+4x+10}}}dx Consider {{x}^{2}}+4x+10=\text{t}\therefore (2x+4)dx=dt \Rightarrow {{I}_{1}}=\int{\dfrac{dt}{t}}=2\sqrt{t}=2\sqrt{{{x}^{2}}+4x+10}\,\,\,\,\,\,\,......\left( 2 \right) {{I}_{2}}=\int{\dfrac{1}{\sqrt{{{x}^{2}}+4x+10}}}dx =\int{\dfrac{1}{\sqrt{\left( {{x}^{2}}+4x+4 \right)+6}}}dx=\int{\dfrac{1}{{{(x+2)}^{2}}+{{(\sqrt{6})}^{2}}}}dx =\log \left| (x+2)\sqrt{{{x}^{2}}+4x+10} \right|...\left( 3 \right) We obtain the below values by using equations (2) and (3) in (1). \int{\dfrac{5x+3}{\sqrt{{{x}^{2}}+4x+10}}}dx=\dfrac{5}{2}\left[ 2\sqrt{{{x}^{2}}+4x+10} \right]-7\log \left| (x+2)\sqrt{{{x}^{2}}+4x+10} \right|+C$$=5\sqrt{{{x}^{2}}+4x+10}-7\log \left| (x+2)\sqrt{{{x}^{2}}+4x+10} \right|+C$

24. $\int{\dfrac{dx}{{{x}^{2}}+2x+2}}$equals

1. $x{{\tan }^{-1}}\left( x+1 \right)+C$

2. ${{\tan }^{-1}}\left( x+1 \right)+C$

3. $\left( x+1 \right){{\tan }^{-1}}x+C$

4. ${{\tan }^{-1}}x+C$

Ans: $\int{\dfrac{dx}{{{x}^{2}}+2x+2}}=\int{\dfrac{dx}{\left( {{x}^{2}}+2x+1 \right)+1}}$

$=\int{\dfrac{1}{{{(x+1)}^{2}}+{{(1)}^{2}}}}dx$

$=\left[ {{\tan }^{-1}}(x+1) \right]+C$

Hence, the right response is is B.

25. $\int{\dfrac{dx}{\sqrt{9x-4{{x}^{2}}}}}$ equals

1. $\dfrac{1}{9}{{\sin }^{-1}}\left( \dfrac{9x-8}{8} \right)+C$

2. $\dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{8x-9}{9} \right)+C$

3. $\dfrac{1}{3}{{\sin }^{-1}}\left( \dfrac{9x-8}{8} \right)+C$

4. $\dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{9x-8}{9} \right)+C$

Ans:

$\int{\dfrac{dx}{\sqrt{9x-4{{x}^{2}}}}}$

$=\int{\dfrac{1}{\sqrt{-4\left( {{x}^{2}}-\dfrac{9}{4}x \right)}}}dx$

By completing square methods,

$=\int{\dfrac{1}{-4\left( {{x}^{2}}-\dfrac{9}{4}x+\dfrac{81}{64}-\dfrac{81}{64} \right)}}dx$

$=\int{\dfrac{1}{\sqrt{-4\left[ {{\left( x-\dfrac{9}{8} \right)}^{2}}-{{\left( \dfrac{9}{8} \right)}^{2}} \right]}}}dx$

$=\dfrac{1}{2}\int{\dfrac{1}{{{\left( \dfrac{9}{8} \right)}^{2}}-{{\left( x-\dfrac{9}{8} \right)}^{2}}}}dx$

$=\dfrac{1}{2}\left[ {{\sin }^{-1}}\left( \dfrac{x-\dfrac{9}{8}}{\dfrac{9}{8}} \right) \right]+C\quad \left( \int{\dfrac{dy}{\sqrt{{{a}^{2}}-{{y}^{2}}}}}={{\sin }^{-1}}\dfrac{y}{a}+C \right)$

Simplifying,

$=\dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{8x-9}{9} \right)+C$

Hence, the right response is B.

### Exercise 7.5

1. $\dfrac{x}{\left( x+1 \right)\left( x+2 \right)}$

Ans:   Let $\dfrac{x}{(x+1)(x+2)}=\dfrac{A}{(x+1)}+\dfrac{B}{(x+2)}$

$\Rightarrow x=A(x+2)+B(x+1)$

We obtain the below values by equating the coefficients of x  and the constant term on both sides.

$\text{A}+\text{B}=1$

$2~\text{A}+\text{B}=0$

On solving, we get

$\text{A}=-1\text{ and B}=2$

$\therefore \dfrac{x}{(x+1)(x+2)}=\dfrac{-1}{(x+1)}+\dfrac{2}{(x+2)}$

$\Rightarrow \int{\dfrac{x}{(x+1)(x+2)}}dx=\int{\dfrac{-1}{(x+1)}}+\dfrac{2}{(x+2)}dx$

Using the logarithm formula of integration,

$=-\log |x+1|+2\log |x+2|+C$

$=\log {{(x+2)}^{2}}-\log |x+1|+C$

Simplifying,

$=\log \dfrac{{{(x+2)}^{2}}}{(x+1)}+C$

2. $\dfrac{1}{{{x}^{2}}-9}$

Ans:  Let $\dfrac{1}{(x+3)(x-3)}=\dfrac{A}{(x+3)}+\dfrac{B}{(x-3)}$

$1=A\left( x-3 \right)+B\left( x+3 \right)$

Equating the coefficients of $\text{x}$and constant term, we get

$A+B=0$

$1=-3A+3B$

$A=-\dfrac{1}{6}\text{ and }B=\dfrac{1}{6}$

$\therefore \dfrac{1}{(x+3)(x-3)}=\dfrac{-1}{6(x+3)}+\dfrac{1}{6(x-3)}$

$\Rightarrow \int{\dfrac{1}{\left( {{x}^{2}}-9 \right)}}dx=\int{\left( \dfrac{-1}{6(x+3)}+\dfrac{1}{6(x-3)} \right)}dx$

Using the logarithm formula of integration,

$=-\dfrac{1}{6}\log |x+3|+\dfrac{1}{6}\log |x-3|+C=\dfrac{1}{6}\log \dfrac{|(x-3)|}{|(x+3)|}+C$

3. $\dfrac{3x-1}{\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)}$

Ans: Let $\dfrac{3x-1}{(x-1)(x-2)(x-3)}=\dfrac{A}{(x-1)}+\dfrac{B}{(x-2)}+\dfrac{C}{(x-3)}$

$3x-1=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)\quad \ldots \left( 1 \right)$

We obtain the below values by equating the coefficients of x , ${{x}^{2}}$ and the constant term on both sides.

$\text{A}+\text{B}+\text{C}=0$

$-5~\text{A}-4~\text{B}-3\text{C}=3$

$6~\text{A}+3~\text{B}+2\text{C}=-1$

Solving these equations, to obtain

$\text{A}=1,~\text{B}=-5,\text{ and C}=4$

$\therefore \dfrac{3x-1}{(x-1)(x-2)(x-3)}=\dfrac{1}{(x-1)}-\dfrac{5}{(x-2)}+\dfrac{4}{(x-3)}$

$\Rightarrow \int{\dfrac{3x-1}{(x-1)(x-2)(x-3)}}dx=\int{\left\{ \dfrac{1}{(x-1)}-\dfrac{5}{(x-2)}+\dfrac{4}{(x-3)} \right\}}dx$

Using the logarithm formula of integration,

$=\log |x-1|-5\log |x-2|+4\log |x-3|+C$

4. $\dfrac{x}{\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)}$

Ans:Let $\dfrac{x}{(x-1)(x-2)(x-3)}=\dfrac{A}{(x-1)}+\dfrac{B}{(x-2)}+\dfrac{C}{(x-3)}$

$x=A\left( x-2 \right)\left( x-3 \right)+B\left( x-1 \right)\left( x-3 \right)+C\left( x-1 \right)\left( x-2 \right)$

We obtain the below values by equating the coefficients of x , ${{x}^{2}}$ and the constant term on both sides.

$A+B+C=0\text{ }$

$-5\text{ }A-4\text{ }B-3\text{ }C=1$

$6\text{ }A+4\text{ }B+2\text{ }C=0$

Solving these equations, to obtain

$A=\dfrac{1}{2},B=2\text{ and }C=\dfrac{3}{2}$

$\therefore \dfrac{x}{(x-1)(x-2)(x-3)}=\dfrac{1}{2(x-1)}-\dfrac{2}{(x-2)}+\dfrac{3}{2(x-3)}$

$\Rightarrow \int{\dfrac{x}{(x-1)(x-2)(x-3)}}dx=\int{\left\{ \dfrac{1}{2(x-1)}-\dfrac{2}{(x-2)}+\dfrac{3}{2(x-3)} \right\}}dx$

Using the logarithm formula of integration,

$=\dfrac{1}{2}\log |x-1|-2\log |x-2|+\dfrac{3}{2}\log |x-3|+C$

5. $\dfrac{2x}{{{x}^{2}}+3x+2}$

Ans:$\dfrac{2x}{{{x}^{2}}+3x+2}=\dfrac{A}{(x+1)}+\dfrac{B}{(x+2)}$

$2\text{ }x=A\left( x+2 \right)+B\left( x+1 \right)$

We obtain the below values by equating the coefficients of x , ${{x}^{2}}$ and the constant term on both sides.

$A+B=2$

$2\text{ }A+B=0$

Solving these equations, we get

$A=-2\text{ and }B=4$

$\therefore \dfrac{2x}{(x+1)(x+2)}=\dfrac{-2}{(x+1)}+\dfrac{4}{(x+2)}$

$\Rightarrow \int{\dfrac{2x}{(x+1)(x+2)}}dx=\int{\left\{ \dfrac{4}{(x+2)}-\dfrac{2}{(x+1)} \right\}}dx$

Using the logarithm formula of integration,

$=4\log |x+2|-2\log |x+1|+C$

6. $\dfrac{1-{{x}^{2}}}{x\left( 1-2x \right)}$

Ans:  It can be seen that the given integrand is not a proper fraction.

Therefore, on dividing $\left( 1-{{x}^{2}} \right)$ by $x(1-2x)$ to obtain, $\dfrac{1-{{x}^{2}}}{x(1+2x)}=\dfrac{1}{2}+\dfrac{1}{2}\left( \dfrac{2-x}{x(1-2x)} \right)$

Let $\dfrac{2-x}{x(1-2x)}=\dfrac{A}{x}+\dfrac{B}{(1-2x)}...\left( 1 \right)$

$\Rightarrow (2-x)=A(1-2x)+Bx$

We obtain the below values by equating the coefficients of x , ${{x}^{2}}$ and the constant term on both sides.

$-2~\text{A}+\text{B}=-1$ and, $A=2$

Solving these equations, to obtain $A=2\text{ and }B=3$

$\therefore \dfrac{2-x}{x(1-2x)}=\dfrac{2}{x}+\dfrac{3}{1-2x}$

Substituting in equation (1), we get

$\dfrac{1-{{x}^{2}}}{x(1+2)}=\dfrac{1}{2}+\dfrac{1}{2}\left\{ \dfrac{2}{x}+\dfrac{3}{(1-2x)} \right\}$

$\Rightarrow \int{\dfrac{1-{{x}^{2}}}{x(1+2)}}dx=\int \dfrac{1}{2}+\dfrac{1}{2}\text{(}\dfrac{2}{x}+\dfrac{3}{(1-2x)})dx$

Using the power rule and logarithm formula of integration,

$=\dfrac{x}{2}+\log |x|+\dfrac{3}{2(-2)}\log |1-2x|+C=\dfrac{x}{2}+\log |x|-\dfrac{3}{4}\log |1-2x|+C$

7. $\dfrac{x}{\left( {{x}^{2}}+1 \right)\left( x-1 \right)}$

Ans: Let $\dfrac{x}{\left( {{x}^{2}}+1 \right)(x-1)}=\dfrac{Ax+B}{\left( {{x}^{2}}+1 \right)}+\dfrac{C}{(x-1)}...\left( 1 \right)$

$x=(Ax+B)(x-1)+C\left( {{x}^{2}}+1 \right)$

$x=A{{x}^{2}}-Ax+Bx-B+C{{x}^{2}}+C$

We obtain the below values by equating the coefficients of x , ${{x}^{2}}$ and the constant term on both sides.

$A+C=0\text{ }$

$-A+B=1$

$-B+C=0$

On solving these equations, to obtain  $A=-\dfrac{1}{2},B=\dfrac{1}{2},and\,\,C=\dfrac{1}{2}$

From equation (1), to obtain

$\therefore \dfrac{x}{\left( {{x}^{2}}+1 \right)(x-1)}=\dfrac{\left( -\dfrac{1}{2}x+\dfrac{1}{2} \right)}{{{x}^{2}}+1}+\dfrac{\dfrac{1}{2}}{(x-1)}$

$\Rightarrow \int{\dfrac{x}{\left( {{x}^{2}}+1 \right)(x-1)}}=-\dfrac{1}{2}\int{\dfrac{x}{{{x}^{2}}+1}}dx+\dfrac{1}{2}\int{\dfrac{1}{{{x}^{2}}+1}}dx+\dfrac{1}{2}\int{\dfrac{1}{x-1}}dx$

$=-\dfrac{1}{4}\int{\dfrac{2x}{{{x}^{2}}+1}}dx+\dfrac{1}{2}{{\tan }^{-1}}x+\dfrac{1}{2}\log |x-1|+C$

Consider $\int{\dfrac{2x}{{{x}^{2}}+1}}dx,lel\left( {{x}^{2}}+1 \right)=t\Rightarrow 2xdx=dt$

$\Rightarrow \int{\dfrac{2x}{{{x}^{2}}+1}}dx=\int{\dfrac{dt}{t}}=\log |t|=\log \left| {{x}^{2}}+1 \right|$

$\therefore \int{\dfrac{x}{\left( {{x}^{2}}+1 \right)(x-1)}}=-\dfrac{1}{4}\log \left| {{x}^{2}}+1 \right|+\dfrac{1}{2}{{\tan }^{-1}}x+\dfrac{1}{2}\log |x-1|+C$

$=\dfrac{1}{2}\log |x-1|-\dfrac{1}{4}\log \left| {{x}^{2}}+1 \right|+\dfrac{1}{2}{{\tan }^{-1}}x+C$

8. .$\dfrac{x}{{{\left( x-1 \right)}^{2}}\left( x+2 \right)}$

Ans: $\dfrac{x}{{{(x-1)}^{2}}(x+2)}$

$\text{ Let }\dfrac{x}{{{(x-1)}^{2}}(x+2)}=\dfrac{A}{(x-1)}+\dfrac{B}{{{(x-1)}^{2}}}+\dfrac{C}{(x+2)}$

$x=A(x-1)(x+2)+B(x+2)+C{{(x-1)}^{2}}$

We obtain the below values by equating the coefficients of x , ${{x}^{2}}$ and the constant term on both sides.

$A+C=0$

$A+B-2\text{ }C=1$

On solving, to obtain

$A=\dfrac{2}{9}\text{ and }C=\dfrac{-2}{9}$

$B=\dfrac{1}{3}$

$\therefore \dfrac{x}{{{(x-1)}^{2}}(x+2)}=\dfrac{2}{9(x-1)}+\dfrac{1}{3{{(x-1)}^{2}}}-\dfrac{2}{9(x\mid 2)}$

$\Rightarrow \int{\dfrac{x}{{{(x-1)}^{2}}(x+2)}}dx=\dfrac{2}{9}\int{\dfrac{1}{(x-1)}}dx+\dfrac{1}{3}\int{\dfrac{1}{{{(x-1)}^{2}}}}dx-\dfrac{2}{9}\int{\dfrac{1}{(x-2)}}dx$

Using the power rule and logarithm formula of integration,

$=\dfrac{2}{9}\log |x-1|+\dfrac{1}{3}\left( \dfrac{-1}{x-1} \right)-\dfrac{2}{9}\log |x+2|+C$

Simplifying,

$=\dfrac{2}{9}\log \left| \dfrac{x-1}{x+2} \right|-\dfrac{1}{3(x-1)}+C$

9. $\dfrac{3x+5}{{{x}^{3}}-{{x}^{2}}-x+1}$

Ans:   $\dfrac{3x+5}{{{x}^{3}}-{{x}^{2}}-x+1}=\dfrac{3x+5}{{{(x-1)}^{2}}(x+1)}$

let  $\dfrac{3x+5}{{{(x-1)}^{2}}(x+1)}=\dfrac{A}{(x+1)}+\dfrac{B}{{{(x-1)}^{2}}}+\dfrac{C}{(x+1)}$

$3x+5=A(x-1)(x+1)+B(x+1)+{{(x-1)}^{2}}$

$3x+5=A{{(x-1)}^{2}}+B(x+1)+C\left( {{x}^{2}}+1-2x \right)\quad \ldots (1)$

We obtain the below values by equating the coefficients of x , ${{x}^{2}}$ and

the constant term on both sides.

$A+C=0\text{ }$

$B-2\text{ }C=3\text{ }$

$-A+B+C=5$

On solving, to obtain   $B=4\,\,A=-\dfrac{1}{2}\,\,and\,\,C=\dfrac{1}{2}$

$\therefore \dfrac{3x+5}{{{(x-1)}^{2}}(x+1)}=\dfrac{-1}{2(x-1)}+\dfrac{4}{{{(x-1)}^{2}}}+\dfrac{1}{2(x+1)}$

$\Rightarrow \int{\dfrac{3x+5}{{{(x-1)}^{2}}(x+1)}}dx=-\dfrac{1}{2}\int{\dfrac{1}{x-1}}dx+4\int{\dfrac{1}{{{(x-1)}^{2}}}}dx+\dfrac{1}{2}\int{\dfrac{1}{(x+1)}}dx$

Using the power rule and logarithm formula of integration,

$=-\dfrac{1}{2}\log |x-1|+4\left( \dfrac{-1}{x-1} \right)+\dfrac{1}{2}\log |x+1|+C$

$=\dfrac{1}{2}\log \left| \dfrac{x+1}{x-1} \right|-\dfrac{4}{(x-1)}+C$

10. $\dfrac{2x-3}{\left( {{x}^{2}}-1 \right)\left( 2x+3 \right)}$

Ans: $\dfrac{2x-3}{\left( {{x}^{2}}-1 \right)(2x+3)}=\dfrac{2x-3}{(x+1)(x-1)(2x+3)}$

Let $\dfrac{2x-3}{(x+1)(x-1)(2x+3)}=\dfrac{A}{(x+1)}+\dfrac{B}{(x-1)}+\dfrac{C}{(2x+3)}$

$\Rightarrow (2x-3)=A(x-1)(2x+3)+B(x+1)(2x+3)+C(x+1)(x-1)$

$\Rightarrow (2x-3)-A\left( 2{{x}^{2}}+x-3 \right)+B\left( 2{{x}^{2}}+5x-3 \right)+C\left( {{x}^{2}}-1 \right)$

$\Rightarrow (2x-3)=(2A+2B+C){{x}^{2}}+(A+5B)x+(-3A+3B-C)$

We obtain the below values by equating the coefficients of x , ${{x}^{2}}$ and the constant term on both sides.

