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.

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

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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

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