Future Study Point

# Class 12 Maths well explained NCERT solutions of exercise 7.2 of chapter 7-Integrals

Here Class 12 Maths well explained NCERT solutions of exercise 7.2 of chapter 7-Integrals are created to clear your concept of methods solving integration of a function. These NCERT  solutions are the solutions of exercise 7.1 of chapter 7-Integrals of Class 12 maths NCERT textbook prescribed by CBSE. All questions of exercise 7.2 of chapter 7-Integrals are solved by the expert by a step by step method. Although there are several methods used for solving integration but the method used in the exercise 7.2 for solving the integral functions is substitution method.

Click for online shopping

Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc

What is the substitution method for calculation of integral function?-Let two functions exist in the way ∫f(x).f'(x) dx, here second function is the derivative of the first function.

Let t = f(x)

Differentiating it both sides

dt/dx = f'(x)

dt = f'(x) dx

Substituting it in the integral

∫t dt = t²/2 +C , now substituting t = f(x) back in the integration

∫f(x).f'(x) dx = [f(x)]²/2

As an example Let’s calculate ∫2x cosx² dx =?

We have to find out

∫2x cosx² dx

here d/dx of x² is 2x, so ,

Let t = x²

dt/dx = 2x⇒ dt = 2x.dx

Substituting t = x² and dt = 2x dx in the integration, we have

∫cost dt = sint + C

Now substituting t = x² back to the integrals

∫cost dt = sin x² + C

Now, you can easily understand the method used here for calculating integration of the function used here in exercise 7.2 of chapter 7-Integrals

## Class 12 Maths NCERT solutions of Chapter 7 Integrals

Exercise 7.1- Integrals

Exercise 7.2-Integral

Exercise 7.3-Integrals

Exercise 7.4 -Integral

## NCERT solutions of exercise 7.2 of chapter 7-Integrals

Integrate the function in exercises 1 to 37.

$\boldsymbol{Q1.\frac{2x}{1+x^{2}}}$

$\boldsymbol{\int \frac{2x}{1+x^{2}}dx}$

Let 1 + x² = t

dt/dx = 2x

dt = 2x dx

We have

$\boldsymbol{\int \frac{dt}{t}}$

$\boldsymbol{=log\left | t \right |+C}$

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

$\boldsymbol{Q2.\frac{\left ( logx \right )^{2}}{x}}$

Ans.

$\boldsymbol{\int \frac{\left ( logx \right )^{2}}{x}dx}$

Since 1/x is the d/dx of  $\boldsymbol{log\left | x \right |}$  and $\boldsymbol{\left ( logx \right )^{2}=\left ( log\left | x \right | \right )^{2}}$

$\boldsymbol{\int \frac{\left ( log\left | x \right | \right )^{2}}{x}dx}$

So, let t = $\boldsymbol{log\left | x \right |}$

$\boldsymbol{\frac{dt}{dx}=\frac{d}{dx}log\left | x \right |}$

$\boldsymbol{\frac{dt}{dx}=\frac{1}{x}}$

$\boldsymbol{dt=\frac{dx}{x}}$

Substituting the value of dt

$\boldsymbol{\int t^{2}dt=\frac{t^{3}}{3}+C}$

Substituting  t =   $\boldsymbol{log\left | x \right |}$  back in the integral

$\boldsymbol{=\frac{1}{3}\left ( log\left | x \right | \right )^{3}}$

Hence the integration of the given function is    $\boldsymbol{\frac{1}{3}\left ( log\left | x \right | \right )^{3}}$

$\boldsymbol{Q3.\frac{1}{x+xlogx}}$

Ans. The given function is

$\boldsymbol{\frac{1}{x +xlogx }}$

$\boldsymbol{\frac{1}{x\left ( 1+logx \right )}}$

Integrating it

$\boldsymbol{\int \frac{1}{x\left ( 1+x \right ) }dx}$

Let 1 + logx =t

dt/dx = 1/x

dt =dx/x

Substituting the value of dt

$\boldsymbol{=\int \frac{dt}{t}}$

$\boldsymbol{=log\left | t \right |+C}$

Substituting the value of t back in the integral

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

Hence the required integration of the given function is $\boldsymbol{=log\left | 1+x \right |+C}$

Q4. sinx.sin(cosx)

Ans. We are given sinx.sin(cosx)

Integrating it

$\boldsymbol{\int sinx.sin\left ( cosx \right )dx}$

Since d/dx of cosx =-sinx, so

Let t = cosx

dt/dx = -sinx

dt = -sinx dx

sinxdx= -dt

Substituting the value of sinx dx

$\boldsymbol{=-\int sint\: dt}$

= -(-cost) + C

= cost + C

Substituting back the value of t= cosx

= cos (cosx) + C

Q5. sin(ax +b) cos(ax +b)

Ans. We are given the function

sin(ax +b) cos(ax +b)

Divuding and multipling  it by 2, we have

$\boldsymbol{=\frac{2sin\left ( ax+b \right )cos\left ( ax+b \right )}{2}}$

$\boldsymbol{=\frac{1}{2}sin2\left ( ax+b \right )}$

Now,integrating it

$\boldsymbol{\frac{1}{2}\int sin2\left ( ax+b \right )dx}$

Let 2(ax+b) = t

dt/dx = 2a

dt/2a =  dx

Substituting the value of dx

$\boldsymbol{=\frac{1}{2}\int \frac{sint}{2a}dt}$

$\boldsymbol{=\frac{1}{4a}\int sint.dt}$

$\boldsymbol{=-\frac{1}{4a}cost+C}$

Subtituting back the value of t

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

$\boldsymbol{Q6.\sqrt{ax+b}}$

Ans. We are given

$\boldsymbol{\sqrt{ax+b}}$

Integrating it

$\boldsymbol{\int \sqrt{ax+b}.dx}$

Let t = ax +b

dt/dx = a

dx = dt/a

Substituting the value of dx and (ax + b)