$2~\text{A}+2~\text{B}+\text{C}=0$

$\text{A}+5~\text{B}=2$

$-3~\text{A}+3~\text{B}-\text{C}=-3$

On solving, to obtain $B=-\dfrac{1}{10},A=\dfrac{5}{2},\text{ and C}=-\dfrac{24}{5}$

$\therefore \dfrac{2x-3}{(x+1)(x-1)(2x+3)}=\dfrac{5}{2(x+1)}-\dfrac{1}{10(x-1)}-\dfrac{24}{5(2x+3)}$

$\Rightarrow \int{\dfrac{2x-3}{\left( {{x}^{2}}-1 \right)(2x+3)}}dx=\dfrac{5}{2}\int{\dfrac{1}{(x+1)}}dx-\dfrac{1}{10}\int{\dfrac{1}{x-1}}dx-\dfrac{24}{5}\int{\dfrac{1}{(2x+3)}}dx$

Using the logarithm formula of integration,

$=\dfrac{5}{2}\log |x+1|-\dfrac{1}{10}\log |x-1|-\dfrac{24}{5\times 2}\log |2x+3|$

Simplifying,

$=\dfrac{5}{2}\log |x+1|-\dfrac{1}{10}\log |x-1|-\dfrac{12}{5}\log |2x+3|+C$

11. $\dfrac{5x}{\left( x+1 \right)\left( {{x}^{2}}-4 \right)}$

Ans: $\dfrac{5x}{(x+1)\left( {{x}^{2}}-4 \right)}=\dfrac{5x}{(x+1)(x+2)(x-2)}$

let $\dfrac{5x}{(x+1)(x+2)(x-2)}=\dfrac{A}{(x+1)}+\dfrac{B}{(x+2)}+\dfrac{C}{(x-2)}$

$5\text{ }x=A\left( x+2 \right)\left( x-2 \right)+B\left( x+1 \right)\left( x-2 \right)+C\left( x+1 \right)\left( x+2 \right)$

We obtain the below values by equating the coefficients of x , ${{x}^{2}}$ and the constant term on both sides.

$A+B+C=0$

$-B+3C=5\text{ and, }-4A-2B+2C=0$

On solving, to obtain

$A=\dfrac{5}{3},B=-\dfrac{5}{2},\text{ and }C=\dfrac{5}{6}$

$\therefore \dfrac{5x}{(x+1)(x+2)(x-2)}=\dfrac{5}{3(x+1)}+-\dfrac{5}{2(x+2)}+\dfrac{5}{6(x-2)}$

$\Rightarrow \int{\dfrac{5x}{(x+1)\left( {{x}^{2}}-4 \right)}}dx=\dfrac{5}{3}\int{\dfrac{1}{(x+1)}}dx-\dfrac{5}{2}\int{\dfrac{1}{(x+2)}}dx+\dfrac{5}{6}\int{\dfrac{1}{(x-2)}}dx$

Using the logarithm formula of integration,

$=\dfrac{5}{3}\log |x+1|-\dfrac{5}{2}\log |x+2|+\dfrac{5}{6}\log |x-2|+C$

12. $\dfrac{{{x}^{2}}+x+1}{{{x}^{2}}-1}$

Ans:  On dividing $\left( {{x}^{3}}+x+1 \right)\,\,\,by\,\,{{x}^{2}}-1,$we get

$\dfrac{{{x}^{3}}+x+1}{{{x}^{2}}-1}=x+\dfrac{2x+1}{{{x}^{2}}-1}$

$\text{ Let }\dfrac{2x+1}{{{x}^{2}}-1}=\dfrac{A}{(x+1)}+\dfrac{B}{(x-1)}$

$2\text{ }x+1=A\left( x-1 \right)+B\left( x+1 \right)$

We obtain the below values by equating the coefficients of x  and the constant term on both sides.

$A+B=2$

$-A+B=1$

On solving, to obtain

$A=\dfrac{1}{2}\text{ and }B=\dfrac{3}{2}\text{ }$

$\therefore \dfrac{{{x}^{3}}+x+1}{{{x}^{2}}-1}=x+\dfrac{1}{2(x+1)}+\dfrac{3}{2(x-1)}\text{ }$

$\Rightarrow \int{\dfrac{{{x}^{3}}+x+1}{{{x}^{2}}+1}}dx=\int{x}dx+\dfrac{1}{2}\int{\dfrac{1}{(x+1)}}dx+\dfrac{3}{2}\int{\dfrac{1}{(x-1)}}dx$

$=\dfrac{{{x}^{2}}}{2}+\log |x+1|+\dfrac{3}{2}\log |x-1|+C$

13. $\dfrac{2}{\left( 1-x \right)\left( 1+{{x}^{2}} \right)}$

Ans: Let $\dfrac{2}{(1-x)\left( 1+{{x}^{2}} \right)}=\dfrac{A}{(1-x)}+\dfrac{Bx+C}{\left( 1+{{x}^{2}} \right)}$

$2=A\left( 1+{{x}^{2}} \right)+(Bx+C)(1-x)$

$2=A+A{{x}^{2}}+Bx-B{{x}^{2}}+C-Cx$

We obtain the below values by equating the coefficients of x ${{x}^{2}}$ and the constant term on both sides.

$\text{A}-\text{B}=0$

$B-C=0$

$A+C=2$

On solving these equations, to obtain

$A=1,B=1\text{, and }C=1$

$\therefore \dfrac{2}{(1-x)\left( 1+{{x}^{2}} \right)}=\dfrac{1}{1-x}+\dfrac{x+1}{1+{{x}^{2}}}$

$\Rightarrow \int{\dfrac{2}{(1-x)\left( 1+{{x}^{2}} \right)}}dx=\int{\dfrac{1}{1-x}}dx+\int{\dfrac{x}{1+{{x}^{2}}}}dx+\int{\dfrac{1}{1+{{x}^{2}}}}dx$

$=-\int{\dfrac{1}{1-x}}dx+\dfrac{1}{2}\int{\dfrac{2x}{1+{{x}^{2}}}}dx+\int{\dfrac{1}{1+{{x}^{2}}}}dx$

$=-\log |x-1|+\dfrac{1}{2}\log \left| 1+{{x}^{2}} \right|+{{\tan }^{-1}}x+C$

14. $\dfrac{3x-1}{{{\left( x+2 \right)}^{2}}}$

Ans:   Let $\dfrac{3x-1}{{{(x+2)}^{2}}}=\dfrac{A}{(x+2)}+\dfrac{B}{{{(x+2)}^{2}}}$

$\Rightarrow 3x-1=A(x+2)+B$

We obtain the below values by equating the coefficients of x  and the constant term on both sides.

$A=3$

$2A+B=-1\Rightarrow B=-7$

$\therefore \dfrac{3x-1}{{{(x+2)}^{2}}}=\dfrac{3}{(x+2)}-\dfrac{7}{{{(x+2)}^{2}}}$

$\Rightarrow \int{\dfrac{3x-1}{{{(x+2)}^{2}}}}dx=3\int{\dfrac{1}{(x+2)}}dx-7\int{\dfrac{x}{{{(x+2)}^{2}}}}dx$

Using the power rule and logarithm formula of integration

$=3\log |x+2|-7\left( \dfrac{-1}{(x+2)} \right)+C$

$=3\log |x+2|+\dfrac{7}{(x+2)}+C$

15. $\dfrac{1}{{{x}^{4}}-1}$

Ans:  $\dfrac{1}{\left( {{x}^{4}}-1 \right)}=\dfrac{1}{\left( {{x}^{2}}-1 \right)\left( {{x}^{2}}+1 \right)}=\dfrac{1}{(x+1)(x-1)\left( 1+{{x}^{2}} \right)}$

Let $\dfrac{1}{(x+1)(x-1)\left( 1+{{x}^{2}} \right)}=\dfrac{A}{(x+1)}+\dfrac{B}{(x-1)}+\dfrac{Cx+D}{\left( {{x}^{2}}+1 \right)}$

$1=A(x-1)\left( 1+{{x}^{2}} \right)+B(x+1)\left( 1+{{x}^{2}} \right)+(Cx+D)\left( {{x}^{2}}-1 \right)$

$1=A\left( {{x}^{3}}+x-{{x}^{2}}-1 \right)+B\left( {{x}^{3}}+x+{{x}^{2}}+1 \right)+C{{x}^{3}}+D{{x}^{2}}-Cx-D$

$1=(A+B+C){{x}^{3}}+(-A+B+D){{x}^{2}}+(A+B-C)x+(-A+B-D)$

We obtain the below values by equating the coefficients of${{\text{x}}^{3}},{{\text{x}}^{2}},\text{x},$and constant term, we get

$A+B+C=0\text{ }$

$-A+B+D=0$

$A+B-C=0\text{ }$

$-A+B-D=1$

$A=-\dfrac{1}{4},B=\dfrac{1}{4},C=0,\text{ and }D=-\dfrac{1}{2}$

$\therefore \dfrac{1}{\left( {{x}^{4}}-1 \right)}=\dfrac{-1}{4(x+1)}+\dfrac{1}{4(x-1)}+\dfrac{1}{2\left( {{x}^{2}}+1 \right)}$

$\Rightarrow \int{\dfrac{1}{{{x}^{4}}-1}}dx=-\dfrac{1}{4}\log |x-1|+\dfrac{1}{4}\log |x-1|-\dfrac{1}{2}{{\tan }^{1}}x+C$

Simplifying,

$=\dfrac{1}{4}\log \left| \dfrac{x-1}{x+1} \right|-\dfrac{1}{2}{{\tan }^{1}}x+C$

16. $\dfrac{1}{x\left( {{x}^{n}}+1 \right)}$ $\text{ [ }$Hint: Multiply numerator and denominator by ${{x}^{n-1}}$ and put ${{x}^{n}}=t$ $\text{]}$

Ans: $\dfrac{1}{x\left( {{x}^{n}}+1 \right)}$

Numerator and denominator are multiplied by ${{x}^{n-1}}$, to obtain

$\dfrac{1}{x\left( {{x}^{n}}+1 \right)}=\dfrac{{{x}^{n-1}}}{{{x}^{n-1}}x\left( {{x}^{n}}+1 \right)}=\dfrac{{{x}^{n-1}}}{{{x}^{n}}\left( {{x}^{n}}+1 \right)}$

Consider  ${{x}^{n}}=t\Rightarrow {{x}^{n-1}}dx=dt$

$\therefore \int{\dfrac{1}{x\left( {{x}^{n}}+1 \right)}}dx=\int{\dfrac{{{x}^{n-1}}}{{{x}^{n}}\left( {{x}^{n}}+1 \right)}}dx=\dfrac{1}{n}\int{\dfrac{1}{t(t+1)}}dt$

$\text{ Let }\dfrac{1}{t(t+1)}=\dfrac{A}{t}+\dfrac{B}{(t+1)}$

$1=A\left( 1+t \right)+B\text{ }t$

We obtain the below values by equating the coefficients of $\text{t}$and constant,

$A=1\text{ and }B=-1$

$\therefore \dfrac{1}{t(t+1)}=\dfrac{1}{t}-\dfrac{1}{(1+t)}$

$\Rightarrow \int{\dfrac{1}{x\left( {{x}^{n}}+1 \right)}}dx=\dfrac{1}{n}\int{\left\{ \dfrac{1}{t}-\dfrac{1}{(1+t)} \right\}}dx$

$=\dfrac{1}{n}[\log |t|-\log |t+1|]+C$

Substitute the value of t,

$=-\dfrac{1}{n}\left[ \log \left| {{x}^{n}} \right|-\log \left| {{x}^{n}}+1 \right| \right]+C$

Simplifying,

$=\dfrac{1}{n}\log \left| \dfrac{{{x}^{u}}}{{{x}^{\prime \prime }}+1} \right|+C$

17. $\dfrac{\cos x}{\left( 1-\sin x \right)\left( 2-\sin x \right)}$ $\text{[}$ hint: Put $\sin x=t$ $\text{]}$

Ans:  $\dfrac{\cos x}{(1-\sin x)(2-\sin x)}\,\,\,\,Put,\sin x=t\Rightarrow \cos xdx=dt$

$\therefore \int{\dfrac{\cos x}{(1-\sin x)(2-\sin x)}}dx=\int{\dfrac{dt}{(1-t)(2-t)}}$

let $\dfrac{1}{(1-t)(2-t)}=\dfrac{A}{(1-t)}+\dfrac{B}{(2-t)}$

$1=A\left( 2-t \right)+B\left( 1-t \right)$

We obtain the below values by equating the coefficients of t and constant,

$-2~\text{A}-\text{B}=0\,\,,and\,\,2~\text{A}+\text{B}=1$

$A=1\text{ and }B=-1$

$\therefore \dfrac{1}{(1-t)(2-t)}=\dfrac{1}{(1-t)}-\dfrac{1}{(2-t)}$

$\Rightarrow \int{\dfrac{\cos x}{(1-\sin x)(2-\sin x)}}dx=\int{\left\{ \dfrac{1}{1-t}-\dfrac{1}{(2-t)} \right\}}dt=-\log |1-t|+\log |2-t|+C$ Simplifying,

$=\log \left| \dfrac{2-t}{1-t} \right|+C$

Substitute the value of t,

$=\log \left| \dfrac{2-\sin x}{1-\sin x} \right|+C$

18. $\dfrac{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+2 \right)}{\left( {{x}^{2}}+3 \right)\left( {{x}^{2}}+4 \right)}$

Ans:  $\dfrac{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+2 \right)}{\left( {{x}^{2}}+3 \right)\left( {{x}^{2}}+4 \right)}=\dfrac{\left( 4{{x}^{2}}+10 \right)}{\left( {{x}^{2}}+3 \right)\left( {{x}^{2}}+4 \right)}$

Let $\dfrac{\left( 4{{x}^{2}}+10 \right)}{\left( {{x}^{2}}+3 \right)\left( {{x}^{2}}+4 \right)}=\dfrac{Ax+B}{\left( {{x}^{2}}+3 \right)}+\dfrac{Cx+D}{\left( {{x}^{2}}+4 \right)}$

$4{{x}^{2}}+10=(Ax+B)\left( {{x}^{2}}+4 \right)+(Cx+D)\left( {{x}^{2}}+3 \right)$

$4{{x}^{2}}+10=A{{x}^{2}}+4Ax+B{{x}^{2}}+4B+C{{x}^{3}}+3Cx+D{{x}^{2}}+3D$

$4{{x}^{2}}+10=(A+C){{x}^{3}}+(B+D){{x}^{2}}+(4A+3C)x+(4B+3D)$

We obtain the below values by equating the coefficients of ${{\text{x}}^{3}},{{\text{x}}^{2}},\text{x}$and constant term,

$A+C=0$

$B+D=4$

$4\text{ }A+3\text{ }C=0$

$4\text{ }B+3\text{ }D=10$

On solving these equations, to obtain $\text{A}=0.\text{B}=-2.\text{C}=0,and\,\,\,\text{D}=6$

$\therefore \dfrac{\left( 4{{x}^{2}}+10 \right)}{\left( {{x}^{2}}+3 \right)\left( {{x}^{2}}+4 \right)}=\dfrac{-2}{\left( {{x}^{2}}+3 \right)}+\dfrac{6}{\left( {{x}^{2}}+4 \right)}$

$\dfrac{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+2 \right)}{\left( {{x}^{2}}+3 \right)\left( {{x}^{2}}+4 \right)}=\left( \dfrac{-2}{\left( {{x}^{2}}+3 \right)}+\dfrac{6}{\left( {{x}^{2}}+4 \right)} \right)$

$\Rightarrow \int{\dfrac{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+2 \right)}{\left( {{x}^{2}}+3 \right)\left( {{x}^{2}}+4 \right)}}dx=\int{\left\{ 1+\dfrac{2}{\left( {{x}^{2}}+3 \right)}-\dfrac{6}{\left( {{x}^{2}}+4 \right)} \right\}}dx$

$=\int{\left\{ 1+\dfrac{2}{{{x}^{2}}+{{(\sqrt{3})}^{2}}}-\dfrac{6}{{{x}^{2}}+{{2}^{2}}} \right\}}$

$=x+2\left( \dfrac{1}{\sqrt{3}}{{\tan }^{-1}}\dfrac{x}{\sqrt{3}} \right)-6\left( \dfrac{1}{2}{{\tan }^{-1}}\dfrac{x}{2} \right)+C$

Simplifying,

$=x+\dfrac{2}{\sqrt{3}}{{\tan }^{-1}}\dfrac{x}{\sqrt{3}}-3{{\tan }^{-1}}\dfrac{x}{2}+C$

19. $\dfrac{2x}{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+3 \right)}$

Ans:  $\dfrac{2x}{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+3 \right)}$

Put ${{x}^{2}}-t\to 2xdx-dt$

$\therefore \int{\dfrac{2x}{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+3 \right)}}dx=\int{\dfrac{dt}{(t+1)(t+3)}}$

$\text{ Let }\dfrac{1}{(t+1)(t+3)}=\dfrac{A}{(t+1)}+\dfrac{B}{(t+3)}$

$I=A\left( t+3 \right)+B\left( t+1 \right)$

We obtain the below values by equating the coefficients of $\text{t}$and

constant,

$1+B=0$and $3A+B=1$

On solving, we get

$A=\dfrac{1}{2}\text{ and }B=-\dfrac{1}{2}$

$\therefore \dfrac{1}{(t+1)(t+3)}=\dfrac{1}{2(t+1)}+\dfrac{1}{2(t+3)}$

$\Rightarrow \int{\dfrac{2x}{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+3 \right)}}dx=\int{\left\{ \dfrac{1}{2(t+1)}-\dfrac{1}{2(t+3)} \right\}}dt$

$=\dfrac{1}{2}\log |(t+1)|-\dfrac{1}{2}\log |t+3|+C$

Simplifying,

$=\dfrac{1}{2}\log \left| \dfrac{t+1}{t+3} \right|+C=\dfrac{1}{2}\log \left| \dfrac{{{x}^{2}}+1}{{{x}^{2}}+3} \right|+C$

20. $\dfrac{1}{x\left( {{x}^{4}}-1 \right)}$

Ans:  $\dfrac{1}{x\left( {{x}^{4}}-1 \right)}$

Numerator and denominator are multiplied by by ${{\text{x}}^{3}},$we get

$\dfrac{1}{x\left( {{x}^{4}}-1 \right)}=\dfrac{{{x}^{3}}}{{{x}^{4}}\left( {{x}^{4}}-1 \right)}$

$\therefore \int{\dfrac{1}{x\left( {{x}^{4}}-1 \right)}}dx=\int{\dfrac{{{x}^{3}}}{{{x}^{4}}\left( {{x}^{4}}-1 \right)}}dx$

Consider ${{\text{x}}^{4}}=\text{t}\Rightarrow 4{{\text{x}}^{3}}\text{dx}=\text{dt}$

$\therefore \int{\dfrac{1}{x\left( {{x}^{4}}-1 \right)}}dx=\dfrac{1}{4}\int{\dfrac{dt}{t(t-1)}}$

