# NCERT Solutions for Exercise 7.10 Class 12 Maths Chapter 7 - Integrals

NCERT solutions for Class 12 Maths chapter 7 exercise 7.10 is more or less similar to the exercise 7.9. Both these exercises cater to questions in which the value of the integrals is found out. Solutions to exercise 7.10 provided below are in detail. Some of the questions are multiple choice based which are also discussed in detail. NCERT solutions for Class 12 Maths chapter 7 exercise 7.10 discussed here are in detail. Students can choose to practice at least a few of the questions before examination. Overall for integral chapters, NCERT book exercise questions are enough to score well. Also students can refer NCERT chapter Integrals for practice.

• Integrals Exercise 7.1

• Integrals Exercise 7.2

• Integrals Exercise 7.3

• Integrals Exercise 7.4

• Integrals Exercise 7.5

• Integrals Exercise 7.6

• Integrals Exercise 7.7

• Integrals Exercise 7.8

• Integrals Exercise 7.9

• Integrals Exercise 7.11

• Integrals Miscellaneous Exercise

## Integrals Class 12 Chapter 7 Exercise: 7.10

### Question:1 Evaluate the integrals in Exercises 1 to 8 using substitution.

$\int_0^1\frac{x}{x^2 +1}dx$

$\int_0^1\frac{x}{x^2 +1}dx$
let $x^2+1 = t \Rightarrow xdx =dt/2$
when x = 0 then t = 1 and when x =1 then t = 2
$\therefore \int_{o}^{1}\frac{x}{x^2+1}dx=\frac{1}{2}\int_{1}^{2}\frac{dt}{t}$
$\\=\frac{1}{2}[\log\left | t \right |]_{1}^{2}\\ =\frac{1}{2}\log 2$

### Question:2 Evaluate the integrals in Exercises 1 to 8 using substitution.

$\int^\frac{\pi}{2}_0\sqrt{\sin\phi}\cos^5\phi d\phi$

$\int^\frac{\pi}{2}_0\sqrt{\sin\phi}\cos^5\phi d\phi$
let $\sin \phi = t \Rightarrow \cos \phi d\phi = dt$
when $\phi =0,t\rightarrow 0$ and $\phi =\pi/2,t\rightarrow 1$

using the above substitution we can evaluate the integral as

$\\\therefore \int_{0}^{1}\sqrt{t}(1-t^2)dt\\ =\int_{0}^{1} t^\frac{1}{2}(1+t^4-2t^2)dt\\ =\int_{0}^{1}t^\frac{1}{2}dt+\int_{0}^{1}t^{9/2}dt-2\int_{0}^{1}t^{5/2}dt\\ =[2t^{3/2}/3+2t^{11/2}/11+4t^{7/2}/7]^1_0\\ =\frac{64}{231}$

### Question:3 Evaluate the integrals in Exercises 1 to 8 using substitution.

$\int_0^1 \sin^{-1}\left(\frac{2x}{1+x^2} \right )dx$

$\int_0^1 \sin^{-1}\left(\frac{2x}{1+x^2} \right )dx$
let
$x = \tan\theta\Rightarrow dx =\sec^2\theta d\theta$
when x = 0 then $\theta= 0$ and when x = 1 then $\theta= \pi/4$

$\\=\int_{0}^{\pi/4}\sin^{-1}(\frac{2\tan\theta}{1+\tan\theta})\sec^2\theta d\theta\\ =\int_{0}^{\pi/4}\sin^{-1}(\sin 2\theta)\sec^2\theta d\theta\\ =\int_{0}^{\pi/4}2\theta \sec^2\theta d\theta\\$
Taking $\theta$ as a first function and $\sec^2\theta$ as a second function, by using by parts method

$\\=2[\theta\int \sec^2\theta d\theta-\int(\frac{d}{d\theta}\theta.\int \sec^2\theta\ d\theta)d\theta]^{\pi/4}_0\\ =2[\theta\tan\theta-\int \tan\theta\ d\theta]^{\pi/4}_0\\ =2[\theta\tan\theta+\log\left | \cos\theta \right |]^{\pi/4}_0\\ =2[\pi/4+\log(1/\sqrt{2})]\\ =\pi/4-\log 2$