$\boldsymbol{=\int \frac{\sqrt{t}}{a}.dt}$

$\boldsymbol{=\frac{1}{a}\int t^{1/2}.dt}$

$\boldsymbol{=\frac{1}{a}.\frac{t^{3/2}}{3/2}+C}$

$\boldsymbol{=\frac{2}{3a}.t^{3/2}+C}$

$\boldsymbol{=\frac{2}{3a}.\left ( ax+b \right )^{3/2}+C}$

$\boldsymbol{Q7.x\sqrt{x+2}}$

Ans. We are given the function

$\boldsymbol{x\sqrt{x+2}}$

Integrating it

$\boldsymbol{\int x\sqrt{x+2}.dx}$

Let t = x + 2⇒ x = t -2

dt/dx = 1⇒ dt = dx

Substituting value of x and dx

$\boldsymbol{=\int \left ( t-2 \right )\sqrt{t}.dt}$

$\boldsymbol{=\int t^{3/2}.dt-2 \int t^{1/2}.dt}$

$\boldsymbol{=\frac{t^{5/2}}{5/2}-2\frac{t^{3/2}}{3/2}+C}$

$\boldsymbol{=\frac{2}{5}t^{5/2}-\frac{4}{3}t^{3/2}+C}$

Substituting back the value of t

$\boldsymbol{=\frac{2}{3}\left ( x+2 \right )^{5/2}-\frac{4}{3}\left ( x+2 \right )^{3/2}+C}$

Therefore required integration of the given function is $\boldsymbol{=\frac{2}{3}\left ( x+2 \right )^{5/2}-\frac{4}{3}\left ( x+2 \right )^{3/2}+C}$

$\boldsymbol{Q8.x\sqrt{1+2x^{2}}}$

Ans. We are given the function

$\boldsymbol{x\sqrt{1+2x^{2}}}$

Integrating it

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

Let t = 1 + 2x²

dt/dx = 4x

xdx = dt/4

Substituting the value of xdx and (1 + 2x²), we have

$\boldsymbol{=\frac{1}{4}\int\sqrt{t} .dt}$

$\boldsymbol{\Rightarrow \frac{1}{4}.\frac{t^{3/2}}{3/2}+C=\frac{1}{6}.t^{3/2}+C}$

Substituting the value of t back

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

Therefore the required value of integration is    $\boldsymbol{=\frac{1}{6}\left ( 1+2x^{2} \right )^{3/2}+C}$

$\boldsymbol{Q9.\left (4x+2 \right )\sqrt{x^{2}+x+1}}$

Ans. We are given the function

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

Rewriting it

$\boldsymbol{2\left (2x+1 \right )\sqrt{x^{2}+x+1}}$

Integrating it

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

Let t = x² + x + 1

dt/dx = 2x + 1

We have dt = (2x +1)dx

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

$\boldsymbol{=2\int t^{1/2}.dt}$

$\boldsymbol{=2\frac{t^{3/2}}{3/2}+C=\frac{4}{3}t^{3/2}+C}$

Substituting back the value of t

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

Class 11 NCERT solutions of Physics and Chemistry

Chapter 1 – Some concepts of the chemistry

Chapter 1-Physical World

You can compensate us

Paytm number 9891436286

The money collected by us will be used for the education of poor students who leaves their study because of a lack of money.

## NCERT Solutions of Science and Maths for Class 9,10,11 and 12

### NCERT Solutions for class 10 maths

CBSE Class 10-Question paper of maths 2021 with solutions

CBSE Class 10-Half yearly question paper of maths 2020 with solutions

CBSE Class 10 -Question paper of maths 2020 with solutions

CBSE Class 10-Question paper of maths 2019 with solutions

### NCERT Solutions for class 11 maths

 Chapter 1-Sets Chapter 9-Sequences and Series Chapter 2- Relations and functions Chapter 10- Straight Lines Chapter 3- Trigonometry Chapter 11-Conic Sections Chapter 4-Principle of mathematical induction Chapter 12-Introduction to three Dimensional Geometry Chapter 5-Complex numbers Chapter 13- Limits and Derivatives Chapter 6- Linear Inequalities Chapter 14-Mathematical Reasoning Chapter 7- Permutations and Combinations Chapter 15- Statistics Chapter 8- Binomial Theorem Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

### NCERT solutions for class 12 maths

 Chapter 1-Relations and Functions Chapter 9-Differential Equations Chapter 2-Inverse Trigonometric Functions Chapter 10-Vector Algebra Chapter 3-Matrices Chapter 11 – Three Dimensional Geometry Chapter 4-Determinants Chapter 12-Linear Programming Chapter 5- Continuity and Differentiability Chapter 13-Probability Chapter 6- Application of Derivation CBSE Class 12- Question paper of maths 2021 with solutions Chapter 7- Integrals Chapter 8-Application of Integrals

Class 12 Solutions of Maths Latest Sample Paper Published by CBSE for 2021-22 Term 2

Class 12 Maths Important Questions-Application of Integrals

Class 12 Maths Important questions on Chapter 7 Integral with Solutions for term 2 CBSE Board 2021-22

Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22

Solutions of class 12  maths question paper 2021 preboard exam CBSE Solution