$\text{ Let }\dfrac{1}{t(t-1)}=\dfrac{A}{t}+\dfrac{B}{(t-1)}$

$1=A(t-1)+Bt\quad \ldots (1)$

We obtain the below values by equating the coefficients of $\text{t}$and constant,

$A=-1\text{ and }B=1$

$\Rightarrow \dfrac{1}{t(t-1)}=\dfrac{-1}{t}+\dfrac{1}{t-1}$

$\Rightarrow \int{\dfrac{1}{x\left( {{x}^{4}}-1 \right)}}dx=\dfrac{1}{4}\int{\left\{ \dfrac{-1}{t}+\dfrac{1}{t-1} \right\}}dt$

Using the logarithm formula of integration,

$=\dfrac{1}{4}[-\log |t|+\log |t-1|]+C$

Simplifying,

$=\dfrac{1}{4}\log \left| \dfrac{t-1}{t} \right|+C=\dfrac{1}{4}\log \left| \dfrac{{{x}^{4}}-1}{{{x}^{4}}} \right|+C$

21. $\dfrac{1}{{{e}^{x}}-1}$ $\text{[}$ hint: Put ${{e}^{x}}=t$ $\text{]}$

Ans:  $\dfrac{1}{\left( {{e}^{x}}-1 \right)}$

Put ${{\text{e}}^{x}}=\text{t }\Rightarrow {{\text{e}}^{x}}\text{dx}=dt$

$\Rightarrow \int{\dfrac{1}{\left( {{e}^{x}}-1 \right)}}dx=\int{\dfrac{1}{t-1}}\times \dfrac{dt}{t}=\int{\dfrac{1}{t(t-1)}}dt$

$\text{ Let }\dfrac{1}{t(t-1)}=\dfrac{A}{t}+\dfrac{B}{t-1}$

$1=A\left( t-1 \right)+B\text{ }t$

We obtain the below values by equating the coefficients of t and constant, $A=-1\text{ and }B=1$

$\therefore \dfrac{1}{t(t-1)}=\dfrac{-1}{t}+\dfrac{1}{t-1}$

$\Rightarrow \int{\dfrac{1}{t(t-1)}}dt=\log \left| \dfrac{t-1}{t} \right|+C$

Substitute the value of t,

$=\log \left| \dfrac{{{e}^{x}}-1}{{{e}^{x}}} \right|+C$

22. $\int{\dfrac{xdx}{\left( x-1 \right)\left( x-2 \right)}}$ equals

1. $\log \left| \dfrac{{{\left( x-1 \right)}^{2}}}{x-2} \right|+C$

2. $\log \left| \dfrac{{{\left( x-2 \right)}^{2}}}{x-1} \right|+C$

3. $\log \left| {{\left( \dfrac{x-1}{x-2} \right)}^{2}} \right|+C$

4. $\log \left| \left( x-1 \right)\left( x-2 \right) \right|+C$

Ans:  Let $\dfrac{x}{(x-1)(x-2)}=\dfrac{A}{(x-1)}+\dfrac{B}{(x-2)}$

$x=A\left( x-2 \right)+B\left( x-1 \right)$

We obtain the below values by equating the coefficients of $\text{x}$and constant,  $\text{A}=-1\,\,\,and\,\,\text{B}=2$

$\therefore \dfrac{x}{(x-1)(x-2)}=-\dfrac{1}{(x1)}+\dfrac{2}{(x-2)}$

$\Rightarrow \int{\dfrac{x}{(x-1)(x-2)}}dx=\int{\left\{ \dfrac{-1}{(x-1)}+\dfrac{2}{(x-2)} \right\}}dx$

Using the logarithm formula of integration,

$=-\log |x-1|+2\log |x-2|+C$

Simplifying,

$=\log \left| \dfrac{{{(x-2)}^{2}}}{x-1} \right|+C$

Thus, the right response is B.

23. $\int{\dfrac{dx}{x\left( {{x}^{2}}+1 \right)}}$ equals

1. $\log \left| x \right|-\dfrac{1}{2}\log \left( {{x}^{2}}+1 \right)+C$

2. $\log \left| x \right|+\dfrac{1}{2}\log \left( {{x}^{2}}+1 \right)+C$

3. $-\log \left| x \right|+\dfrac{1}{2}\log \left( {{x}^{2}}+1 \right)+C$

4. $\dfrac{1}{2}\log \left| x \right|+\log \left( {{x}^{2}}+1 \right)+C$

Ans: Let $\dfrac{1}{x\left( {{x}^{2}}+1 \right)}-\dfrac{A}{x},\dfrac{Bx+C}{{{x}^{2}}+1}$

$1=A\left( {{x}^{2}}+1 \right)+(Bx+C)x$

We obtain the below values by equating the coefficients of ${{\text{x}}^{2}},\text{x},$     and constant term,