### $\int_0^2x\sqrt{x+2}$ . (Put ${x+2} = t^2$ )

Let $x+2 = t^2\Rightarrow dx =2tdt$
when x = 0 then t = $\sqrt{2}$ and when x=2 then t = 2

$I=\int_{0}^{2}x\sqrt{x+2}dx$

$\\=2\int_{\sqrt{2}}^{2}(t^2-2)t^2dt\\ =2\int_{\sqrt{2}}^{2}(t^4-2t^2)dt\\ =2[t^5/5-\frac{2}{3}t^3]^2_{\sqrt{2}}\\ =2[\frac{32}{5}-\frac{16}{3}-\frac{4\sqrt{2}}{5}+\frac{4\sqrt{2}}{3}]\\ =\frac{16\sqrt{2}(\sqrt{2}+1)}{15}$

### Question:5 Evaluate the integrals in Exercises 1 to 8 using substitution.

$\int_0^{\frac{\pi}{2}}\frac{\sin x}{1 + \cos^2 x}dx$

$\int_0^{\frac{\pi}{2}}\frac{\sin x}{1 + \cos^2 x}dx =I$
let $\cos x =t\Rightarrow -\sin x dx = dt$
when x=0 then t = 1 and when x= $\pi/2$ then t = 0

$\\I=\int_{1}^{0}\frac{dt}{1+t^2}\\ =[\tan ^{-1}t]^0_1\\ =\pi/4$

### Question:6 Evaluate the integrals in Exercises 1 to 8 using substitution.

$\int_0^2\frac{dx}{x + 4 - x^2}$

By adjusting, the denominator can also be written as $(\frac{\sqrt{17}}{2})^2-(x-\frac{1}{2})^2 =x+4-x^2$
Now,
$\Rightarrow \int_{0}^{2}\frac{dx}{(\frac{\sqrt{17}}{2})^2-(x-\frac{1}{2})^2}$
let $x-1/2 = t\Rightarrow dx=dt$
when x= 0 then t =-1/2 and when x =2 then t = 3/2

$\\\Rightarrow\int_{-1/2}^{3/2}\frac{dt}{(\frac{\sqrt{17}}{2})^2-t^2}\\ =\frac{1}{2.\frac{\sqrt{17}}{2}}\log\frac{\frac{\sqrt{17}}{2}+t}{\frac{\sqrt{17}}{2}-t}\\ =\frac{1}{\sqrt{17}}[\log\frac{\sqrt{17}/2+3/2}{\sqrt{17}/2-3/2}-\log\frac{\sqrt{17}/2-1/2}{\sqrt{17}/2+1/2}]\\ =\frac{1}{\sqrt{17}}[\log\frac{\sqrt{17}+3}{\sqrt{17}-3/}.\frac{\sqrt{17}+1}{\sqrt{17}+1}]$
$\\ =\frac{1}{\sqrt{17}}[\log (\frac{17+3+4\sqrt{17}}{17+3-4\sqrt{17}})]\\ =\frac{1}{\sqrt{17}}[\log (\frac{5+\sqrt{17}}{5-\sqrt{17}})]$
On rationalisation, we get

$=\frac{1}{\sqrt{17}}\log \frac{21+5\sqrt{17}}{4}$

### Question:7 Evaluate the integrals in Exercises 1 to 8 using substitution.

$\int_{-1}^1\frac{dx}{x^2 +2x + 5}$

$\int_{-1}^1\frac{dx}{x^2 +2x + 5}$
the Dr can be written as $x^2+2x+5 = (x+1)^2+2^2$
and put x+1 = t then dx =dt

when x= -1 then t = 0 and when x = 1 then t = 2

$\\\Rightarrow \int_{0}^{2}\frac{dt}{t^2+2^2}\\ =\frac{1}{2}[\tan^{-1}\frac{t}{2}]^2_0\\ =\frac{1}{2}( \pi/4)\\ =\frac{\pi}{8}$

