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# NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 - Integrals

Last updated date: 14th Sep 2024
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## NCERT Solutions for Maths Class 12 Exercise 7.2 Chapter 7 Integrals - FREE PDF Download

This section dives deeper into a powerful technique called Integration by Parts. You'll learn the formula for integrating products of functions and gain practice solving these integrals through illustrative problems. Integration by Parts is a fundamental concept in Calculus, so get ready to sharpen your skills! Class 12 Maths Ex 7.2 contains all questions and answers for the current NCERT Syllabus.

Table of Content
1. NCERT Solutions for Maths Class 12 Exercise 7.2 Chapter 7 Integrals - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 7 Exercise 7.2 Class 12 | Vedantu
3. Some Properties of Indefinite Integral
4. Some Basic Integral Formulas
5. NCERT Solutions for Class 12 Maths Chapter 7 – Exercise 7.2 Questions
6. Class 12 Maths Chapter 7: Exercises Breakdown
7. CBSE Class 12 Maths Chapter 7 Other Study Materials
8. Chapter-Specific NCERT Solutions for Class 12 Maths
FAQs

## Glance on NCERT Solutions Maths Chapter 7 Exercise 7.2 Class 12 | Vedantu

• This exercise likely focuses on practicing integration techniques you learned in Chapter 7.

• Integration is the process of finding the function whose derivative is a given function.

• Integration by substitution technique involves substituting a new variable for a part of the function that makes integration easier.

• Integration by parts is used to integrate the product of two functions. It involves integrating the product of one function times the derivative of the other function, minus the integral of the first function times the derivative of the second function.

• The exercise also include applications of integration, such as finding areas and volumes under curves, finding the work done by a force over a distance, or finding moments and centers of mass.

## Some Properties of Indefinite Integral

(i) ∫[f(x) + g(x)] dx = ∫f(x) dx + ∫g(x) dx

(ii) For any real number k, ∫k f(x) dx = k∫f(x)dx.

(iii) In general, if f1, f2,………, fn are functions and k1, k2,…, kn are real numbers, then

∫[k1f1(x) + k2 f2(x)+…+ knfn(x)] dx = k1 ∫f1(x) dx + k2 ∫ f2(x) dx+…+ kn ∫fn(x) dx

## Some Basic Integral Formulas

• ∫ 1 dx = x + C

• ∫ a dx = ax + C

• ∫ xn dx = ((xn+1)/(n+1))+C ; n≠1

• ∫ sin x dx = – cos x + C

• ∫ cos x dx = sin x + C

• ∫ sec2x dx = tan x + C

• ∫ csc2x dx = -cot x + C

• ∫ sec x (tan x) dx = sec x + C

• ∫ csc x ( cot x) dx = – csc x + C

• ∫ (1/x) dx = ln |x| + C

• ∫ ex dx = ex+ C

• ∫ ax dx = (ax/ln a) + C ; a>0,  a≠1

Competitive Exams after 12th Science

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

Exercise 7.2

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

Ans:

Let  $1+{{x}^{2}}=\text{t}$

$\therefore 2\text{xdx}=\text{dt}$

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

$=\log |t|+C$

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

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

where C is an arbitrary constant.

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

Ans:

Let  $\log x=t$

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

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

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

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

where C is an arbitrary constant.

3. Integrate  $\dfrac{1}{x+x\log x}$

Ans:

The given function can be written as

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

Let  $1+\log x=t$

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

$\Rightarrow \int{\dfrac{1}{x(1+\log x)}}dx=\int_{t}^{1}{d}t$

$=\log |t|+C$

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

where C is an arbitrary constant.

4. Integrate  $\sin x.\sin \left( \cos x \right)$

Ans:

Let  $\cos x=t$

$\therefore -\sin xdx=t$

$\Rightarrow \int{\sin }x\cdot \sin (\cos x)dx=-\int{\sin }tdt$

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

$=\cos t+C$

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

where C is an arbitrary constant.

5. Integrate  $\sin \left( ax+b \right)\cos \left( ax+b \right)$

Ans:

The given function can be rewritten as

$\sin (ax+b)\cos (ax+b)=\dfrac{2\sin (ax+b)\cos (ax+b)}{2}=\dfrac{\sin 2(ax+b)}{2}$

let  $2(ax+b)=t$

$\therefore 2adx=dt$

$\Rightarrow \int{\dfrac{\sin 2(ax+b)}{2}}dx=\dfrac{1}{2}\int{\dfrac{\sin tdt}{2a}}$

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

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

where C is an arbitrary constant.

6. Integrate $\sqrt{ax+b}$

Ans:

Let  $ax+b=t$

$\Rightarrow adx=dt$

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

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

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

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

where C is an arbitrary constant.

7. Integrate  $x\sqrt{x+2}$

Ans:

Let  $x+2=t$

$\therefore dx=dt$

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

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

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

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

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

where C is an arbitrary constant.