$A+B=0$, $C=0$$A=1 On solving these equations, to obtain A=1,B=-1,\text{ and }C=0 \therefore \dfrac{1}{x\left( {{x}^{2}},1 \right)}=\dfrac{1}{x}+\dfrac{-x}{{{x}^{2}}+1} \Rightarrow \int{\dfrac{1}{x\left( {{x}^{2}}+1 \right)}}dx=\int{\left\{ \dfrac{1}{x}-\dfrac{x}{{{x}^{2}}+1} \right\}}dx =\log |x|-\dfrac{1}{2}\log \left| {{x}^{2}}+1 \right|+C Thus, the right response is \text{A}. ### Exercise 7.6 1. x\sin x Ans: Let I=\int{x}\sin xdx Consider \text{u}=\text{x} and \text{v}=\sin \text{x}and integrating by parts, to obtain I=\int{x}\sin xdx-\int{\left\{ \left( \dfrac{d}{dx}x \right)\int{\sin }xdx \right\}}dx =x(-\cos x)-\int{1}.(-\cos x)dx =-x\cos x+\sin x+C 2. x\sin 3x Ans: Let \text{I}=\int{x}\sin 3xdx Consider \text{u}=\text{x} and \text{v}=\sin 3\text{x} and integrating by parts, to obtain I=x\int{\sin }3xdx-\int{\left\{ \left( \dfrac{d}{dx}x \right)\int{\sin }3xdx \right\}} =x\left( \dfrac{-\cos 3x}{3} \right)-\int{1}\cdot \left( \dfrac{-\cos 3x}{3} \right)dx =\dfrac{-x\cos 3x}{3}+\dfrac{1}{3}\int{\cos }3xdx=\dfrac{-x\cos 3x}{3}+\dfrac{1}{9}\sin 3x+C 3. {{x}^{2}}{{e}^{x}} Ans: Let I=\int{{{x}^{2}}}{{e}^{x}}dx Consider \text{u}={{\text{x}}^{2}}\,\,and\,\,\text{v}={{\text{e}}^{x}} I={{x}^{2}}\int{{{e}^{x}}}dx-\int{\left\{ \left( \dfrac{d}{dx}{{x}^{2}} \right)\int{{{e}^{x}}}dx \right\}}dx ={{x}^{2}}{{e}^{x}}-\int{2}x-{{e}^{x}}dx ={{x}^{2}}{{e}^{x}}-2\int{x}\cdot {{e}^{x}}dx Again using integration by parts, to obtain ={{x}^{2}}{{e}^{x}}-2\left[ x\cdot \int{{{e}^{x}}}dx-\int{\left\{ \left( \dfrac{d}{dx}{{x}^{2}} \right)\int{{{e}^{x}}}dx \right\}}dx \right] ={{x}^{2}}{{e}^{x}}-2\left[ x{{e}^{x}}-\int{{{e}^{x}}}dx \right] Simplifying, ={{x}^{2}}{{e}^{x}}-2\left[ x{{e}^{x}}-{{e}^{x}} \right] ={{x}^{2}}{{e}^{x}}-2x{{e}^{x}}+2{{e}^{x}}+C ={{e}^{x}}\left( {{x}^{2}}-2x+2 \right)+C 4. x\log x Ans: Let I=\int{x}\log xdx Consider \text{u}=\log \text{x}\,\,\,\,and\,\,\,\text{v}=\text{x} and integrating by parts, to obtain I=\log x\int{x}dx-\int{\left\{ \left( \dfrac{d}{dx}\log x \right)\int{x}dx \right\}}dx =\log x\cdot \dfrac{{{x}^{2}}}{2}-\int{\dfrac{1}{x}}\cdot \dfrac{{{x}^{2}}}{2}dx =\dfrac{{{x}^{2}}\log x}{2}\cdot \sqrt{\dfrac{x}{2}}dx=\dfrac{{{x}^{2}}\log x}{2}-\dfrac{{{x}^{2}}}{4}+C 5. x\log 2x Ans: Let I=\int{x}\log 2xdx Consider u=\log 2x and v=xand integrating by parts, to obtain I=\log 2x\int{x}dx-\int{\left\{ \left( \dfrac{d}{dx}2\log x \right)\int{x}dx \right\}}dx =\log 2x\cdot \dfrac{{{x}^{2}}}{2}-\int{\dfrac{2}{2x}}\cdot \dfrac{{{x}^{2}}}{2}dx =\dfrac{{{x}^{2}}\log 2x}{2}-\int{\dfrac{x}{2}}dx Integrating using the power rule =\dfrac{{{x}^{2}}\log 2x}{2}-\dfrac{{{x}^{2}}}{4}+C 6. {{x}^{2}}\log x Ans: Let I=\int{{{x}^{2}}}\log xdx Consider u=\log xand v={{x}^{2}} and integrating by parts, to obtain I=\log x\int{{{x}^{2}}}dx-\int{\left\{ \left( \dfrac{d}{dx}\log x \right)\int{{{x}^{2}}}dx \right\}}dx =\log x\left( \dfrac{{{x}^{3}}}{3} \right)-\int{\dfrac{1}{x}}\cdot \dfrac{{{x}^{3}}}{3}dx Integrating using the power rule =\dfrac{{{x}^{3}}\log x}{3}-\int{\dfrac{{{x}^{2}}}{3}}dx=\dfrac{{{x}^{3}}\log x}{3}-\dfrac{{{x}^{3}}}{9}+C 7. x{{\sin }^{-1}}x Ans: Let I=\int{x}{{\sin }^{-1}}xdx Consider u={{\sin }^{-1}}x\,\,and\,\,\,v=x and integrating by parts, to obtain I={{\sin }^{-1}}x\int{x}dx\int{\left\{ \left( \dfrac{d}{dx}{{\sin }^{-1}}x \right)\int{x}dx \right\}}dx ={{\sin }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)-\int{\dfrac{1}{\sqrt{1-{{x}^{2}}}}}\cdot \dfrac{{{x}^{2}}}{2}dx =\dfrac{{{x}^{2}}{{\sin }^{-1}}x}{2}+\dfrac{1}{2}\int{\dfrac{-{{x}^{2}}}{\sqrt{1-{{x}^{2}}}}}dx Adding and subtracting by 1 =\dfrac{{{x}^{2}}{{\sin }^{-1}}x}{2}+\dfrac{1}{2}\int{\left\{ \dfrac{1-{{x}^{2}}}{\sqrt{1-{{x}^{2}}}}-\dfrac{1}{\sqrt{1-{{x}^{2}}}} \right\}}dx Simplifying, =\dfrac{{{x}^{2}}{{\sin }^{-1}}x}{2}+\dfrac{1}{2}\int{\left\{ \sqrt{1-{{x}^{2}}}-\dfrac{1}{\sqrt{1-{{x}^{2}}}} \right\}}dx =\dfrac{{{x}^{2}}{{\sin }^{-1}}x}{2}+\dfrac{1}{2}\left\{ \int{\sqrt{1-{{x}^{2}}}}dx-\int{\dfrac{1}{\sqrt{1-{{x}^{2}}}}}dx \right\} =\dfrac{{{x}^{2}}{{\sin }^{-1}}x}{2}+\dfrac{1}{2}\left\{ \dfrac{x}{2}\sqrt{1-{{x}^{2}}}+\dfrac{1}{2}{{\sin }^{-1}}x-{{\sin }^{-1}}x \right\}+C Simplifying, =\dfrac{{{x}^{2}}{{\sin }^{-1}}x}{2}+\dfrac{x}{4}\sqrt{1-{{x}^{2}}}+\dfrac{1}{4}{{\sin }^{-1}}x-\dfrac{1}{2}{{\sin }^{-1}}x+C=\dfrac{1}{4}\left( 2{{x}^{2}}-1 \right){{\operatorname{in}}^{-1}}x+\dfrac{x}{4}\sqrt{1-{{x}^{2}}}+C 8. x{{\tan }^{-1}}x Ans: Let I=\int{x}{{\tan }^{-1}}xdx Consider \text{u}={{\tan }^{-1}}\text{x} and \text{v}=\text{x}and integrating by parts, to obtain I={{\tan }^{-1}}x\int{x}dx-\int{\left\{ \left( \dfrac{d}{dx}{{\tan }^{-1}}x \right)\int{x}dx \right\}}dx ={{\tan }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)\int{\dfrac{1}{1+{{x}^{2}}}}\cdot \dfrac{{{x}^{2}}}{2}dx=\dfrac{{{x}^{2}}{{\tan }^{-1}}x}{2}-\dfrac{1}{2}\int{\dfrac{{{x}^{2}}}{1+{{x}^{2}}}}dx Adding and subtracting by -1 =\dfrac{{{x}^{2}}{{\tan }^{-1}}x}{2}-\dfrac{1}{2}\int{\left( \dfrac{{{x}^{2}}+1}{1+{{x}^{2}}}-\dfrac{1}{1+{{x}^{2}}} \right)}dx=\dfrac{{{x}^{2}}{{\tan }^{-1}}x}{2}-\dfrac{1}{2}\int{\left( 1-\dfrac{1}{1+{{x}^{2}}} \right)}dx Simplifying, =\dfrac{{{x}^{2}}{{\operatorname{lan}}^{-1}}x}{2}-\dfrac{1}{2}\left( x-{{\tan }^{-1}}x \right)+C=\dfrac{{{x}^{2}}}{2}{{\tan }^{-1}}x-\dfrac{x}{2}+\dfrac{1}{2}{{\tan }^{-1}}x+C 9. x{{\cos }^{-1}}x Ans:Let I=\int{x}{{\cos }^{-1}}xdx Taking u={{\cos }^{-1}}x and \text{v}=\text{x}and integrating by parts, to obtain I={{\cos }^{-1}}x\int{x}dx-\int{\left\{ \left( \dfrac{d}{dx}{{\cos }^{-1}}x \right)\int{x}dx \right\}}dx ={{\cos }^{-1}}x\dfrac{{{x}^{2}}}{2}-\int{\dfrac{-1}{\sqrt{1-{{x}^{2}}}}}\cdot \dfrac{{{x}^{2}}}{2}dx Adding and subtracting by -1 =\dfrac{{{x}^{2}}{{\cos }^{-1}}x}{2}-\dfrac{1}{2}\int{\dfrac{1-{{x}^{2}}-1}{\sqrt{1-{{x}^{2}}}}}dx Simplifying, =\dfrac{{{x}^{2}}{{\cos }^{-1}}x}{2}-\dfrac{1}{2}\int{\left\{ \sqrt{1-{{x}^{2}}}+\left( \dfrac{-1}{\sqrt{1-{{x}^{2}}}} \right) \right\}}dx =\dfrac{{{x}^{2}}{{\cos }^{-1}}x}{2}-\dfrac{1}{2}\int{\sqrt{1-{{x}^{2}}}}dx-\dfrac{1}{2}\int{\left( \dfrac{-1}{\sqrt{1-{{x}^{2}}}} \right)}dx =\dfrac{{{x}^{2}}{{\cos }^{-1}}x}{2}-\dfrac{1}{2}{{I}_{1}}-\dfrac{1}{2}{{\cos }^{-1}}x....\left( 1 \right) Where {{I}_{1}}=\int{\sqrt{1-{{x}^{2}}}}dx \Rightarrow {{I}_{1}}=x\int{\sqrt{1-{{x}^{2}}}}-\int{\dfrac{d}{dx}}\sqrt{1-{{x}^{2}}}\int{x}dx\Rightarrow {{I}_{1}}=x\sqrt{1-{{x}^{2}}}-\int{\dfrac{-{{x}^{2}}}{\sqrt{1-{{x}^{2}}}}}dx \Rightarrow {{I}_{1}}=x\sqrt{1-{{x}^{2}}}-\int{\dfrac{1-{{x}^{2}}-1}{\sqrt{1-{{x}^{2}}}}}dx\Rightarrow {{I}_{1}}=x\sqrt{1-{{x}^{2}}}-\left\{ \int{\sqrt{1-{{x}^{2}}}}dx+\int{\dfrac{-dx}{\sqrt{1-{{x}^{2}}}}} \right\} \Rightarrow {{I}_{1}}=x\sqrt{1-{{x}^{2}}}-\left\{ {{I}_{1}}+{{\cos }^{-1}}x \right\}\Rightarrow 2{{I}_{1}}=x\sqrt{1-{{x}^{2}}}-{{\cos }^{-1}}x \therefore {{I}_{1}}=\dfrac{x}{2}\sqrt{1-{{x}^{2}}}-\dfrac{1}{2}{{\cos }^{-1}}x Substituting in (1), we get I=\dfrac{x{{\cos }^{-1}}x}{2}-\dfrac{1}{2}\left( \dfrac{x}{2}\sqrt{1-{{x}^{2}}}-\dfrac{1}{2}{{\cos }^{-1}}x \right)-\dfrac{1}{2}{{\cos }^{-1}}x Simplifying, =\dfrac{\left( 2{{x}^{2}}-1 \right)}{4}{{\cos }^{-1}}x-\dfrac{x}{4}\sqrt{1-{{x}^{2}}}+C 10. {{\left( {{\sin }^{-1}}x \right)}^{2}} Ans: Let I=\int{{{\left( {{\sin }^{-1}}x \right)}^{2}}}\cdot 1dx Consider \text{u}={{\left( {{\sin }^{-1}}\text{x} \right)}^{2}} and \text{v}=1and integrating by parts, to obtain I=\int{\left( {{\sin }^{-1}}x \right)}\cdot \int{1}dx-\int{\left\{ \dfrac{d}{dx}{{\left( {{\sin }^{-1}}x \right)}^{2}}\cdot \int{1}.dx \right\}}dx ={{\left( {{\sin }^{-1}}x \right)}^{2}}x-\int{\dfrac{2{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}}\cdot xdx =x{{\left( {{\sin }^{-1}}x \right)}^{2}}+\int{{{\sin }^{-1}}}x\cdot \left( \dfrac{-2x}{\sqrt{1-{{x}^{2}}}} \right)dx =x{{\left( {{\sin }^{-1}}x \right)}^{2}}+\left[ {{\sin }^{-1}}x\int{\dfrac{-2x}{\sqrt{1-{{x}^{2}}}}}dx-\int{\left\{ \left( \dfrac{d}{dx}{{\sin }^{-1}}x \right)\int{\dfrac{-2x}{\sqrt{1-{{x}^{2}}}}}dx \right\}}dx \right] =x{{\left( {{\sin }^{-1}}x \right)}^{2}}+\left[ {{\sin }^{-1}}x\cdot 2\sqrt{1-{{x}^{2}}}-\int{\dfrac{1}{\sqrt{1-{{x}^{2}}}}}\cdot 2\sqrt{1-{{x}^{2}}}dx \right] =x{{\left( {{\sin }^{-1}}x \right)}^{2}}+2\sqrt{1-{{x}^{2}}}{{\sin }^{-1}}x-\int{2}dx =x{{\left( {{\sin }^{-1}}x \right)}^{2}}+2\sqrt{1-{{x}^{2}}}{{\sin }^{-1}}x-2x+C 11. \dfrac{x{{\cos }^{-1}}x}{\sqrt{1-{{x}^{2}}}} Ans: Let I=\int{\dfrac{x{{\cos }^{-1}}x}{\sqrt{1-{{x}^{2}}}}}dx Multiplying and dividing by 2 I=\dfrac{-1}{2}\int{\dfrac{-2x}{\sqrt{1-{{x}^{2}}}}}\cdot {{\cos }^{-1}}xdx Consider \text{u}={{\cos }^{-1}}\text{x} and \text{v}=\left( \dfrac{-2x}{\sqrt{1-{{x}^{2}}}} \right)and integrating by parts, to obtain I=\dfrac{-1}{2}\left[ {{\cos }^{-1}}x\int{\dfrac{-2x}{\sqrt{1-{{x}^{2}}}}}dx-\int{\left\{ \left( \dfrac{d}{dx}{{\cos }^{-1}}x \right)\int{\dfrac{-2x}{\sqrt{1-{{x}^{2}}}}}dx \right\}}dx \right] =\dfrac{-1}{2}\left[ {{\cos }^{-1}}x\cdot 2\sqrt{1-{{x}^{2}}}-\int{\dfrac{-1}{\sqrt{1-{{x}^{2}}}}}\cdot 2\sqrt{1-{{x}^{2}}}dx \right]=\dfrac{-1}{2}\left[ 2\sqrt{1-{{x}^{2}}}{{\cos }^{-1}}x+\int{2}dx \right] Simplifying, =\dfrac{-1}{2}\left[ 2\sqrt{1-{{x}^{2}}}{{\cos }^{-1}}x+2x \right]+C =-\left[ \sqrt{1-{{x}^{2}}}{{\cos }^{-1}}x+x \right]+C 12. x{{\sec }^{2}}x Ans:Let I=\int{x}{{\sec }^{2}}xdx Consider\text{u}=\text{x} and \text{v}={{\sec }^{2}}\text{x} and integrating by parts, to obtain I=x\int{{{\sec }^{2}}}xdx-\int{\left\{ \left\{ \dfrac{d}{dx}x \right\}\int{{{\sec }^{2}}}xdx \right\}}dx =x\tan x-\int{1}\cdot \tan xdx =x\tan x+\log |\cos x|+C 13. {{\tan }^{-1}}x Ans: Let I=\int{1}\cdot {{\tan }^{-1}}xdx Consider \text{u}={{\tan }^{-1}}\text{x} and \text{v}=1 and integrating by parts, to obtain I={{\tan }^{-1}}x\int{1}dx-\int{\left\{ \left( \dfrac{d}{dx}{{\tan }^{-1}}x \right)\int{1}.dx \right\}}dx={{\tan }^{-1}}xx-\int{\dfrac{1}{1+{{x}^{2}}}}xd =x{{\tan }^{-1}}x-\dfrac{1}{2}\int{\dfrac{2x}{1+{{x}^{2}}}}dx =x{{\tan }^{-1}}x-\dfrac{1}{2}\log \left| 1+{{x}^{2}} \right|+C =x{{\tan }^{-1}}x-\dfrac{1}{2}\log \left( 1+{{x}^{2}} \right)+C 14. x{{\left( \log x \right)}^{2}}dx Ans: I=\int{x}{{(\log x)}^{2}}dx Consideru={{(\log x)}^{2}} and v=1 and integrating by parts, to obtain I={{(\log )}^{2}}\int{x}dx-\int{\left[ \left\{ {{\left( \dfrac{d}{dx}\log x \right)}^{2}} \right\}\int{x}dx \right]}dx =\dfrac{{{x}^{2}}}{2}{{(\log x)}^{2}}-\left[ \int{2}\log x\cdot \dfrac{1}{x}\cdot \dfrac{{{x}^{2}}}{2}dx \right] =\dfrac{{{x}^{2}}}{2}{{(\log x)}^{2}}-\int{x}\log xdx Again using integration by parts, to obtain I=\dfrac{{{x}^{2}}}{2}{{(\log x)}^{2}}-\left[ \log x\int{x}dx-\int{\left\{ \left( \dfrac{d}{dx}\log x \right)\int{x}dx \right\}}dx \right] =\dfrac{{{x}^{2}}}{2}{{(\log x)}^{2}}-\left[ \dfrac{{{x}^{2}}}{2}-\log x-\int{\dfrac{1}{x}}\cdot \dfrac{{{x}^{2}}}{2}dx \right] =\dfrac{{{x}^{2}}}{2}{{(\log x)}^{2}}-\dfrac{{{x}^{2}}}{2}\log x+\dfrac{1}{2}\int{x}dx=\dfrac{{{x}^{2}}}{2}{{(\log x)}^{2}}-\dfrac{{{x}^{2}}}{2}\log x+\dfrac{{{x}^{2}}}{4}+C 15. \left( {{x}^{2}}+1 \right)\log x Ans: Let I=\int{\left( {{x}^{2}}+1 \right)}\log xdx=\int{{{x}^{2}}}\log xdx+\int{\log }xdx Let \text{I}={{\text{I}}_{1}}+{{\text{I}}_{2}}\ldots (1) Where,{{I}_{1}}=\int{{{x}^{2}}}\log xdx\,\,\,\,\,\,and\,\,\,{{\text{I}}_{2}}=\int{\log }xdx {{I}_{1}}=\int{{{x}^{2}}}\log xdx Consider\text{u}=\log \text{x} and v=x^2 and integrating by parts, to obtain {{I}_{1}}=\log x-\int{{{x}^{2}}}dx-\int{\left\{ \left( \dfrac{d}{dx}\log x \right)\int{{{x}^{2}}}dx \right\}}dx =\log x\cdot \dfrac{{{x}^{3}}}{3}-\int{\dfrac{1}{x}}\cdot \dfrac{{{x}^{3}}}{3}dx=\dfrac{{{x}^{3}}}{3}\log x-\dfrac{1}{3}\int{{{x}^{2}}}dx =\dfrac{{{x}^{3}}}{3}\log x-\dfrac{{{x}^{3}}}{9}+{{C}_{1}}\quad \ldots (2) {{I}_{2}}=\int{\log }xdx Consider \text{u}=\log \text{x} and \text{v}=1and integrating by parts, to obtain {{I}_{2}}=\log x\int{1}.dx-\int{\left\{ \left( \dfrac{d}{dx}\log x \right)\int{1}.dx \right\}} =\log x\cdot x-\int{\dfrac{1}{x}}xdx =x\log x-x..\left( 3 \right) Using equations (2) and (3) in (1), we get I=\dfrac{{{x}^{3}}}{3}\log x-\dfrac{{{x}^{3}}}{9}+{{C}_{1}}+x\log x-x+{{C}_{2}} =\dfrac{{{x}^{3}}}{3}\log x-\dfrac{{{x}^{3}}}{9}+x\log x-x+\left( {{C}_{1}}+{{C}_{2}} \right) =\left( \dfrac{{{x}^{3}}}{3}+x \right)\log x-\dfrac{{{x}^{3}}}{9}-x+C 16. {{e}^{x}}\left( \sin x+\cos x \right) Ans: ConsiderI=\int{{{e}^{x}}}(\sin x+\cos x)dx Considerf(x)=\sin x {{f}^{\prime }}(x)=\cos x I=\int{{{e}^{x}}}\left\{ f(x)+{{f}^{\prime }}(x) \right\}dx Since, \int{{{e}^{x}}}\left\{ f(x)+{{f}^{\prime }}(x) \right\}dx={{e}^{x}}f(x)+C \therefore I={{e}^{x}}\sin x+C 17. \dfrac{x{{e}^{x}}}{{{\left( 1+x \right)}^{2}}} Ans: Consider I=\int{\dfrac{x{{e}^{x}}}{{{(1+x)}^{2}}}}dx=\int{{{e}^{x}}}\left\{ \dfrac{x}{{{(1+x)}^{2}}} \right\}dx =\int{{{e}^{x}}}\left\{ \dfrac{1+x-1}{{{(1+x)}^{2}}} \right\}dx=\int{{{e}^{x}}}\left\{ \dfrac{1}{1+x}-\dfrac{1}{{{(1+x)}^{2}}} \right\}dx Here, f(x)=\dfrac{1}{1+x}\quad {{f}^{\prime }}(x)=\dfrac{-1}{{{(1+x)}^{2}}} \Rightarrow \int{\dfrac{x{{e}^{x}}}{{{(1+x)}^{2}}}}dx=\int{{{e}^{x}}}\left\{ f(x)+{{f}^{\prime }}(x) \right\}dx Since, \int{{{e}^{x}}}\left\{ f(x)+{{f}^{\prime }}(x) \right\}dx={{e}^{x}}f(x)+C \therefore \int{\dfrac{x{{e}^{x}}}{{{(1+x)}^{2}}}}dx=\dfrac{{{e}^{x}}}{1+x}+C 18. Integrate the function - {{e}^{x}}\left( \dfrac{1+\sin x}{1+\cos x} \right) Ans: First simplify –{{e}^{x}}\left( \dfrac{1+\sin x}{1+\cos x} \right) It is known that – 1+\sin x={{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}+2\sin \dfrac{x}{2}\cos \dfrac{x}{2} 1+\cos x=2{{\cos }^{2}}\dfrac{x}{2} \therefore {{e}^{x}}\left( \dfrac{1+\sin x}{1+\cos x} \right)={{e}^{x}}\left( \dfrac{{{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}+2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}{2{{\cos }^{2}}\dfrac{x}{2}} \right) ={{e}^{x}}\left( \dfrac{{{\left( \sin \dfrac{x}{2}+\cos \dfrac{x}{2} \right)}^{2}}}{2{{\cos }^{2}}\dfrac{x}{2}} \right) =\dfrac{1}{2}{{e}^{x}}\left( \dfrac{{{\left( \sin \dfrac{x}{2}+\cos \dfrac{x}{2} \right)}^{2}}}{{{\cos }^{2}}\dfrac{x}{2}} \right) =\dfrac{1}{2}{{e}^{x}}{{\left( \dfrac{\sin \dfrac{x}{2}+\cos \dfrac{x}{2}}{\cos \dfrac{x}{2}} \right)}^{2}} =\dfrac{1}{2}{{e}^{x}}{{\left( \dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}+\dfrac{\cos \dfrac{x}{2}}{\cos \dfrac{x}{2}} \right)}^{2}} =\dfrac{1}{2}{{e}^{x}}{{\left( \tan \dfrac{x}{2}+1 \right)}^{2}} =\dfrac{1}{2}{{e}^{x}}\left( {{\tan }^{2}}\dfrac{x}{2}+1+2\tan \dfrac{x}{2} \right) But, 1+{{\tan }^{2}}\dfrac{x}{2}={{\sec }^{2}}\dfrac{x}{2} =\dfrac{1}{2}{{e}^{x}}\left( {{\sec }^{2}}\dfrac{x}{2}+2\tan \dfrac{x}{2} \right) ={{e}^{x}}\left( \dfrac{1}{2}{{\sec }^{2}}\dfrac{x}{2}+\tan \dfrac{x}{2} \right) \Rightarrow {{e}^{x}}\left( \dfrac{1+\sin x}{1+\cos x} \right)={{e}^{x}}\left( \dfrac{1}{2}{{\sec }^{2}}\dfrac{x}{2}+\tan \dfrac{x}{2} \right) It is known that, \int{{{e}^{x}}\left\{ f(x)+f'(x) \right\}dx=}{{e}^{x}}f(x)+C If we say, f(x)=\tan \dfrac{x}{2}\Rightarrow f'(x)=\dfrac{1}{2}{{\sec }^{2}}\dfrac{x}{2} Thus, we get – \int{{{e}^{x}}\left( \dfrac{1+\sin x}{1+\cos x} \right)}dx={{e}^{x}}\tan \dfrac{x}{2}+C 19. Integrate the function - {{e}^{x}}\left( \dfrac{1}{x}-\dfrac{1}{{{x}^{2}}} \right) Ans: Say, I=\int{{{e}^{x}}\left( \dfrac{1}{x}-\dfrac{1}{{{x}^{2}}} \right)dx} Suppose, f(x)=\dfrac{1}{x}\Rightarrow f'(x)=-\dfrac{1}{{{x}^{2}}} It is known that, \int{{{e}^{x}}\left\{ f(x)+f'(x) \right\}dx=}{{e}^{x}}f(x)+C Thus, we get – I=\int{{{e}^{x}}\left( \dfrac{1}{x}-\dfrac{1}{{{x}^{2}}} \right)dx}=\dfrac{{{e}^{x}}}{x}+C 20. Integrate the function - \dfrac{(x-3){{e}^{x}}}{{{(x-1)}^{3}}} Ans: \int{{{e}^{x}}\dfrac{(x-3)}{{{(x-1)}^{3}}}dx=\int{{{e}^{x}}\left[ \dfrac{(x-1-2)}{{{(x-1)}^{3}}} \right]dx}} =\int{{{e}^{x}}\left[ \dfrac{(x-1)}{{{(x-1)}^{3}}}-\dfrac{2}{{{(x-1)}^{3}}} \right]dx} =\int{{{e}^{x}}\left[ \dfrac{1}{{{(x-1)}^{2}}}-\dfrac{2}{{{(x-1)}^{3}}} \right]dx} Suppose, f(x)=\dfrac{1}{{{(x-1)}^{2}}}\Rightarrow f'(x)=-\dfrac{2}{{{(x-1)}^{3}}} It is known that, \int{{{e}^{x}}\left\{ f(x)+f'(x) \right\}dx=}{{e}^{x}}f(x)+C Thus, \int{{{e}^{x}}\dfrac{(x-3)}{{{(x-1)}^{3}}}dx=\dfrac{{{e}^{x}}}{{{(x-1)}^{2}}}+C} 21. Integrate the function - {{e}^{2x}}\sin x Ans: Say, I=\int{{{e}^{2x}}\sin xdx} Perform Integration by parts – \int{uv}dx=u\int{vdx}-\int{\left( u'\int{vdx} \right)dx} With –u=\sin x\text{ }v={{e}^{2x}} I=\int{{{e}^{2x}}\sin x}dx=\sin x\int{{{e}^{2x}}dx}-\int{\left[ \left( \dfrac{d}{dx}\sin x \right)\int{{{e}^{2x}}dx} \right]dx} =\sin x\dfrac{{{e}^{2x}}}{2}-\int{\left[ \left( \cos x \right)\dfrac{{{e}^{2x}}}{2} \right]dx} =\sin x\dfrac{{{e}^{2x}}}{2}-\dfrac{1}{2}\int{\left( {{e}^{2x}}\cos x \right)dx} Perform Integration by parts for – \int{\left( {{e}^{2x}}\cos x \right)dx} =\sin x\dfrac{{{e}^{2x}}}{2}-\dfrac{1}{2}\left\{ \cos x\int{{{e}^{2x}}dx}-\int{\left[ \left( \dfrac{d}{dx}\cos x \right)\int{{{e}^{2x}}dx} \right]dx} \right\} =\sin x\dfrac{{{e}^{2x}}}{2}-\dfrac{1}{2}\left\{ \cos x\dfrac{{{e}^{2x}}}{2}-\int{\left[ \left( -\sin x \right)\dfrac{{{e}^{2x}}}{2} \right]dx} \right\} =\sin x\dfrac{{{e}^{2x}}}{2}-\dfrac{1}{2}\left\{ \cos x\dfrac{{{e}^{2x}}}{2}+\dfrac{1}{2}\int{(\sin x){{e}^{2x}}dx} \right\} =\sin x\dfrac{{{e}^{2x}}}{2}-\dfrac{{{e}^{2x}}\cos x}{4}-\dfrac{1}{4}\left\{ \int{(\sin x){{e}^{2x}}dx} \right\} But, I=\int{{{e}^{2x}}\sin xdx} \Rightarrow I=\sin x\dfrac{{{e}^{2x}}}{2}-\dfrac{{{e}^{2x}}\cos x}{4}-\dfrac{1}{4}I \Rightarrow I+\dfrac{1}{4}I=\sin x\dfrac{{{e}^{2x}}}{2}-\dfrac{{{e}^{2x}}\cos x}{4} \Rightarrow \dfrac{5}{4}I=\dfrac{{{e}^{2x}}\sin x}{2}-\dfrac{{{e}^{2x}}\cos x}{4} \Rightarrow \dfrac{5}{4}I=\dfrac{2{{e}^{2x}}\sin x}{4}-\dfrac{{{e}^{2x}}\cos x}{4} \Rightarrow 5I={{e}^{2x}}(2\sin x-\cos x) Thus, we get – I=\dfrac{{{e}^{2x}}}{5}(2\sin x-\cos x)+C. 22. Integrate the function - {{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{3}}} \right) Ans: Say, x=\tan \theta \text{ }\Rightarrow \text{dx=se}{{\text{c}}^{2}}\theta d\theta \therefore {{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{3}}} \right)={{\sin }^{-1}}\left( \dfrac{2\tan \theta }{1+{{\tan }^{3}}\theta } \right) But, \sin 2\theta =\dfrac{2\tan \theta }{1+{{\tan }^{3}}\theta } \Rightarrow {{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{3}}} \right)={{\sin }^{-1}}\left( \dfrac{2\tan \theta }{1+{{\tan }^{3}}\theta } \right)=\therefore {{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{3}}} \right)={{\sin }^{-1}}\left( \sin 2\theta \right)=2\theta Therefore, \int{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{3}}} \right)dx}=\int{2\theta \text{se}{{\text{c}}^{2}}\theta d\theta } =2\int{\theta \text{se}{{\text{c}}^{2}}\theta d\theta } Perform Integration by parts – \int{uv}dx=u\int{vdx}-\int{\left( u'\int{vdx} \right)dx} With –u=\theta \text{ }v={{\sec }^{2}}\theta 2\int{\theta \text{se}{{\text{c}}^{2}}\theta d\theta }=2\left\{ \theta \int{{{\sec }^{2}}\theta d\theta }-\int{\left[ \left( \dfrac{d}{d\theta }\theta \right)\int{{{\sec }^{2}}\theta d\theta } \right]d\theta } \right\} =2\left\{ \theta \tan \theta -\int{\left[ \tan \theta \right]d\theta } \right\} =2\left\{ \theta \tan \theta -(-\log |\cos \theta |) \right\}+C =2\left\{ \theta \tan \theta +\log |\cos \theta | \right\}+C Replace \theta ={{\tan }^{-1}}x =2\left\{ {{\tan }^{-1}}x\tan ({{\tan }^{-1}}x)+\log |\cos ({{\tan }^{-1}}x)| \right\}+C It is known that – {{\tan }^{-1}}x={{\cos }^{-1}}\dfrac{1}{\sqrt{1+{{x}^{2}}}} =2\left\{ {{\tan }^{-1}}x(x)+\log |\cos ({{\cos }^{-1}}\dfrac{1}{\sqrt{1+{{x}^{2}}}})| \right\}+C =2\left\{ x{{\tan }^{-1}}x+\log |\dfrac{1}{\sqrt{1+{{x}^{2}}}}| \right\}+C =2\left\{ x{{\tan }^{-1}}x+\log {{(1+{{x}^{2}})}^{-\dfrac{1}{2}}} \right\}+C Here, \log {{m}^{n}}=n\log m =2\left\{ x{{\tan }^{-1}}x-\dfrac{1}{2}\log (1+{{x}^{2}}) \right\}+C =2x{{\tan }^{-1}}x-\log (1+{{x}^{2}})+C Thus, \int{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{3}}} \right)dx}=2x{{\tan }^{-1}}x-\log (1+{{x}^{2}})+C 23. Choose the correct answer: \int{{{x}^{2}}{{e}^{{{x}^{3}}}}}dx equals 1. \dfrac{1}{3}{{e}^{{{x}^{3}}}}+C 2. \dfrac{1}{3}{{e}^{{{x}^{2}}}}+C 3. \dfrac{1}{2}{{e}^{{{x}^{3}}}}+C 4. \dfrac{1}{2}{{e}^{{{x}^{2}}}}+C Ans: Say, I=\int{{{x}^{2}}{{e}^{{{x}^{3}}}}}dx Suppose, t={{x}^{3}}\Rightarrow dt=3{{x}^{2}}dx Rewriting the equation – I=\int{{{x}^{2}}{{e}^{{{x}^{3}}}}}dx=\dfrac{1}{3}\int{{{e}^{t}}}dt \Rightarrow I=\dfrac{1}{3}\int{{{e}^{t}}}dt=\dfrac{1}{3}{{e}^{t}}+C Replacing t={{x}^{3}} \Rightarrow I=\dfrac{1}{3}{{e}^{{{x}^{3}}}}+C The correct option is A. 24. Choose the correct answer: \int{{{e}^{x}}\sec x(1+\tan x)}dx 1. {{e}^{x}}\cos x+C 2. {{e}^{x}}\sec x+C 3. {{e}^{x}}\sin x+C 4. {{e}^{x}}\tan x+C Ans: Say, I=\int{{{e}^{x}}\sec x(1+\tan x)}dx \Rightarrow I=\int{{{e}^{x}}(\sec x+\sec x\tan x)}dx Suppose, f(x)=\sec x\Rightarrow f'(x)=\sec x\tan x It is known that, \int{{{e}^{x}}\left\{ f(x)+f'(x) \right\}dx=}{{e}^{x}}f(x)+C \Rightarrow I=\int{{{e}^{x}}(\sec x+\sec x\tan x)}dx={{e}^{x}}\sec x+C Thus, I={{e}^{x}}\sec x+C The correct option is B. ### Exercise 7.7 1. Integrate the function - \sqrt{4-{{x}^{2}}} Ans: Say, I=\int{\sqrt{4-{{x}^{2}}}dx}=\int{\sqrt{{{2}^{2}}-{{x}^{2}}}dx} It is known that – \int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx=\dfrac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\dfrac{{{a}^{2}}}{2}{{\sin }^{-1}}\dfrac{x}{a}+C} \Rightarrow I=\int{\sqrt{{{2}^{2}}-{{x}^{2}}}dx=\dfrac{x}{2}\sqrt{{{2}^{2}}-{{x}^{2}}}+\dfrac{{{2}^{2}}}{2}{{\sin }^{-1}}\dfrac{x}{2}+C} \Rightarrow I=\dfrac{x}{2}\sqrt{4-{{x}^{2}}}+2{{\sin }^{-1}}\dfrac{x}{2}+C Thus, \int{\sqrt{4-{{x}^{2}}}dx}=\dfrac{x}{2}\sqrt{4-{{x}^{2}}}+2{{\sin }^{-1}}\dfrac{x}{2}+C 2. Integrate the function - \sqrt{1-4{{x}^{2}}} Ans: Say, I=\int{\sqrt{1-4{{x}^{2}}}dx}=\int{\sqrt{{{1}^{2}}-{{(2x)}^{2}}}dx} Let, 2x=t\Rightarrow 2dx=dt x=\dfrac{t}{2}\Rightarrow dx=\dfrac{dt}{2} So, we get – I=\int{\sqrt{{{1}^{2}}-{{\left[ 2(\dfrac{t}{2}) \right]}^{2}}}\dfrac{dt}{2}}=\dfrac{1}{2}\int{\sqrt{{{1}^{2}}-{{\left[ t \right]}^{2}}}dt} \Rightarrow I=\dfrac{1}{2}\int{\sqrt{{{1}^{2}}-{{\left[ t \right]}^{2}}}dt} It is known that – \int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx=\dfrac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\dfrac{{{a}^{2}}}{2}{{\sin }^{-1}}\dfrac{x}{a}+C} \Rightarrow I=\dfrac{1}{2}\left[ \dfrac{t}{2}\sqrt{1-{{t}^{2}}}+\dfrac{1}{2}{{\sin }^{-1}}t \right]+C \Rightarrow I=\left[ \dfrac{t}{4}\sqrt{1-{{t}^{2}}}+\dfrac{1}{4}{{\sin }^{-1}}t \right]+C Replace – t=2x \Rightarrow I=\left[ \dfrac{2x}{4}\sqrt{1-{{(2x)}^{2}}}+\dfrac{1}{4}{{\sin }^{-1}}2x \right]+C \Rightarrow I=\left[ \dfrac{x}{2}\sqrt{1-4{{x}^{2}}}+\dfrac{1}{4}{{\sin }^{-1}}2x \right]+C Thus,\int{\sqrt{1-4{{x}^{2}}}dx}=\dfrac{x}{2}\sqrt{1-4{{x}^{2}}}+\dfrac{1}{4}{{\sin }^{-1}}2x+C 3. Integrate the function - \sqrt{{{x}^{2}}+4x+6} Ans: First simplify –{{x}^{2}}+4x+6 {{x}^{2}}+4x+6={{x}^{2}}+4x+4+2 =({{x}^{2}}+4x+4)+2={{(x+2)}^{2}}+{{(\sqrt{2})}^{2}} \Rightarrow \sqrt{{{x}^{2}}+4x+6}=\sqrt{{{(x+2)}^{2}}+{{(\sqrt{2})}^{2}}} \therefore \int{\sqrt{{{x}^{2}}+4x+6}dx}=\int{\sqrt{{{(x+2)}^{2}}+{{(\sqrt{2})}^{2}}}dx} It is known that – \int{\sqrt{{{x}^{2}}+{{a}^{2}}}dx=\dfrac{x}{2}\sqrt{{{x}^{2}}+{{a}^{2}}}+\dfrac{{{a}^{2}}}{2}\log \left| x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right|+C} \Rightarrow \int{\sqrt{{{(x+2)}^{2}}+{{(\sqrt{2})}^{2}}}dx}=\dfrac{x+2}{2}\sqrt{{{(x+2)}^{2}}+{{(\sqrt{2})}^{2}}}+\dfrac{{{(\sqrt{2})}^{2}}}{2}\log \left| (x+2)+\sqrt{{{(x+2)}^{2}}+{{(\sqrt{2})}^{2}}} \right|+C$$=\dfrac{x+2}{2}\sqrt{{{x}^{2}}+4x+6}+\dfrac{2}{2}\log \left| (x+2)+\sqrt{{{x}^{2}}+4x+6} \right|+C$