### Question:8 Evaluate the integrals in Exercises 1 to 8 using substitution.

$\int_1^2\left(\frac{1}{x} - \frac{1}{2x^2} \right )e^{2x}dx$

$\int_1^2\left(\frac{1}{x} - \frac{1}{2x^2} \right )e^{2x}dx$
let $2x =t \Rightarrow 2dx =dt$
when x = 1 then t = 2 and when x = 2 then t= 4

$\\=\frac{1}{2}\int_{2}^{4}(\frac{2}{t}-\frac{2}{t^2})e^tdt\\$
let
$\frac{1}{t} = f(t)\Rightarrow f'(t)=-\frac{1}{t^2}$
$\Rightarrow \int_{2}^{4}(\frac{1}{t}-\frac{1}{t^2})e^tdt =\int_{2}^{}4e^t[f(t)+f'(t)]dt$
$\\=[e^tf(t)]^4_2\\ =[e^t.\frac{1}{t}]^4_2\\ =\frac{e^4}{4}-\frac{e^2}{2}\\ =\frac{e^2(e^2-2)}{4}$

### Question:9 Choose the correct answer in Exercises 9 and 10.

The value of the integral $\int_{\frac{1}{3}}^1\frac{(x-x^3)^\frac{1}{3}}{x^4}dx$ is

(A) 6

(B) 0

(C) 3

(D) 4

The value of integral is (A) = 6

$\int_{\frac{1}{3}}^1\frac{(x-x^3)^\frac{1}{3}}{x^4}dx$
$\int_{\frac{1}{3}}^1\frac{(\frac{1}{x^2}-1)^\frac{1}{3}}{x^3}dx\\$
let
$\frac{1}{x^2}-1 = t\Rightarrow \frac{dx}{x^3}=-dt/2$
now, when x = 1/3, t = 8 and when x = 1 , t = 0

therefore

$\\=-\frac{1}{2}\int_{8}^{0}t^{1/3}dt\\ =-\frac{1}{2}.\frac{3}{4}[t^4/3]^0_8\\ =-\frac{3}{8}[-2^4]\\ =6$

### Question:10 Choose the correct answer in Exercises 9 and 10.

If $f(x) = \int_0^x t \sin t dt$ , then $f'(x)$ is

(A) $\cos x + x\sin x$

(B) $x\sin x$

(C) $x\cos x$

(D) $\sin x + x\cos x$

The correct answer is (B) = $x\sin x$

$f(x) = \int_0^x t \sin t dt$
by using by parts method,
$\\=t\int \sin t dt - \int (\frac{d}{dt}t\int \sin t dt)dt\\ =[t(-\cos t )+\sin t]^x_0$

$f(x)= -x\cos x+sinx$
So, $f'(x)= -\cos x+x\sin x+\cos x\\ =x\sin x$

## More About NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.10

The NCERT Class 12 Maths chapter Integrals has sufficient questions to practice for the exam. Students need not refer to any other book to score well in Board examinations. Exercise 7.10 Class 12 Maths has similar questions asked in earlier exercises. NCERT solutions for Class 12 Maths chapter 7 exercise 7.10 has some multiple choice questions also which are generally asked in JEE Main and NEET.

## Benefits of NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.10

• The Class 12th Maths chapter 7 exercise provided here is of high quality which can be referred directly by the student .

• NCERT syllabus Exercise 7.10 Class 12 Maths provides some good questions which can be asked in the examination.

• Class 12 Maths chapter 7 exercise 7.10 solutions can be found similar to previous exercises. Hence students can skip some of the similar questions.

## Also see-

• NCERT Exemplar Solutions Class 12 Maths Chapter 7

• NCERT Solutions for Class 12 Maths Chapter 7

## NCERT Solutions Subject Wise

• NCERT Solutions Class 12 Chemistry

• NCERT Solutions for Class 12 Physics

• NCERT Solutions for Class 12 Biology

• NCERT Solutions for Class 12 Mathematics

## Subject Wise NCERT Exemplar Solutions

• NCERT Exemplar Class 12 Maths

• NCERT Exemplar Class 12 Physics

• NCERT Exemplar Class 12 Chemistry

• NCERT Exemplar Class 12 Biology

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