8. Integrate  $x\sqrt{1+2{{x}^{2}}}$

Ans:

Let  $1+2{{x}^{2}}=t$

$\therefore 4\text{xdx}=\text{dt}$

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

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

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

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

where C is an arbitrary constant.

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

Ans:

Let  ${{\text{x}}^{2}}+\text{x}+1=\text{t}$

$\therefore (2\text{x}+1)\text{dx}=\text{dt}$

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

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

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

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

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

where C is an arbitrary constant.

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

Ans:

The given function can be rewritten as

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

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

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

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

$=2\log |t|+C$

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

where C is an arbitrary constant.

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

Ans:

Let  $x+4=t$

$\therefore dx=dt$

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

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

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

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

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

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

$=\dfrac{2}{3}{{(x+4)}^{\dfrac{1}{2}}}(x+4-12)+C$

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

where C is an arbitrary constant.

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

Ans:

Let  ${{x}^{3}}-1=t$

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

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

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

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

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

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

where C is an arbitrary constant.

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

Ans:

Let  $2+3{{x}^{3}}=t$

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

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

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

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

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

where C is an arbitrary constant.

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

Ans:

Let  $\log \text{x}=\text{t}$

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

$\Rightarrow \int{\dfrac{1}{x{{(\log x)}^{m}}}}dx=\int{\dfrac{dt}{{{(t)}^{m}}}}$

$=\dfrac{{{t}^{-m+1}}}{-m+1}+C$

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

where C is an arbitrary constant.

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

Ans:

Let  $9-4{{x}^{2}}=t$

$\therefore -8xdx=dt$

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

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

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

where C is an arbitrary constant.

16. Integrate  ${{e}^{2x+3}}$

Ans:

Let  $2x+3=t$

$\therefore 2dx=dt$

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

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

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

where C is an arbitrary constant.

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

Ans:

Let  ${{\text{x}}^{2}}=\text{t}$

$\therefore 2\text{xdx}=\text{dt}$

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

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

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

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

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

where C is an arbitrary constant.

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

Ans:

Let  ${{\tan }^{-1}}x=t$

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

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

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

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

where C is an arbitrary constant.

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

Ans:

Let  ${{e}^{2x}}+{{e}^{-2x}}=t$

$\Rightarrow 2{{e}^{2x}}-2{{e}^{-2x}}dx=dt$

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

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

$=\dfrac{1}{2}\int{\dfrac{1}{t}}d$

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

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

where C is an arbitrary constant.

21. Integrate  ${{\tan }^{2}}\left( 2x-3 \right)$

Ans:

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

Let  $2\text{x}-3=\text{t}$

$\therefore 2\text{dx}=\text{dt}$

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

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

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

$=\dfrac{1}{2}\tan t-x+C$

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

where C is an arbitrary constant.

22. Integrate ${{\sec }^{2}}\left( 7-4x \right)$

Ans:

Let  $7-4x=t$

$\therefore -4~\text{d}x=dt$

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

$=\dfrac{-1}{4}(\tan t)+C$

$=\dfrac{-1}{4}\tan (7-4x)+C$

where C is an arbitrary constant.

23. Integrate  $\dfrac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}$

Ans:

Let  ${{\sin }^{-1}}x=t$

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

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

$=\dfrac{{{t}^{2}}}{2}+C$

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

where C is an arbitrary constant.

24. Integrate  $\dfrac{2\cos x-3\sin x}{6\cos x+4\sin x}$

Ans:

The given function is

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

Let  $3\cos x+2\sin x=t$

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

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

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

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

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

where C is an arbitrary constant.

25. Integrate  $\dfrac{1}{{{\cos }^{2}}x{{\left( 1-\tan x \right)}^{2}}}$

Ans:

The given function is,

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

Let  $(1-\tan x)=t$

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

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

$=-\int{{{t}^{-2}}}dt$

$=+\dfrac{1}{t}+C$

$=\dfrac{1}{\left( 1-\tan x \right)}+C$

where C is an arbitrary constant.

26. Integrate  $\dfrac{\cos \sqrt{x}}{\sqrt{x}}$

Ans:

let  $\sqrt{x}=t$

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

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

where C is an arbitrary constant.

27. Integrate  $\sqrt{\sin 2x}\cos 2x$

Ans:

Let  $\sin 2\text{x}=\text{t}$

So,  $2\cos 2\text{xdx}=\text{dt}$

$\Rightarrow \int{\sqrt{\sin 2x}}\cos 2xdx$

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

$=\dfrac{1}{2}\int{\sqrt{t}}dt$

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

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

where C is an arbitrary constant.

28. Integrate  $\dfrac{\cos x}{\sqrt{1+\sin x}}$

Ans:

Let  $1+\sin \text{x}=\text{t}$

$\therefore \cos \text{xdx}=\text{dt}$

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

$=\dfrac{{{t}^{\dfrac{1}{2}}}}{\dfrac{1}{2}}+C$

$=2\sqrt{t}+\text{C}$

$=2\sqrt{1+\sin x}+C$

where C is an arbitrary constant.