Thus, $\int{\sqrt{{{x}^{2}}+4x+6}dx}=\dfrac{x+2}{2}\sqrt{{{x}^{2}}+4x+6}+\log \left| (x+2)+\sqrt{{{x}^{2}}+4x+6} \right|+C$

4. Integrate the function - $\sqrt{{{x}^{2}}+4x+1}$

Ans: First simplify –${{x}^{2}}+4x+1$

${{x}^{2}}+4x+1={{x}^{2}}+4x+4-3$

$=({{x}^{2}}+4x+4)-3={{(x+2)}^{2}}-{{(\sqrt{3})}^{2}}$

$\Rightarrow \sqrt{{{x}^{2}}+4x+1}=\sqrt{{{(x+2)}^{2}}-{{(\sqrt{3})}^{2}}}$

$\therefore \int{\sqrt{{{x}^{2}}+4x+1}dx}=\int{\sqrt{{{(x+2)}^{2}}-{{(\sqrt{3})}^{2}}}dx}$

It is known that – $\int{\sqrt{{{x}^{2}}-{{a}^{2}}}dx=\dfrac{x}{2}\sqrt{{{x}^{2}}-{{a}^{2}}}-\dfrac{{{a}^{2}}}{2}\log \left| x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right|+C}$

$\Rightarrow \int{\sqrt{{{(x+2)}^{2}}-{{(\sqrt{3})}^{2}}}dx}=\dfrac{x+2}{2}\sqrt{{{(x+2)}^{2}}-{{(\sqrt{3})}^{2}}}-\dfrac{{{(\sqrt{3})}^{2}}}{2}\log \left| (x+2)+\sqrt{{{(x+2)}^{2}}-{{(\sqrt{3})}^{2}}} \right|+C$$=\dfrac{x+2}{2}\sqrt{{{x}^{2}}+4x+1}-\dfrac{3}{2}\log \left| (x+2)+\sqrt{{{x}^{2}}+4x+1} \right|+C Thus, \int{\sqrt{{{x}^{2}}+4x+1}dx}=\dfrac{x+2}{2}\sqrt{{{x}^{2}}+4x+1}-\dfrac{3}{2}\log \left| (x+2)+\sqrt{{{x}^{2}}+4x+1} \right|+C 5. Integrate the function - \sqrt{1-4x-{{x}^{2}}} Ans: First simplify –1-4x-{{x}^{2}} 1-4x-{{x}^{2}}=1-4x-{{x}^{2}}-4+4=1+4-({{x}^{2}}+4x+4) =5-({{x}^{2}}+4x+4)={{(\sqrt{5})}^{2}}-{{(x+2)}^{2}} \Rightarrow \sqrt{1-4x-{{x}^{2}}}=\sqrt{{{(\sqrt{5})}^{2}}-{{(x+2)}^{2}}} \therefore \int{\sqrt{1-4x-{{x}^{2}}}dx}=\int{\sqrt{{{(\sqrt{5})}^{2}}-{{(x+2)}^{2}}}dx} It is known that – \int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx=\dfrac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\dfrac{{{a}^{2}}}{2}{{\sin }^{-1}}\dfrac{x}{a}+C} \Rightarrow \int{\sqrt{{{(\sqrt{5})}^{2}}-{{(x+2)}^{2}}}dx}=\dfrac{x+2}{2}\sqrt{{{(\sqrt{5})}^{2}}-{{(x+2)}^{2}}}+\dfrac{{{(\sqrt{5})}^{2}}}{2}{{\sin }^{-1}}\dfrac{x+2}{\sqrt{5}}+C$$=\dfrac{x+2}{2}\sqrt{1-4x-{{x}^{2}}}+\dfrac{5}{2}{{\sin }^{-1}}\dfrac{x+2}{\sqrt{5}}+C$

Thus, $\int{\sqrt{1-4x-{{x}^{2}}}dx}=\dfrac{x+2}{2}\sqrt{1-4x-{{x}^{2}}}+\dfrac{5}{2}{{\sin }^{-1}}\dfrac{x+2}{\sqrt{5}}+C$

6. Integrate the function - $\sqrt{{{x}^{2}}+4x+5}$

Ans: First simplify –${{x}^{2}}+4x-5$

${{x}^{2}}+4x-5={{x}^{2}}+4x-5+4-4=({{x}^{2}}+4x+4)-5-4$

$=({{x}^{2}}+4x+4)-9={{(x+2)}^{2}}-{{(3)}^{2}}$

$\Rightarrow \sqrt{{{x}^{2}}+4x-5}=\sqrt{{{(x+2)}^{2}}-{{(3)}^{2}}}$

$\therefore \int{\sqrt{{{x}^{2}}+4x-5}dx}=\int{\sqrt{{{(x+2)}^{2}}-{{(3)}^{2}}}dx}$

It is known that – $\int{\sqrt{{{x}^{2}}-{{a}^{2}}}dx=\dfrac{x}{2}\sqrt{{{x}^{2}}-{{a}^{2}}}-\dfrac{{{a}^{2}}}{2}\log \left| x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right|+C}$

$\Rightarrow \int{\sqrt{{{(x+2)}^{2}}-{{(3)}^{2}}}dx}=\dfrac{x+2}{2}\sqrt{{{(x+2)}^{2}}-{{(3)}^{2}}}-\dfrac{{{(3)}^{2}}}{2}\log \left| (x+2)+\sqrt{{{(x+2)}^{2}}-{{(3)}^{2}}} \right|+C$$=\dfrac{x+2}{2}\sqrt{{{x}^{2}}+4x-5}-\dfrac{9}{2}\log \left| (x+2)+\sqrt{{{x}^{2}}+4x-5} \right|+C Thus, \int{\sqrt{{{x}^{2}}+4x-5}dx}=\dfrac{x+2}{2}\sqrt{{{x}^{2}}+4x-5}-\dfrac{9}{2}\log \left| (x+2)+\sqrt{{{x}^{2}}+4x-5} \right|+C 7. Integrate the function - \sqrt{1+3x-{{x}^{2}}} Ans: First simplify –1+3x-{{x}^{2}} 1+3x-{{x}^{2}}=1-{{x}^{2}}+3x+\dfrac{9}{4}-\dfrac{9}{4}=1+\dfrac{9}{4}-({{x}^{2}}-3x+\dfrac{9}{4}) =\dfrac{9+4}{4}-({{x}^{2}}-3x+\dfrac{9}{4})=\left( \dfrac{13}{4} \right)-({{x}^{2}}-3x+\dfrac{9}{4})={{\left( \dfrac{\sqrt{13}}{2} \right)}^{2}}-{{\left( x-\dfrac{3}{2} \right)}^{2}} \Rightarrow \sqrt{1+3x-{{x}^{2}}}=\sqrt{{{\left( \dfrac{\sqrt{13}}{2} \right)}^{2}}-{{\left( x-\dfrac{3}{2} \right)}^{2}}} \therefore \int{\sqrt{1+3x-{{x}^{2}}}dx}=\int{\sqrt{{{\left( \dfrac{\sqrt{13}}{2} \right)}^{2}}-{{\left( x-\dfrac{3}{2} \right)}^{2}}}dx} It is known that – \int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx=\dfrac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\dfrac{{{a}^{2}}}{2}{{\sin }^{-1}}\dfrac{x}{a}+C} \Rightarrow \int{\sqrt{{{\left( \dfrac{\sqrt{13}}{2} \right)}^{2}}-{{\left( x-\dfrac{3}{2} \right)}^{2}}}dx}=\dfrac{\left( x-\dfrac{3}{2} \right)}{2}\sqrt{{{\left( \dfrac{\sqrt{13}}{2} \right)}^{2}}-{{\left( x-\dfrac{3}{2} \right)}^{2}}}+\dfrac{{{\left( \dfrac{\sqrt{13}}{2} \right)}^{2}}}{2}{{\sin }^{-1}}\dfrac{\left( x-\dfrac{3}{2} \right)}{\left( \dfrac{\sqrt{13}}{2} \right)}+C$$=\dfrac{2x-3}{4}\sqrt{1+3x-{{x}^{2}}}+\dfrac{13}{8}{{\sin }^{-1}}\dfrac{2x-3}{\sqrt{13}}+C$

Thus, $\int{\sqrt{1+3x-{{x}^{2}}}dx}=\dfrac{2x-3}{4}\sqrt{1+3x-{{x}^{2}}}+\dfrac{13}{8}{{\sin }^{-1}}\dfrac{2x-3}{\sqrt{13}}+C$

8. Integrate the function - $\sqrt{{{x}^{2}}+3x}$

Ans: First simplify –${{x}^{2}}+3x$

${{x}^{2}}+x={{x}^{2}}+3x+\dfrac{9}{4}-\dfrac{9}{4}=({{x}^{2}}+3x+\dfrac{9}{4})-\dfrac{9}{4}$

$={{\left( x+\dfrac{3}{2} \right)}^{2}}-\left( \dfrac{9}{4} \right)={{\left( x+\dfrac{3}{2} \right)}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}}$

$\Rightarrow \sqrt{{{x}^{2}}+3x}=\sqrt{{{\left( x+\dfrac{3}{2} \right)}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}}}$

$\therefore \int{\sqrt{{{x}^{2}}+3x}dx}=\int{\sqrt{{{\left( x+\dfrac{3}{2} \right)}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}}}dx}$

It is known that – $\int{\sqrt{{{x}^{2}}-{{a}^{2}}}dx=\dfrac{x}{2}\sqrt{{{x}^{2}}-{{a}^{2}}}-\dfrac{{{a}^{2}}}{2}\log \left| x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right|+C}$

$\Rightarrow \int{\sqrt{{{\left( x+\dfrac{3}{2} \right)}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}}}dx}=\dfrac{\left( x+\dfrac{3}{2} \right)}{2}\sqrt{{{\left( x+\dfrac{3}{2} \right)}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}}}-\dfrac{{{\left( \dfrac{3}{2} \right)}^{2}}}{2}\log \left| \left( x+\dfrac{3}{2} \right)+\sqrt{{{\left( x+\dfrac{3}{2} \right)}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}}} \right|+C$$=\dfrac{2x+3}{4}\sqrt{{{x}^{2}}+3x}-\dfrac{9}{8}\log \left| \left( x+\dfrac{3}{2} \right)+\sqrt{{{x}^{2}}+3x} \right|+C Thus, \int{\sqrt{{{x}^{2}}+3x}dx}=\dfrac{2x+3}{4}\sqrt{{{x}^{2}}+3x}-\dfrac{9}{8}\log \left| \left( x+\dfrac{3}{2} \right)+\sqrt{{{x}^{2}}+3x} \right|+C 9. Integrate the function - \sqrt{1+\dfrac{{{x}^{2}}}{9}} Ans: First simplify –1+\dfrac{{{x}^{2}}}{9} 1+\dfrac{{{x}^{2}}}{9}=\dfrac{1}{9}(9+{{x}^{2}})=\dfrac{1}{9}({{3}^{2}}+{{x}^{2}}) \Rightarrow \sqrt{1+\dfrac{{{x}^{2}}}{9}}=\sqrt{\dfrac{1}{9}({{3}^{2}}+{{x}^{2}})}=\dfrac{1}{3}\sqrt{({{3}^{2}}+{{x}^{2}})} \therefore \int{\sqrt{1+\dfrac{{{x}^{2}}}{9}}dx}=\int{\dfrac{1}{3}\sqrt{({{3}^{2}}+{{x}^{2}})}dx}=\dfrac{1}{3}\int{\sqrt{({{3}^{2}}+{{x}^{2}})}dx} It is known that – \int{\sqrt{{{x}^{2}}+{{a}^{2}}}dx=\dfrac{x}{2}\sqrt{{{x}^{2}}+{{a}^{2}}}+\dfrac{{{a}^{2}}}{2}\log \left| x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right|+C} \Rightarrow \dfrac{1}{3}\int{\sqrt{({{3}^{2}}+{{x}^{2}})}dx}=\dfrac{1}{3}\left\{ \dfrac{x}{2}\sqrt{{{(x)}^{2}}+{{(3)}^{2}}}+\dfrac{{{(3)}^{2}}}{2}\log \left| (x)+\sqrt{{{(x)}^{2}}+{{(3)}^{2}}} \right| \right\}+C$$=\dfrac{1}{3}\left\{ \dfrac{x}{2}\sqrt{{{x}^{2}}+9}+\dfrac{9}{2}\log \left| x+\sqrt{{{x}^{2}}+9} \right| \right\}+C$

