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

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

Edited By Ramraj Saini | Updated on Dec 03, 2023 10:15 PM IST | #CBSE Class 12th

## NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.10

NCERT Solutions for Exercise 7.10 Class 12 Maths Chapter 7 Integrals are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. 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.

12th class Maths exercise 7.10 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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## Integrals Class 12 Chapter 7 Exercise: 7.10

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

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

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

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

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

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.

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

$\int_1^2\left(\frac{1}{x} - \frac{1}{2x^2} \right )e^{2x}dx$
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.
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## Key Features Of NCERT Solutions for Exercise 7.10 Class 12 Maths Chapter 7

• Comprehensive Coverage: The solutions encompass all the topics covered in ex 7.10 class 12, ensuring a thorough understanding of the concepts.
• Step-by-Step Solutions: In this class 12 maths ex 7.10, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
• Accuracy and Clarity: Solutions for class 12 ex 7.10 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
• Conceptual Clarity: In this 12th class maths exercise 7.10 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
• Inclusive Approach: Solutions for ex 7.10 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
• Relevance to Curriculum: The solutions for class 12 maths ex 7.10 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

## Subject Wise NCERT Exemplar Solutions

Happy learning!!!

1. What is the weightage of Exercise 7.10 Class 12 Maths ?

Alteat 1 questions will be there in the board examination. Hence students can expect 5 marks minimum. It can go upto 10 marks also.

2. Integrals without upper and lower limits are known as ………... ?

Integrals without upper and lower limits are known as Indefinite integrals.

3. Give the number of questions discussed in Exercise 7.10 Class 12 Maths ?

There are 10 questions discussed in Exercise 7.10 Class 12 Maths

4. How many multiple choice questions are there in Exercise 7.10 Class 12 Maths ?

There are 2 Multiple choice questions in Exercise 7.10 Class 12 Maths

5. Are questions repeated in Board examination from this chapter?

Exact questions are rare to observe but questions are repeated based on the same concepts.

6. How important is Exercise 7.10 Class 12 Maths for NEET and JEE?

In JEE mains and NEET, questions are asked on similar lines of Exercise 7.10 Class 12 Maths .

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### Questions related to CBSE Class 12th

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If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

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I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

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