29. Integrate  $\cot x\log \sin x$

Ans:

Let $\log \sin x=t$

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

$\therefore \cot xdx=dt$

$\Rightarrow \int{\cot }x\log \sin xdx=\int{t}dt$

$=\dfrac{{{t}^{2}}}{2}+C$

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

where C is an arbitrary constant.

30. Integrate  $\dfrac{\sin x}{1+\cos x}$

Ans:

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

$=-\log |t|+C$

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

where C is an arbitrary constant.

31. Integrate  $\dfrac{\sin x}{{{\left( 1+\cos x \right)}^{2}}}$

Ans:

Let  $1+\cos \text{x}=\text{t}$

$\therefore -\sin \text{xdx}=\text{dt}$

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

$=-\int{{{t}^{-2}}}dt$

$=\dfrac{1}{t}+C$

$=\dfrac{1}{1+\cos x}+C$

where C is an arbitrary constant.

32. Integrate $\dfrac{1}{1+\cot x}$

Ans:

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

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

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

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

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

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

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

Let  $\sin x+\cos x=t\Rightarrow (\cos x-\sin x)dx=dt$

$\therefore I=\dfrac{x}{2}+\dfrac{1}{2}\int{\dfrac{-(dt)}{t}}$

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

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

where C is an arbitrary constant.

33. Integrate  $\dfrac{1}{1-\tan x}$

Ans:

Let  $I=\int{\dfrac{1}{1-\tan x}}dx$

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

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

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

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

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

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

Put  $\cos x-\sin x=t\Rightarrow (-\sin x-\cos x)dx=dt$

$\therefore I=\dfrac{x}{2}+\dfrac{1}{2}\int{\dfrac{-(dt)}{t}}$

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

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

where C is an arbitrary constant.

34. Integrate  $\dfrac{\sqrt{\tan x}}{\sin x\cos x}$

Ans:

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

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

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

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

Let  $\tan x=t\Rightarrow {{\sec }^{2}}xdx=dt$

$\therefore I=\int{\dfrac{dt}{\sqrt{t}}}$

$=2\sqrt{t}+C$

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

where C is an arbitrary constant.

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

Ans:

Let  $1+\log x=t$

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

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

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

$=\dfrac{{{(1+\log x)}^{3}}}{3}+C$

where C is an arbitrary constant.

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

Ans:

The given function can be rewritten as

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

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

Let  $(x+\log x)=t$

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

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

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

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

$=\dfrac{1}{3}{{(x+\log x)}^{3}}+C$

where C is an arbitrary constant.

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

Ans:

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

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

Let  ${{\tan }^{-1}}t=u$

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

From (1), we obtain

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

$=\dfrac{1}{4}(-\cos u)+C$

$=-\dfrac{1}{4}\cos \left( {{\tan }^{-1}}t \right)+C$

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

where C is an arbitrary constant.

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

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

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

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

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

Ans:

Let  ${{x}^{10}}+{{10}^{x}}=t$

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

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

$=\log t+C$

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

Hence, the correct Answer is D.

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

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

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

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

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

Ans:

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

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

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

$=\int{{{\sec }^{2}}}xdx+\int{{{\operatorname{cosec}}^{2}}}dx$

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

Hence, the correct Answer is B.

## Conclusion

The NCERT Solutions for Class 12 Maths Exercise 7.2 Chapter 7 - Integrals, Ex 7.2 class 12 Maths, provided by Vedantu, offers comprehensive guidance for mastering integration techniques. This exercise focuses on the application of various integration methods, including substitution and integration by parts, to solve complex integrals. Key topics include Integration using substitution. Integration by parts. Solving integrals involving trigonometric functions and logarithms. By solving these questions, students can solidify their understanding and improve their problem-solving skills in integration​

## Class 12 Maths Chapter 7: Exercises Breakdown

 Chapter 7 - Integrals Exercises in PDF Format Exercise 7.1 22 Questions & Solutions Exercise 7.3 24 Questions & Solutions Exercise 7.4 25 Questions & Solutions Exercise 7.5 23 Questions & Solutions Exercise 7.6 24 Questions & Solutions Exercise 7.7 11 Questions & Solutions Exercise 7.8 22 Questions & Solutions Exercise 7.9 10 Questions & Solutions Exercise 7.10 21 Questions & Solutions

## Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 - Integrals

1. What are the three most prominent and useful methods of integration in calculus?

The three most prominent methods of integration are:

• Substitution method

• Partial Fractions method

• Integration by parts

Class 12 maths ch 7 ex 7.2 covers problems of integration of different types. Each method has its own set of rules which must be understood thoroughly to be able to do the integration.

2. What are the different applications of definite integrals?

Integration is applied to find the value of many things in maths, physics, economics. A few of the important applications of definite integrals are:

• Area of a curve-bound region

• The volume of a solid for which the cross-section is known

• Length of an arc

• Pressure of fluids

• Electric circuit and integration

• Moment of inertia