$\dfrac{x}{6}{\sqrt{x^2+9}}+\dfrac{3}{2}\log \left| x+\sqrt{{{x}^{2}}+9} \right| +C$
Thus, $\int{\sqrt{1+\dfrac{{{x}^{2}}}{9}}dx}=\dfrac{x}{6}\sqrt{{{x}^{2}}+9}+\dfrac{3}{2}\log \left| x+\sqrt{{{x}^{2}}+9} \right|+C$

10. Choose the correct answer: $\int{\sqrt{1+{{x}^{2}}}dx}$ is equal to –

1. $\dfrac{x}{2}\sqrt{1+{{x}^{2}}}+\dfrac{1}{2}\log \left| (x+\sqrt{1+{{x}^{2}}}) \right|+C$

2. $\dfrac{2}{3}{{(1+{{x}^{2}})}^{\dfrac{3}{2}}}+C$

3. $\dfrac{2}{3}x{{(1+{{x}^{2}})}^{\dfrac{3}{2}}}+C$

4. $\dfrac{{{x}^{2}}}{2}\sqrt{1+{{x}^{2}}}+\dfrac{1}{2}{{x}^{2}}\log \left| (x+\sqrt{1+{{x}^{2}}}) \right|+C$

Ans: It is known that – $\int{\sqrt{{{x}^{2}}+{{a}^{2}}}dx=\dfrac{x}{2}\sqrt{{{x}^{2}}+{{a}^{2}}}+\dfrac{{{a}^{2}}}{2}\log \left| x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right|+C}$

Thus,  $\int{\sqrt{{{x}^{2}}+{{1}^{2}}}dx=\dfrac{x}{2}\sqrt{{{x}^{2}}+{{1}^{2}}}+\dfrac{{{1}^{2}}}{2}\log \left| x+\sqrt{{{x}^{2}}+{{1}^{2}}} \right|+C}$

$\int{\sqrt{{{x}^{2}}+1}dx=\dfrac{x}{2}\sqrt{{{x}^{2}}+1}+\dfrac{1}{2}\log \left| x+\sqrt{{{x}^{2}}+1} \right|+C}$

The correct answer is option A.

11. Choose the correct answer: $\int{\sqrt{{{x}^{2}}-8x+7}dx}$ is equal to –

1. $\dfrac{1}{2}(x-4)\sqrt{{{x}^{2}}-8x+7}+9\log \left| (x-4+\sqrt{{{x}^{2}}-8x+7}) \right|+C$

2. $\dfrac{1}{2}(x+4)\sqrt{{{x}^{2}}-8x+7}+9\log \left| (x+4+\sqrt{{{x}^{2}}-8x+7}) \right|+C$

3. $\dfrac{1}{2}(x-4)\sqrt{{{x}^{2}}-8x+7}-3\sqrt{2}\log \left| (x-4+\sqrt{{{x}^{2}}-8x+7}) \right|+C$

4. $\dfrac{1}{2}(x-4)\sqrt{{{x}^{2}}-8x+7}+\dfrac{9}{2}\log \left| (x-4+\sqrt{{{x}^{2}}-8x+7}) \right|+C$

Ans: First simplify –${{x}^{2}}-8x+7$

${{x}^{2}}-8x+7+9-9={{x}^{2}}-8x+16-9=({{x}^{2}}-8x+16)-9$

$={{(x-4)}^{2}}-{{(3)}^{2}}$

$\Rightarrow \sqrt{{{x}^{2}}-8x+7}=\sqrt{{{(x-4)}^{2}}-{{(3)}^{2}}}$

$\therefore \int{\sqrt{{{x}^{2}}-8x+7}dx}=\int{\sqrt{{{(x-4)}^{2}}-{{(3)}^{2}}}dx}$

It is known that – $\int{\sqrt{{{x}^{2}}-{{a}^{2}}}dx=\dfrac{x}{2}\sqrt{{{x}^{2}}-{{a}^{2}}}-\dfrac{{{a}^{2}}}{2}\log \left| x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right|+C}$

$\Rightarrow \int{\sqrt{{{(x-4)}^{2}}-{{(3)}^{2}}}dx}=\dfrac{(x-4)}{2}\sqrt{{{(x-4)}^{2}}-{{(3)}^{2}}}-\dfrac{{{(3)}^{2}}}{2}\log \left| (x-4)+\sqrt{{{(x-4)}^{2}}-{{(3)}^{2}}} \right|+C$$=\dfrac{x-4}{2}\sqrt{{{x}^{2}}-8x+7}-\dfrac{9}{2}\log \left| (x-4)+\sqrt{{{x}^{2}}-8x+7} \right|+C$

Thus, $\int{\sqrt{{{x}^{2}}-8x+7}dx}=\dfrac{x-4}{2}\sqrt{{{x}^{2}}-8x+7}-\dfrac{9}{2}\log \left| (x-4)+\sqrt{{{x}^{2}}-8x+7} \right|+C$

The correct answer is option D

### Exercise 7.8

1. Evaluate the definite integral– $\int\limits_{-1}^{1}{(x+1)dx}$

Ans: The second fundamental theorem of integral calculus states that – $\int\limits_{a}^{b}{f(x)dx}=F(b)-F(a)$

Here, $\int{(x+1)dx}=\dfrac{{{x}^{2}}}{2}+x$

So, $\int\limits_{-1}^{1}{(x+1)dx=}\left[ \dfrac{{{x}^{2}}}{2}+x \right]_{-1}^{1}$

$=\left[ \dfrac{{{1}^{2}}}{2}+1 \right]-\left[ \dfrac{{{(-1)}^{2}}}{2}+(-1) \right]$

$=\left[ \dfrac{1}{2}+1 \right]-\left[ \dfrac{1}{2}-1 \right]$

$=\dfrac{1}{2}+1-\dfrac{1}{2}+1=2$

Thus, $\int\limits_{-1}^{1}{(x+1)dx=}2$

2. Evaluate the definite integral– $\int\limits_{2}^{3}{\dfrac{1}{x}dx}$

Ans: The second fundamental theorem of integral calculus states that – $\int\limits_{a}^{b}{f(x)dx}=F(b)-F(a)$

Here, $\int{\dfrac{1}{x}dx}=\log \left| x \right|$

So, $\int\limits_{2}^{3}{\dfrac{1}{x}dx}=\left[ \log \left| x \right| \right]_{2}^{3}$

$=\left[ \log \left| 3 \right| \right]-\left[ \log \left| 2 \right| \right]$

$=\log \dfrac{3}{2}$

Thus, $\int\limits_{2}^{3}{\dfrac{1}{x}dx}=\log \dfrac{3}{2}$

3. Evaluate the definite integral– $\int\limits_{1}^{2}{(4{{x}^{3}}-5{{x}^{2}}+6x+9)dx}$

Ans: The second fundamental theorem of integral calculus states that – $\int\limits_{a}^{b}{f(x)dx}=F(b)-F(a)$

Here, $\int{(4{{x}^{3}}-5{{x}^{2}}+6x+9)dx}=4\left( \dfrac{{{x}^{4}}}{4} \right)-5\left( \dfrac{{{x}^{3}}}{3} \right)+6\left( \dfrac{{{x}^{2}}}{2} \right)+9x$

$={{x}^{4}}-\dfrac{5{{x}^{3}}}{3}+3{{x}^{2}}+9x$

So,  $\int\limits_{1}^{2}{(4{{x}^{3}}-5{{x}^{2}}+6x+9)dx}=\left[ {{x}^{4}}-\dfrac{5{{x}^{3}}}{3}+3{{x}^{2}}+9x \right]_{1}^{2}$

$=\left[ {{2}^{4}}-\dfrac{5{{(2)}^{3}}}{3}+3{{(2)}^{2}}+9(2) \right]-\left[ {{1}^{4}}-\dfrac{5{{(1)}^{3}}}{3}+3{{(1)}^{2}}+9(1) \right]$

$=\left[ 16-\dfrac{40}{3}+12+18 \right]-\left[ 1-\dfrac{5}{3}+3+9 \right]$

$=\left[ 46-\dfrac{40}{3} \right]-\left[ 13-\dfrac{5}{3} \right]$

$=46-\dfrac{40}{3}-13+\dfrac{5}{3}$

$=33-\dfrac{35}{3}$

$=\dfrac{99-35}{3}$

$=\dfrac{64}{3}$

Thus, $\int\limits_{1}^{2}{(4{{x}^{3}}-5{{x}^{2}}+6x+9)dx}=\dfrac{64}{3}$

4. Evaluate the definite integral– $\int\limits_{0}^{\dfrac{\pi }{4}}{\sin 2xdx}$

Ans: The second fundamental theorem of integral calculus states that – $\int\limits_{a}^{b}{f(x)dx}=F(b)-F(a)$

Here, $\int{\sin 2xdx}=\dfrac{-\cos 2x}{2}$

So, $\int\limits_{0}^{\dfrac{\pi }{4}}{\sin 2xdx}=\left[ \dfrac{-\cos 2x}{2} \right]_{0}^{\dfrac{\pi }{4}}$

$=\left[ \dfrac{-\cos 2\left( \dfrac{\pi }{4} \right)}{2} \right]-\left[ \dfrac{-\cos 0}{2} \right]$

$=\left[ \dfrac{-\cos \left( \dfrac{\pi }{2} \right)}{2} \right]+\left[ \dfrac{1}{2} \right]$

$=\left[ \dfrac{0}{2} \right]+\left[ \dfrac{1}{2} \right]$

$=\dfrac{1}{2}$

Thus, $\int\limits_{0}^{\dfrac{\pi }{4}}{\sin 2xdx}=\dfrac{1}{2}$

5. Evaluate the definite integral– $\int\limits_{0}^{\dfrac{\pi }{2}}{\cos 2xdx}$

Ans: The second fundamental theorem of integral calculus states that – $\int\limits_{a}^{b}{f(x)dx}=F(b)-F(a)$

Here, $\int{\cos 2xdx}=\dfrac{\sin 2x}{2}$

So, $\int\limits_{0}^{\dfrac{\pi }{2}}{\cos 2xdx}=\left[ \dfrac{\sin 2x}{2} \right]_{0}^{\dfrac{\pi }{2}}$

$=\left[ \dfrac{\sin 2\left( \dfrac{\pi }{2} \right)}{2} \right]-\left[ \dfrac{\sin 0}{2} \right]$

$=\left[ \dfrac{\sin \pi }{2} \right]+\left[ \dfrac{0}{2} \right]$

$=0+0$

$=0$

Thus,  $\int\limits_{0}^{\dfrac{\pi }{2}}{\cos 2xdx}=0$

6. Evaluate the definite integral– $\int\limits_{4}^{5}{{{e}^{x}}dx}$

Ans: The second fundamental theorem of integral calculus states that – $\int\limits_{a}^{b}{f(x)dx}=F(b)-F(a)$

Here, $\int{{{e}^{x}}dx}={{e}^{x}}$

So, $\int\limits_{4}^{5}{{{e}^{x}}dx}=\left[ {{e}^{x}} \right]_{4}^{5}$

$=\left[ {{e}^{5}} \right]-\left[ {{e}^{4}} \right]$

$={{e}^{4}}(e-1)$

Thus, $\int\limits_{4}^{5}{{{e}^{x}}dx}={{e}^{4}}(e-1)$

7. $\int\limits_{0}^{\dfrac{\pi }{4}}{\tan xdx}$

Ans: We know that,

$\int{\tan xdx}=-\log \left| \cos x \right|+C$

Therefore, by second fundamental theorem of calculus

$\int{\tan x dx} = \left[ -\log \left| \cos x \right| \right]_{0}^{\dfrac{\pi}{4}}$

$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{4}}{\tan xdx}=\left[ -\log \left| \cos \dfrac{\pi }{4} \right|+\log \left| \cos 0 \right| \right]$

$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{4}}{\tan xdx}=\left[ -\log \left| \dfrac{1}{\sqrt{2}} \right|+\log \left| 1 \right| \right]$

$\therefore \int\limits_{0}^{\dfrac{\pi }{4}}{\tan xdx}=\dfrac{1}{2}\log 2$

8. $\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}}{\cos ecxdx}$

Ans: We know that,

$\int{\cos ecxdx}=\log \left| \cos ecx-\cot x \right|+C$

Therefore, by second fundamental theorem of calculus

$\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}}{\cos ecxdx}=\left[ \log \left| \cos ecx-\cot x \right| \right]_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}}$

$\Rightarrow \int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}}{\cos ecxdx}=\left[ \log \left| \cos ec\dfrac{\pi }{4}-\cot \dfrac{\pi }{4} \right|-\log \left| \cos ec\dfrac{\pi }{6}-\cot \dfrac{\pi }{6} \right| \right]$

$\Rightarrow \int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}}{\cos ecxdx}=\left[ \log \left| \sqrt{2}-1 \right|-\log \left| 2-\sqrt{3} \right| \right]$

$\therefore \int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}}{\cos ecxdx}=\log \left( \dfrac{\sqrt{2}-1}{2-\sqrt{3}} \right)$

9. $\int\limits_{0}^{1}{\dfrac{1}{\sqrt{1-{{x}^{2}}}}dx}$

Ans: We know that,

$\int{\dfrac{1}{\sqrt{1-{{x}^{2}}}}dx}={{\sin }^{-1}}x+C$

Therefore, by second fundamental theorem of calculus

$\int\limits_{0}^{1}{\dfrac{1}{\sqrt{1-{{x}^{2}}}}dx}=\left[ {{\sin }^{-1}}x \right]_{0}^{1}$

$\Rightarrow \int\limits_{0}^{1}{\dfrac{1}{\sqrt{1-{{x}^{2}}}}dx}=\left[ {{\sin }^{-1}}1-{{\sin }^{-1}}0 \right]$

$\Rightarrow \int\limits_{0}^{1}{\dfrac{1}{\sqrt{1-{{x}^{2}}}}dx}=\left[ \dfrac{\pi }{2}-0 \right]$

$\therefore \int\limits_{0}^{1}{\dfrac{1}{\sqrt{1-{{x}^{2}}}}dx}=\dfrac{\pi }{2}$

10. $\int\limits_{0}^{1}{\dfrac{1}{1+{{x}^{2}}}dx}$

Ans: We know that,

$\int{\dfrac{1}{1+{{x}^{2}}}dx}={{\tan }^{-1}}x+C$

Therefore, by second fundamental theorem of calculus

$\int\limits_{0}^{1}{\dfrac{1}{1+{{x}^{2}}}dx}=\left[ {{\tan }^{-1}}x \right]_{0}^{1}$

$\Rightarrow \int\limits_{0}^{1}{\dfrac{1}{1+{{x}^{2}}}dx}=\left[ {{\tan }^{-1}}1-{{\tan }^{-1}}0 \right]$

$\Rightarrow \int\limits_{0}^{1}{\dfrac{1}{1+{{x}^{2}}}dx}=\left[ \dfrac{\pi }{4}-0 \right]$

$\therefore \int\limits_{0}^{1}{\dfrac{1}{1+{{x}^{2}}}dx}=\dfrac{\pi }{4}$

11. $\int\limits_{2}^{3}{\dfrac{1}{{{x}^{2}}-1}dx}$

Ans: We know that,

$\int{\dfrac{1}{{{x}^{2}}-1}dx}=\dfrac{1}{2}\log \left| \dfrac{x-1}{x+1} \right|+C$

Therefore, by second fundamental theorem of calculus

$\int\limits_{2}^{3}{\dfrac{1}{{{x}^{2}}-1}dx}=\dfrac{1}{2}\left[ \log \left| \dfrac{x-1}{x+1} \right| \right]_{2}^{3}$

$\Rightarrow \int\limits_{2}^{3}{\dfrac{1}{{{x}^{2}}-1}dx}=\dfrac{1}{2}\left[ \log \left| \dfrac{3-1}{3+1} \right|-\log \left| \dfrac{2-1}{2+1} \right| \right]$

$\Rightarrow \int\limits_{2}^{3}{\dfrac{1}{{{x}^{2}}-1}dx}=\dfrac{1}{2}\left[ \log \dfrac{1}{2}-\log \dfrac{1}{3} \right]$

$\therefore \int\limits_{2}^{3}{\dfrac{1}{{{x}^{2}}-1}dx}=\dfrac{1}{2}\log \dfrac{3}{2}$

12. $\int\limits_{0}^{\dfrac{\pi }{4}}{{{\cos }^{2}}xdx}$

Ans: We know that,

$\int{{{\cos }^{2}}xdx}=\int{\left( \dfrac{1+\cos 2x}{2} \right)dx}$

$\Rightarrow \int{{{\cos }^{2}}xdx}=\dfrac{x}{2}+\dfrac{\sin 2x}{4}$

Therefore, by second fundamental theorem of calculus

$\int\limits_{0}^{\dfrac{\pi }{4}}{{{\cos }^{2}}xdx}=\left[ \dfrac{x}{2}+\dfrac{\sin 2x}{4} \right]_{0}^{\dfrac{\pi }{4}}$

$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{4}}{{{\cos }^{2}}xdx}=\dfrac{1}{2}\left[ \dfrac{\pi }{2}-\dfrac{\sin \pi }{2}-0-\dfrac{\sin 0}{2} \right]$

$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{4}}{{{\cos }^{2}}xdx}=\dfrac{1}{2}\left[ \dfrac{\pi }{2}+0-0-0 \right]$

$\therefore \int\limits_{0}^{\dfrac{\pi }{4}}{{{\cos }^{2}}xdx}=\dfrac{\pi }{4}$

13. $\int\limits_{2}^{3}{\dfrac{xdx}{{{x}^{2}}+1}}$

Ans: We know that,

$\int{\dfrac{xdx}{{{x}^{2}}+1}}=\dfrac{1}{2}\int{\left( \dfrac{2x}{{{x}^{2}}+1} \right)dx}$

$\Rightarrow \int{\dfrac{xdx}{{{x}^{2}}+1}}=\dfrac{1}{2}\log \left( 1+{{x}^{2}} \right)$

Therefore, by second fundamental theorem of calculus

$\int\limits_{2}^{3}{\dfrac{xdx}{{{x}^{2}}+1}}=\left[ \log \left( 1+{{3}^{2}} \right)-\log \left( 1+{{2}^{2}} \right) \right]_{2}^{3}$

$\Rightarrow \int\limits_{2}^{3}{\dfrac{xdx}{{{x}^{2}}+1}}=\dfrac{1}{2}\left[ \log 10-\log 5 \right]$

$\Rightarrow \int\limits_{2}^{3}{\dfrac{xdx}{{{x}^{2}}+1}}=\dfrac{1}{2}\log \dfrac{10}{5}$

$\therefore \int\limits_{2}^{3}{\dfrac{xdx}{{{x}^{2}}+1}}=\dfrac{1}{4}\log 2$

14. $\int\limits_{0}^{1}{\dfrac{2x+3}{5{{x}^{2}}+1}dx}$

Ans: Solving $\int{\dfrac{2x+3}{5{{x}^{2}}+1}dx}$ ,

$\int{\dfrac{2x+3}{5{{x}^{2}}+1}dx}=\dfrac{1}{5}\int{\dfrac{5\left( 2x+3 \right)}{5{{x}^{2}}+1}dx}$

$\int{\dfrac{2x+3}{5{{x}^{2}}+1}dx}=\dfrac{1}{5}\int{\dfrac{10x+15}{5{{x}^{2}}+1}dx}$

$\int{\dfrac{2x+3}{5{{x}^{2}}+1}dx}=\dfrac{1}{5}\int{\dfrac{10x}{5{{x}^{2}}+1}dx}+3\int{\dfrac{1}{5{{x}^{2}}+1}dx}$

$\int{\dfrac{2x+3}{5{{x}^{2}}+1}dx}=\dfrac{1}{5}\int{\dfrac{10x}{5{{x}^{2}}+1}dx}+3\int{\dfrac{1}{5\left( {{x}^{2}}+\dfrac{1}{5} \right)}dx}$

$\int{\dfrac{2x+3}{5{{x}^{2}}+1}dx}=\dfrac{1}{5}\log \left( 5{{x}^{2}}+1 \right)+\dfrac{3}{\sqrt{5}}{{\tan }^{-1}}\left( \sqrt{5} \right)x$

Therefore, by second fundamental theorem of calculus

$\int\limits_{0}^{1}{\dfrac{2x+3}{5{{x}^{2}}+1}dx}=\left\{ \dfrac{1}{5}\log \left( 5+1 \right)+\dfrac{3}{\sqrt{5}}{{\tan }^{-1}}\left( \sqrt{5} \right)x \right\}-\left\{ \dfrac{1}{5}\log \left( 1 \right)+\dfrac{3}{\sqrt{5}}{{\tan }^{-1}}0 \right\}$

$\therefore \int\limits_{0}^{1}{\dfrac{2x+3}{5{{x}^{2}}+1}dx}=\dfrac{1}{5}\log 6+\dfrac{3}{\sqrt{5}}{{\tan }^{-1}}\sqrt{5}$

15. $\int\limits_{0}^{1}{x{{e}^{{{x}^{2}}}}dx}$

Ans: Let ${{x}^{2}}=t$ ,

Differentiating it we get,

$2xdx=dt$

Therefore, the integral becomes,

$\dfrac{1}{2}\int\limits_{0}^{1}{{{e}^{t}}dt}$

$\dfrac{1}{2}\int\limits_{0}^{1}{{{e}^{t}}dt}=\left[ \dfrac{1}{2}{{e}^{t}} \right]_{0}^{1}$

$\dfrac{1}{2}\int\limits_{0}^{1}{{{e}^{t}}dt}=\dfrac{1}{2}e-\dfrac{1}{2}{{e}^{0}}$

$\dfrac{1}{2}\left( e-1 \right)$

16. $\int\limits_{1}^{2}{\dfrac{5{{x}^{2}}}{{{x}^{2}}+4x+3}dx}$

Ans: The given integral can be written as

$\int\limits_{1}^{2}{\dfrac{5{{x}^{2}}}{{{x}^{2}}+4x+3}dx}=\int\limits_{1}^{2}{\left\{ 5-\dfrac{20x+15}{{{x}^{2}}+4x+3} \right\}dx}$

$\int\limits_{1}^{2}{\dfrac{5{{x}^{2}}}{{{x}^{2}}+4x+3}dx}=\left[ 5x \right]_{1}^{2}-\int\limits_{1}^{2}{\dfrac{20x+15}{{{x}^{2}}+4x+3}dx}$                             …(1)

Solving $\int\limits_{1}^{2}{\dfrac{20x+15}{{{x}^{2}}+4x+3}dx}$ ,

Let $20x+15=A\dfrac{d}{dx}\left( {{x}^{2}}+4x+3 \right)+B$

Equating the coefficients of $x$ and constant term we get,

$A=10,B=-25$

Let ${{x}^{2}}+4x+3=t$

Differentiating it we get,

$\left( 2x+4 \right)dx=dt$

Therefore, the integral becomes

$10\int{\dfrac{dt}{t}-25\int{\dfrac{dx}{{{\left( x+2 \right)}^{2}}-{{1}^{2}}}}}$

$10\int{\dfrac{dt}{t}-25\int{\dfrac{dx}{{{\left( x+2 \right)}^{2}}-{{1}^{2}}}}}=10\log t-25\left[ \dfrac{1}{2}\log \left( \dfrac{x+2-1}{x+2+1} \right) \right]$

$\Rightarrow \int\limits_{1}^{2}{\dfrac{20x+15}{{{x}^{2}}+4x+3}dx}=\left[ 10\log \left( {{x}^{2}}+4x+3 \right)-25\left[ \dfrac{1}{2}\log \left( \dfrac{x+1}{x+3} \right) \right] \right]_{1}^{2}$

$\Rightarrow \int\limits_{1}^{2}{\dfrac{20x+15}{{{x}^{2}}+4x+3}dx}=10\log 15-10\log 8-25\left[ \dfrac{1}{2}\log \dfrac{3}{5}-\dfrac{1}{2}\log \dfrac{2}{4} \right]$

$\Rightarrow \int\limits_{1}^{2}{\dfrac{20x+15}{{{x}^{2}}+4x+3}dx}=10\log 5+10\log 3-10\log 4-10\log 2-\dfrac{25}{2}\left[ \log 3-\log 5-\log 2+\log 4 \right]$

$\Rightarrow \int\limits_{1}^{2}{\dfrac{20x+15}{{{x}^{2}}+4x+3}dx}=\dfrac{45}{2}\log 5-\dfrac{45}{2}\log 4-\dfrac{5}{2}\log 3+\dfrac{5}{2}\log 2$

$\therefore \int\limits_{1}^{2}{\dfrac{20x+15}{{{x}^{2}}+4x+3}dx}=\dfrac{45}{2}\log \dfrac{5}{4}-\dfrac{5}{2}\log \dfrac{3}{2}$

Substituting it in (1) we get,

$\int\limits_{1}^{2}{\dfrac{5{{x}^{2}}}{{{x}^{2}}+4x+3}dx}=5-\left[ \dfrac{45}{2}\log \dfrac{5}{4}-\dfrac{5}{2}\log \dfrac{3}{2} \right]$

$\therefore \int\limits_{1}^{2}{\dfrac{5{{x}^{2}}}{{{x}^{2}}+4x+3}dx}=5-\dfrac{5}{2}\left[ 9\log \dfrac{5}{4}-\log \dfrac{3}{2} \right]$

17. $\int\limits_{0}^{\dfrac{\pi }{4}}{\left( 2{{\sec }^{2}}x+{{x}^{3}}+2 \right)dx}$

Ans: We know that,

$\int{\left( 2{{\sec }^{2}}x+{{x}^{3}}+2 \right)dx}=2\tan x+\dfrac{{{x}^{4}}}{4}+2x$

Therefore, by second fundamental theorem of calculus

$\int{\left( 2{{\sec }^{2}}x+{{x}^{3}}+2 \right)dx}=\left[ 2\tan x+\dfrac{{{x}^{4}}}{4}+2x \right]_{0}^{\dfrac{\pi }{4}}$

$\Rightarrow \int{\left( 2{{\sec }^{2}}x+{{x}^{3}}+2 \right)dx}=\left[ 2\tan \dfrac{\pi }{4}+\dfrac{1}{4}{{\left( \dfrac{\pi }{4} \right)}^{2}}+2\left( \dfrac{\pi }{4} \right)-\left( 2\tan 0+0+0 \right) \right]$

$\Rightarrow \int{\left( 2{{\sec }^{2}}x+{{x}^{3}}+2 \right)dx}=2\tan \dfrac{\pi }{4}+\dfrac{{{\pi }^{4}}}{{{4}^{5}}}+\dfrac{\pi }{2}$

$\therefore \int{\left( 2{{\sec }^{2}}x+{{x}^{3}}+2 \right)dx}=2+\dfrac{\pi }{2}+\dfrac{{{\pi }^{4}}}{1024}$

18. $\int\limits_{0}^{\pi }{\left( {{\sin }^{2}}\dfrac{x}{2}-{{\cos }^{2}}\dfrac{x}{2} \right)dx}$

Ans: We know that,

$\int\limits_{0}^{\pi }{\left( {{\sin }^{2}}\dfrac{x}{2}-{{\cos }^{2}}\dfrac{x}{2} \right)dx}=-\int\limits_{0}^{\pi }{\left( {{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2} \right)dx}$

$\Rightarrow -\int\limits_{0}^{\pi }{\left( {{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2} \right)dx}=-\int\limits_{0}^{\pi }{\cos xdx}$

$\int{\cos xdx}=\sin x+C$

Therefore, by second fundamental theorem of calculus

$\int\limits_{0}^{\pi }{\cos xdx}=\sin \pi -sin0$

$\therefore \int\limits_{0}^{\pi }{\cos xdx}=0$

19. $\int\limits_{0}^{2}{\dfrac{6x+3}{{{x}^{2}}+4}dx}$

Ans: Solving the integral we get,

$\int{\dfrac{6x+3}{{{x}^{2}}+4}dx}=3\int{\dfrac{2x+1}{{{x}^{2}}+4}dx}$

$\int{\dfrac{6x+3}{{{x}^{2}}+4}dx}=3\int{\dfrac{2x}{{{x}^{2}}+4}dx}+3\int{\dfrac{1}{{{x}^{2}}+4}dx}$

$\therefore \int{\dfrac{6x+3}{{{x}^{2}}+4}dx}=3\log \left( {{x}^{2}}+4 \right)+\dfrac{3}{2}{{\tan }^{-1}}\dfrac{x}{2}$

Therefore, by second fundamental theorem of calculus

$\int\limits_{0}^{2}{\dfrac{6x+3}{{{x}^{2}}+4}dx}=\left[ 3\log \left( {{x}^{2}}+4 \right)+\dfrac{3}{2}{{\tan }^{-1}}\dfrac{x}{2} \right]_{0}^{2}$

$\Rightarrow \int\limits_{0}^{2}{\dfrac{6x+3}{{{x}^{2}}+4}dx}=\left[ 3\log \left( {{2}^{2}}+4 \right)+\dfrac{3}{2}{{\tan }^{-1}}\dfrac{2}{2}-3\log \left( {{0}^{2}}+4 \right)-\dfrac{3}{2}{{\tan }^{-1}}\dfrac{0}{2} \right]$

$\Rightarrow \int\limits_{0}^{2}{\dfrac{6x+3}{{{x}^{2}}+4}dx}=3\log 8+\dfrac{3}{2}{{\tan }^{-1}}1-3\log 4-\dfrac{3}{2}{{\tan }^{-1}}0$

$\Rightarrow \int\limits_{0}^{2}{\dfrac{6x+3}{{{x}^{2}}+4}dx}=3\log 8+\dfrac{3}{2}\left( \dfrac{\pi }{4} \right)-3\log 4-0$

$\Rightarrow \int\limits_{0}^{2}{\dfrac{6x+3}{{{x}^{2}}+4}dx}=3\log \dfrac{8}{4}+\dfrac{3\pi }{8}$

$\therefore \int\limits_{0}^{2}{\dfrac{6x+3}{{{x}^{2}}+4}dx}=3\log 2+\dfrac{3\pi }{8}$

20. $\int\limits_{0}^{1}{\left( x{{e}^{x}}+\sin \dfrac{\pi x}{4} \right)dx}$

Ans: Solving the integral we get,

$\int\limits_{0}^{1}{\left( x{{e}^{x}}+\sin \dfrac{\pi x}{4} \right)dx}=x\int{{{e}^{x}}dx}-\int{\left\{ \left( \dfrac{d}{dx}x \right)\int{{{e}^{x}}dx} \right\}dx}+\left\{ \dfrac{-\cos \dfrac{\pi x}{4}}{\dfrac{\pi }{4}} \right\}$

$\int\limits_{0}^{1}{\left( x{{e}^{x}}+\sin \dfrac{\pi x}{4} \right)dx}=x{{e}^{x}}-\int{{{e}^{x}}dx}-\dfrac{4}{\pi }\cos \dfrac{x}{4}$

$\therefore \int\limits_{0}^{1}{\left( x{{e}^{x}}+\sin \dfrac{\pi x}{4} \right)dx}=x{{e}^{x}}-{{e}^{x}}-\dfrac{4}{\pi }\cos \dfrac{x}{4}$

Therefore, by second fundamental theorem of calculus

$\int\limits_{0}^{1}{\left( x{{e}^{x}}+\sin \dfrac{\pi x}{4} \right)dx}=\left( 1{{e}^{1}}-{{e}^{1}}-\dfrac{4}{\pi }\cos \dfrac{\pi }{4} \right)-\left( 0{{e}^{0}}-{{e}^{0}}-\dfrac{4}{\pi }\cos 0 \right)$

$\therefore \int\limits_{0}^{1}{\left( x{{e}^{x}}+\sin \dfrac{\pi x}{4} \right)dx}=\left( 1+\dfrac{4}{\pi }-\dfrac{2\sqrt{2}}{\pi } \right)$

21. $\int\limits_{1}^{\sqrt{3}}{\dfrac{1}{1+{{x}^{2}}}dx}$

1. $\dfrac{\pi }{3}$

2. $\dfrac{2\pi }{3}$

3. $\dfrac{\pi }{6}$

4. $\dfrac{\pi }{12}$

Ans: Solving the integral we get,

$\int\limits_{1}^{\sqrt{3}}{\dfrac{1}{1+{{x}^{2}}}dx}={{\tan }^{-1}}x$

Therefore, by second fundamental theorem of calculus

$\int\limits_{1}^{\sqrt{3}}{\dfrac{1}{1+{{x}^{2}}}dx}={{\tan }^{-1}}\sqrt{3}-{{\tan }^{-1}}1$

$\Rightarrow \int\limits_{1}^{\sqrt{3}}{\dfrac{1}{1+{{x}^{2}}}dx}=\dfrac{\pi }{3}-\dfrac{\pi }{4}$

$\therefore \int\limits_{1}^{\sqrt{3}}{\dfrac{1}{1+{{x}^{2}}}dx}=\dfrac{\pi }{12}$

Thus, the correct option is (D)

22. $\int\limits_{0}^{\dfrac{2}{3}}{\dfrac{1}{4+9{{x}^{2}}}dx}$

1. $\dfrac{\pi }{6}$

2. $\dfrac{\pi }{12}$

3. $\dfrac{\pi }{24}$

4. $\dfrac{\pi }{4}$

Ans: Solving the integral we get,

$\int{\dfrac{1}{4+9{{x}^{2}}}dx}=\int{\dfrac{1}{{{2}^{2}}+{{\left( 3x \right)}^{2}}}dx}$

Let $3x=t$ ,

Differentiating it we get,

$3dx=dt$

$\therefore \int{\dfrac{1}{{{2}^{2}}+{{\left( 3x \right)}^{2}}}dx}=\dfrac{1}{6}{{\tan }^{-1}}\left( \dfrac{3x}{2} \right)$

Therefore, by second fundamental theorem of calculus

$\int\limits_{0}^{\dfrac{2}{3}}{\dfrac{1}{4+9{{x}^{2}}}dx}=\dfrac{1}{6}{{\tan }^{-1}}\left( \dfrac{3}{2}\cdot \dfrac{2}{3} \right)-\dfrac{1}{6}{{\tan }^{-1}}0$

$\int\limits_{0}^{\dfrac{2}{3}}{\dfrac{1}{4+9{{x}^{2}}}dx}=\dfrac{1}{6}\cdot \dfrac{\pi }{4}$

$\therefore \int\limits_{0}^{\dfrac{2}{3}}{\dfrac{1}{4+9{{x}^{2}}}dx}=\dfrac{\pi }{24}$

Thus, the correct answer is (C).

### Exercise 7.9

Solve the following integrals.

1. $\int\limits_{0}^{1}{\dfrac{x}{{{x}^{2}}+1}dx}$

Ans: Let ${{x}^{2}}+1$

Differentiating $2xdx=dt$ ,

Therefore, the integral becomes

$\int\limits_{0}^{1}{\dfrac{x}{{{x}^{2}}+1}dx}=\dfrac{1}{2}\int\limits_{1}^{2}{\dfrac{dt}{t}}$

$\Rightarrow \int\limits_{0}^{1}{\dfrac{x}{{{x}^{2}}+1}dx}=\dfrac{1}{2}\left[ \log \left| t \right| \right]_{1}^{2}$

$\Rightarrow \int\limits_{0}^{1}{\dfrac{x}{{{x}^{2}}+1}dx}=\dfrac{1}{2}\left[ \log 2-\log 1 \right]$

$\Rightarrow \int\limits_{0}^{1}{\dfrac{x}{{{x}^{2}}+1}dx}=\dfrac{1}{2}\log 2$

2. $\int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\sin \phi }{{\cos }^{5}}\phi d\phi }$

Ans: The integral can be written as:

$\int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\sin \phi }{{\cos }^{5}}\phi d\phi }=\int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\sin \phi }{{\cos }^{4}}\phi \cos \phi d\phi }$

$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\sin \phi }{{\cos }^{5}}\phi d\phi }=\dfrac{64}{231}$

Let $\sin \phi =t$

Differentiating it we get,

$\cos \phi d\phi =dt$

Therefore, the integral becomes

$\int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\sin \phi }{{\cos }^{5}}\phi d\phi }=\int\limits_{0}^{1}{\sqrt{t}{{\left( 1-{{t}^{2}} \right)}^{2}}dt}$

$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\sin \phi }{{\cos }^{5}}\phi d\phi }=\int\limits_{0}^{1}{\sqrt{t}\left( 1+{{t}^{4}}-2{{t}^{2}} \right)dt}$

$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\sin \phi }{{\cos }^{5}}\phi d\phi }=\int\limits_{0}^{1}{\left( {{t}^{\dfrac{1}{2}}}+{{t}^{\dfrac{9}{2}}}-2{{t}^{\dfrac{5}{2}}} \right)dt}$

$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\sin \phi }{{\cos }^{5}}\phi d\phi }=\left[ \dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}+\dfrac{{{t}^{\dfrac{11}{2}}}}{\dfrac{11}{2}}-\dfrac{2{{t}^{\dfrac{7}{2}}}}{\dfrac{7}{2}} \right]_{0}^{1}$

$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\sin \phi }{{\cos }^{5}}\phi d\phi }=\dfrac{2}{3}+\dfrac{2}{11}-\dfrac{4}{7}$

$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\sin \phi }{{\cos }^{5}}\phi d\phi }=\dfrac{154+42-132}{231}$

$\therefore \int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\sin \phi }{{\cos }^{5}}\phi d\phi }=\dfrac{64}{231}$

3. $\int\limits_{0}^{1}{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)dx}$

Ans: Let $x=\tan \theta$

Differentiating it we get,

$dx={{\sec }^{2}}\theta d\theta$

Therefore, the integral becomes

$\int\limits_{0}^{1}{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)dx}=\int\limits_{0}^{\dfrac{\pi }{4}}{{{\sin }^{-1}}\left( \dfrac{2\tan \theta }{1+{{\left( \tan \theta \right)}^{2}}} \right){{\sec }^{2}}\theta d\theta }$

$\int\limits_{0}^{1}{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)dx}=\int\limits_{0}^{\dfrac{\pi }{4}}{{{\sin }^{-1}}\left( \sin 2\theta \right){{\sec }^{2}}\theta d\theta }$

$\int\limits_{0}^{1}{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)dx}=2\int\limits_{0}^{\dfrac{\pi }{4}}{\theta {{\sec }^{2}}\theta d\theta }$

Let $u=\theta$

And $v={{\sec }^{2}}\theta$

Using integration by parts we get,

$\int\limits_{0}^{1}{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)dx}=2\left[ \theta \int{{{\sec }^{2}}\theta d\theta }-\int{\left\{ \left( \dfrac{d}{d\theta }\theta \right)\int{{{\sec }^{2}}\theta d\theta } \right\}d\theta } \right]_{0}^{\dfrac{\pi }{4}}$

$\Rightarrow \int\limits_{0}^{1}{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)dx}=2\left[ \theta \tan \theta -\int{\tan \theta d\theta } \right]_{0}^{\dfrac{\pi }{4}}$

$\Rightarrow \int\limits_{0}^{1}{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)dx}=2\left[ \theta \tan \theta -\log \left| \cos \theta \right| \right]_{0}^{\dfrac{\pi }{4}}$

$\Rightarrow \int\limits_{0}^{1}{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)dx}=2\left[ \dfrac{\pi }{4}\tan \dfrac{\pi }{4}-\log \left| \cos \dfrac{\pi }{4} \right|-\log \left| \cos 0 \right| \right]$

$\Rightarrow \int\limits_{0}^{1}{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)dx}=2\left[ \dfrac{\pi }{4}-\log \left| \dfrac{1}{\sqrt{2}} \right|-\log 1 \right]$

$\therefore \int\limits_{0}^{1}{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)dx}=\dfrac{\pi }{2}-\log 2$

4. $\int\limits_{0}^{2}{x\sqrt{x+2}dx}$

(Put $x+2={{t}^{2}}$)

Ans: Let $x+2={{t}^{2}}$

Differentiating it, $dx=2tdt$

Therefore, the integral becomes

$\int\limits_{0}^{2}{x\sqrt{x+2}dx}=\int\limits_{\sqrt{2}}^{2}{2{{t}^{2}}\left( {{t}^{2}}-2 \right)dt}$

$\Rightarrow \int\limits_{0}^{2}{x\sqrt{x+2}dx}=2\left[ \dfrac{{{t}^{5}}}{5}-\dfrac{2{{t}^{3}}}{3} \right]_{\sqrt{2}}^{2}$

$\Rightarrow \int\limits_{0}^{2}{x\sqrt{x+2}dx}=2\left[ \dfrac{35}{5}-\dfrac{16}{3}-\dfrac{4\sqrt{2}}{5}+\dfrac{4\sqrt{2}}{3} \right]$

$\therefore \int\limits_{0}^{2}{x\sqrt{x+2}dx}=\dfrac{16\sqrt{2}\left( \sqrt{2}+1 \right)}{15}$

5. $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x}{1+{{\cos }^{2}}x}dx}$

Ans: Let $\cos x=t$

Differentiating it, $-\sin xdx=dt$

Therefore, the integral becomes

$\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x}{1+{{\cos }^{2}}x}dx}=-\int\limits_{1}^{0}{\dfrac{dt}{1+{{t}^{2}}}}$

$\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x}{1+{{\cos }^{2}}x}dx}=-\left[ {{\tan }^{-1}}t \right]_{1}^{0}$

$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x}{1+{{\cos }^{2}}x}dx}=-\left[ {{\tan }^{-1}}0-{{\tan }^{-1}}1 \right]$

$\therefore \int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x}{1+{{\cos }^{2}}x}dx}=\dfrac{\pi }{4}$

6. $\int\limits_{0}^{2}{\dfrac{1}{x+4-{{x}^{2}}}dx}$

Ans: The integral can be written as

$\int\limits_{0}^{2}{\dfrac{1}{x+4-{{x}^{2}}}dx}=\int\limits_{0}^{2}{\dfrac{dx}{-\left( {{\left( x-\dfrac{1}{2} \right)}^{2}}-\dfrac{17}{4} \right)}}$

$\int\limits_{0}^{2}{\dfrac{1}{x+4-{{x}^{2}}}dx}=\int\limits_{0}^{2}{\dfrac{dx}{\left( {{\left( \dfrac{\sqrt{17}}{2} \right)}^{2}}-{{\left( x-\dfrac{1}{2} \right)}^{2}} \right)}}$

Let $x-\dfrac{1}{2}=t$

Differentiating it we get,

$dx=dt$

Therefore, the integral becomes

$\int\limits_{0}^{2}{\dfrac{1}{x+4-{{x}^{2}}}dx}=\int\limits_{\dfrac{-1}{2}}^{\dfrac{3}{2}}{\dfrac{dt}{\left( {{\left( \dfrac{\sqrt{17}}{2} \right)}^{2}}-{{\left( t \right)}^{2}} \right)}}$

$\Rightarrow \int\limits_{\dfrac{-1}{2}}^{\dfrac{3}{2}}{\dfrac{dt}{\left( {{\left( \dfrac{\sqrt{17}}{2} \right)}^{2}}-{{\left( t \right)}^{2}} \right)}}=\left[ \dfrac{1}{2\left( \dfrac{\sqrt{17}}{2} \right)}\log \dfrac{\left( \dfrac{\sqrt{17}}{2} \right)+t}{\left( \dfrac{\sqrt{17}}{2} \right)-t} \right]_{\dfrac{-1}{2}}^{\dfrac{3}{2}}$

$\Rightarrow \int\limits_{\dfrac{-1}{2}}^{\dfrac{3}{2}}{\dfrac{dt}{\left( {{\left( \dfrac{\sqrt{17}}{2} \right)}^{2}}-{{\left( t \right)}^{2}} \right)}}=\dfrac{1}{\sqrt{17}}\left[ \log \dfrac{\left( \dfrac{\sqrt{17}}{2} \right)+\dfrac{3}{2}}{\left( \dfrac{\sqrt{17}}{2} \right)-\dfrac{3}{2}}-\log \dfrac{\left( \dfrac{\sqrt{17}}{2} \right)-\dfrac{1}{2}}{\left( \dfrac{\sqrt{17}}{2} \right)+\dfrac{1}{2}} \right]$

$\Rightarrow \int\limits_{\dfrac{-1}{2}}^{\dfrac{3}{2}}{\dfrac{dt}{\left( {{\left( \dfrac{\sqrt{17}}{2} \right)}^{2}}-{{\left( t \right)}^{2}} \right)}}=\dfrac{1}{\sqrt{17}}\left[ \log \dfrac{\sqrt{17}+3}{\sqrt{17}-3}-\log \dfrac{\sqrt{17}-1}{\sqrt{17}+1} \right]$

$\Rightarrow \int\limits_{\dfrac{-1}{2}}^{\dfrac{3}{2}}{\dfrac{dt}{\left( {{\left( \dfrac{\sqrt{17}}{2} \right)}^{2}}-{{\left( t \right)}^{2}} \right)}}=\dfrac{1}{\sqrt{17}}\left[ \log \dfrac{\sqrt{17}+3}{\sqrt{17}-3}\times \dfrac{\sqrt{17}+1}{\sqrt{17}-1} \right]$

$\Rightarrow \int\limits_{\dfrac{-1}{2}}^{\dfrac{3}{2}}{\dfrac{dt}{\left( {{\left( \dfrac{\sqrt{17}}{2} \right)}^{2}}-{{\left( t \right)}^{2}} \right)}}=\dfrac{1}{\sqrt{17}}\left[ \log \dfrac{25+17+10\sqrt{17}}{8} \right]$

$\therefore \int\limits_{0}^{2}{\dfrac{1}{x+4-{{x}^{2}}}dx}=\dfrac{1}{\sqrt{17}}\left[ \log \dfrac{21+5\sqrt{17}}{4} \right]$

7. $\int\limits_{-1}^{1}{\dfrac{1}{{{x}^{2}}+2x+5}dx}$

Ans: The integral can be written as

$\int\limits_{-1}^{1}{\dfrac{1}{{{x}^{2}}+2x+5}dx}=\int\limits_{-1}^{1}{\dfrac{1}{{{\left( x-1 \right)}^{2}}+{{2}^{2}}}dx}$

Let $x+1=t$

Differentiating it we get,

$dx=dt$

Therefore, the integral becomes

$\int\limits_{-1}^{1}{\dfrac{1}{{{x}^{2}}+2x+5}dx}=\int\limits_{0}^{2}{\dfrac{dt}{{{t}^{2}}+{{2}^{2}}}}$

$\Rightarrow \int\limits_{-1}^{1}{\dfrac{1}{{{x}^{2}}+2x+5}dx}=\left[ \dfrac{1}{2}{{\tan }^{-1}}\dfrac{t}{2} \right]_{0}^{2}$

$\Rightarrow \int\limits_{-1}^{1}{\dfrac{1}{{{x}^{2}}+2x+5}dx}=\left[ \dfrac{1}{2}{{\tan }^{-1}}1-\dfrac{1}{2}{{\tan }^{-1}}0 \right]$

$\therefore \int\limits_{-1}^{1}{\dfrac{1}{{{x}^{2}}+2x+5}dx}=\dfrac{\pi }{8}$

8. $\int\limits_{1}^{2}{\left( \dfrac{1}{x}-\dfrac{1}{2{{x}^{2}}} \right){{e}^{2x}}dx}$

Ans: Let $2x=t$

Differentiating it $2dx=dt$ ,

Therefore, the integral becomes

$\int\limits_{1}^{2}{\left( \dfrac{1}{x}-\dfrac{1}{2{{x}^{2}}} \right){{e}^{2x}}dx}=\dfrac{1}{2}\int\limits_{2}^{1}{\left( \dfrac{2}{t}-\dfrac{2}{{{t}^{2}}} \right){{e}^{t}}dt}$

$\Rightarrow \int\limits_{1}^{2}{\left( \dfrac{1}{x}-\dfrac{1}{2{{x}^{2}}} \right){{e}^{2x}}dx}=\int\limits_{2}^{1}{\left( \dfrac{1}{t}-\dfrac{1}{{{t}^{2}}} \right){{e}^{t}}dt}$

Let $\dfrac{1}{t}=f\left( t \right)$

Then, $f'\left( t \right)=-\dfrac{1}{{{t}^{2}}}$

$\int\limits_{2}^{1}{\left( \dfrac{1}{t}-\dfrac{1}{{{t}^{2}}} \right){{e}^{t}}dt}=\int\limits_{2}^{1}{\left[ f\left( t \right)+f'\left( t \right) \right]{{e}^{t}}dt}$

$\Rightarrow \int\limits_{2}^{1}{\left( \dfrac{1}{t}-\dfrac{1}{{{t}^{2}}} \right){{e}^{t}}dt}=\left[ {{e}^{t}}f\left( t \right) \right]_{2}^{4}$

$\Rightarrow \int\limits_{2}^{1}{\left( \dfrac{1}{t}-\dfrac{1}{{{t}^{2}}} \right){{e}^{t}}dt}=\left[ \dfrac{{{e}^{t}}}{t} \right]_{2}^{4}$

$\Rightarrow \int\limits_{2}^{1}{\left( \dfrac{1}{t}-\dfrac{1}{{{t}^{2}}} \right){{e}^{t}}dt}=\dfrac{{{e}^{4}}}{4}-\dfrac{{{e}^{2}}}{2}$

$\therefore \int\limits_{1}^{2}{\left( \dfrac{1}{x}-\dfrac{1}{2{{x}^{2}}} \right){{e}^{2x}}dx}=\dfrac{{{e}^{2}}\left( {{e}^{2}}-2 \right)}{4}$

9. The value of the integral $\int\limits_{\dfrac{1}{3}}^{1}{\dfrac{{{\left( x-{{x}^{3}} \right)}^{\dfrac{1}{3}}}}{{{x}^{4}}}dx}$

1. $6$

2. $0$

3. $3$

4. $4$

Ans: Let $x=\sin \theta$

Differentiating it, $dx=\cos \theta d\theta$

Therefore, the integral becomes

$\int\limits_{\dfrac{1}{3}}^{1}{\dfrac{{{\left( x-{{x}^{3}} \right)}^{\dfrac{1}{3}}}}{{{x}^{4}}}dx}=\int\limits_{{{\sin }^{-1}}\dfrac{1}{3}}^{\dfrac{\pi }{2}}{\dfrac{{{\left( \sin \theta -{{\sin }^{3}}\theta \right)}^{\dfrac{1}{3}}}}{{{\sin }^{4}}\theta }\cos \theta d\theta }$

$\Rightarrow \int\limits_{{{\sin }^{-1}}\dfrac{1}{3}}^{\dfrac{\pi }{2}}{\dfrac{{{\left( \sin \theta -{{\sin }^{3}}\theta \right)}^{\dfrac{1}{3}}}}{{{\sin }^{4}}\theta }\cos \theta d\theta }=\int \limits_{{{\sin }^{-1}}\dfrac{1}{3}}^{\dfrac{\pi }{2}}{\dfrac{{{\left( \sin \theta \right)}^{\dfrac{1}{3}}}{{\left( \cos \theta \right)}^{\dfrac{2}{3}}}}{{{\sin }^{4}}\theta }\cos \theta d\theta }$

$\Rightarrow \int\limits_{{{\sin }^{-1}}\dfrac{1}{3}}^{\dfrac{\pi }{2}}{\dfrac{{{\left( \sin \theta -{{\sin }^{3}}\theta \right)}^{\dfrac{1}{3}}}}{{{\sin }^{4}}\theta }\cos \theta d\theta }=\int\limits_{{{\sin }^{-1}}\dfrac{1}{3}}^{\dfrac{\pi }{2}}{\dfrac{{{\left( \cos \theta \right)}^{\dfrac{5}{3}}}}{{{\sin }^{\dfrac{5}{3}}}\theta }\cos e{{c}^{2}}\theta d\theta }$

$\Rightarrow \int\limits_{{{\sin }^{-1}}\dfrac{1}{3}}^{\dfrac{\pi }{2}}{\dfrac{{{\left( \sin \theta -{{\sin }^{3}}\theta \right)}^{\dfrac{1}{3}}}}{{{\sin }^{4}}\theta }\cos \theta d\theta }=\int\limits_{{{\sin }^{-1}}\dfrac{1}{3}}^{\dfrac{\pi }{2}}{{{\left( \cot \theta \right)}^{\dfrac{5}{3}}}\cos e{{c}^{2}}\theta d\theta }$

Let $\cot \theta =t$

Differentiating it, $-\cos e{{c}^{2}}\theta d\theta =dt$

Therefore, the integral becomes

$\int\limits_{\dfrac{1}{3}}^{1}{\dfrac{{{\left( x-{{x}^{3}} \right)}^{\dfrac{1}{3}}}}{{{x}^{4}}}dx}=\int\limits_{2\sqrt{2}}^{0}{{{t}^{\dfrac{5}{3}}}dt}$

$\Rightarrow \int\limits_{\dfrac{1}{3}}^{1}{\dfrac{{{\left( x-{{x}^{3}} \right)}^{\dfrac{1}{3}}}}{{{x}^{4}}}dx}=\dfrac{3}{8}{{\sqrt{8}}^{\dfrac{8}{3}}}$

$\Rightarrow \int\limits_{\dfrac{1}{3}}^{1}{\dfrac{{{\left( x-{{x}^{3}} \right)}^{\dfrac{1}{3}}}}{{{x}^{4}}}dx}=\dfrac{3}{8}\times 16$

$\therefore \int\limits_{\dfrac{1}{3}}^{1}{\dfrac{{{\left( x-{{x}^{3}} \right)}^{\dfrac{1}{3}}}}{{{x}^{4}}}dx}=6$

Thus, the correct answer is A.

10. If $f\left( x \right)=\int\limits_{0}^{x}{t\sin tdt}$ ,then $f'\left( x \right)$ is

1. $\cos x+x\sin x$

2. $x\sin x$

3. $x\cos x$

4. $\sin x+x\cos x$

Ans: Given $f\left( x \right)=\int\limits_{0}^{x}{t\sin tdt}$

Using Integration by parts, we get

$f\left( x \right)=t\int\limits_{0}^{x}{\sin tdt}-\int\limits_{0}^{x}{\left\{ \left( \dfrac{d}{dt}t \right)\int{\sin tdt} \right\}dt}$

$\Rightarrow f\left( x \right)=\left[ t\left( -\cos t \right) \right]_{0}^{x}-\int\limits_{0}^{x}{-\cot tdt}$

$\Rightarrow f\left( x \right)=\left[ -t\left( \cos t \right)+\sin t \right]_{0}^{x}$

$\Rightarrow f\left( x \right)=-x\cos x+\sin x$

Therefore,

$f'\left( x \right)=-\left[ -x\sin x+\cos x \right]+\cos x$

$\therefore f'\left( x \right)=x\sin x$

Thus, the correct answer is B.

### Exercise 7.10

Solve the following integrals.

1. $\int\limits_{0}^{\dfrac{\pi }{2}}{{{\cos }^{2}}xdx}$

Ans: Given $I=\int\limits_{0}^{\dfrac{\pi }{2}}{{{\cos }^{2}}xdx}$                     …(1)

We know that,

$\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$

Therefore, the integral becomes

$I=\int\limits_{0}^{\dfrac{\pi }{2}}{{{\cos }^{2}}\left( \dfrac{\pi }{2}-x \right)dx}$

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{{{\sin }^{2}}xdx}$                          …(2)

Adding equation (1) and (2),

$2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)dx}$

$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{1dx}$

$\Rightarrow 2I=\left[ x \right]_{0}^{\dfrac{\pi }{2}}$

$\Rightarrow 2I=\dfrac{\pi }{2}$

$\Rightarrow I=\dfrac{\pi }{4}$

$\therefore \int\limits_{0}^{\dfrac{\pi }{2}}{{{\cos }^{2}}xdx}=\dfrac{\pi }{4}$

2. \$\int\lim