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

Question:2 Integrate the functions $\frac{(\log x{)}^2}{x}$

Answer:

Given to integrate $\frac{(\log x{)}^2}{x}$ function,

Let us assume $\log |x|=t$

we get, $\frac{1}{x}dx=dt$

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

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

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

Question:3 Integrate the functions $\frac{1}{x+x\log x}$

Answer:

Given to integrate $\frac{1}{x+x\log x}$ function,

Let us assume $1+\log x=t$

we get, $\frac{1}{x}dx=dt$

$⟹\int \frac{1}{x(1+\log x)}dx=\int \frac{1}{t}dt$

$=\log |t|+C$

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

Question:4 Integrate the functions $\sin x\sin (\cos x)$

Answer:

Given to integrate $\sin x\sin (\cos x)$ function,

Let us assume $\cos x=t$

we get, $-\sin xdx=dt$

$⟹\int \sin x\sin (\cos x)dx=-\int \sin tdt$

$=-(-\cos t)+C$

$=\cos t+C$

Back substituting the value of t we get,

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

Question:5 Integrate the functions $\sin (ax+b)\cos (ax+b)$

Answer:

Given to integrate $\sin (ax+b)\cos (ax+b)$ function,

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

Let us assume $2(ax+b)=t$

we get, $2adx=dt$

$\int \frac{\sin 2(ax+b)}{2}dx=\frac{1}{2}\int \frac{\sin t}{2a}dt$

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

Now, by back substituting the value of t,

$=\frac{-1}{4a}[cos2(ax+b)]+C$

Question:6 Integrate the functions $\sqrt{ax+b}$

Answer:

Given to integrate $\sqrt{ax+b}$ function,

Let us assume $(ax+b)=t$

we get, $adx=dt$

$dx=\frac{1}{a}dt$

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

Now, by back substituting the value of t,

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

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

Question:7 Integrate the functions $x\sqrt{x+2}$

Answer:

Given function $x\sqrt{x+2}$ ,

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

Assume the $(x+2)=t$ 19634

$\therefore dx=dt$

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

$=\int (t-2)\sqrt{t}dt$

$=\int (t^{\frac{3}{2}}-2t^{\frac{1}{2}})dt$

$=\int t^{\frac{3}{2}}dt-2\int t^{\frac{1}{2}}dt$

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

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

Back substituting the value of t in the above equation.

or, $\frac{2}{5}(x+2{)}^{\frac{5}{2}}-\frac{4}{3}(x+2{)}^{\frac{3}{2}}+C$ , where C is any constant value.

Question:8 Integrate the functions $x\sqrt{1+2x^2}$

Answer:

Given function $x\sqrt{1+2x^2}$ ,

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

Assume the $1+2x^2=t$

$\therefore 4xdx=dt$

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

Or $=\frac{1}{4}\int t^{\frac{1}{2}}dt=\frac{1}{4}\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)+C$

$=\frac{1}{6}(1+2x^2{)}^{\frac{3}{2}}+C$ , where C is any constant value.

Question:9 Integrate the functions $(4x+2)\sqrt{x^2+x+1}$

Answer:

Given function $(4x+2)\sqrt{x^2+x+1}$ ,

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

Assume the $1+x+x^2=t$

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

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

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

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

Now, back substituting the value of t in the above equation,

$=\frac{4}{3}(1+x+x^2{)}^{\frac{3}{2}}+C$ , where C is any constant value.

Question:10 Integrate the functions $\frac{1}{x-\sqrt{x}}$

Answer:

Given function $\frac{1}{x-\sqrt{x}}$ ,

$\int \frac{1}{x-\sqrt{x}}dx$

Can be written in the form:

$\frac{1}{x-\sqrt{x}}=\frac{1}{\sqrt{x}(\sqrt{x}-1)}$

Assume the $(\sqrt{x}-1)=t$

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

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

$=2\log |t|+C$

$=2\log |\sqrt{x}-1|+C$ , where C is any constant value.

Question:11 Integrate the functions $\frac{x}{\sqrt{x+4}}$ , x > 0

Answer:

Given function $\frac{x}{\sqrt{x+4}}$ ,

$\int \frac{x}{\sqrt{x+4}}dx$

Assume the $x+4=t$ so, $x=t-4$

$\therefore dx=dt$

$\int \frac{x}{\sqrt{x+4}}dx=\int \frac{t-4}{\sqrt{t}}dt$

$\int t^{\frac{1}{2}}dt-4\int t^{\frac{-1}{2}}dt$

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

$=\frac{2}{3}(x+4{)}^{\frac{3}{2}}-16(x+4{)}^{\frac{1}{2}}+C$

, where C is any constant value.

Question:12 Integrate the functions $(x^3-1{)}^{1/3}{x}^5$

Answer:

Given function $(x^3-1{)}^{1/3}{x}^5$ ,

$\int (x^3-1{)}^{1/3}{x}^{5}dx$

Assume the $x^3-1=t$

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

$⟹\int (x^3-1{)}^{\frac{1}{3}}{x}^{5}dx=\int (x^3-1{)}^{\frac{1}{3}}{x}^3.x^{2}dx$

$=\int t^{\frac{1}{3}}(t+1)\frac{dt}{3}$

$=\frac{1}{3}\int (t^{\frac{4}{3}}+t^{\frac{1}{3}})dt$

$=\frac{1}{3}[\frac{t^{\frac{7}{3}}}{\frac{7}{3}}+\frac{t^{\frac{4}{3}}}{\frac{4}{3}}]+C$

$=\frac{1}{3}[\frac{3}{7}{t}^{\frac{7}{3}}+\frac{3}{4}{t}^{\frac{4}{3}}]+C$

$=\frac{1}{7}(x^3-1{)}^{\frac{7}{3}}+\frac{1}{4}(x^3-1{)}^{\frac{4}{3}}+C$ , where C is any constant value.

Question:13 Integrate the functions $\frac{x^2}{(2+3x^3{)}^3}$

Answer:

Given function $\frac{x^2}{(2+3x^3{)}^3}$ ,

$\int \frac{x^2}{(2+3x^3{)}^3}dx$

Assume the $2+3x^3=t$

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

$⟹\int \frac{x^2}{(2+3x^2)}dx=\frac{1}{9}\int \frac{dt}{t^3}$

$=\frac{1}{9}\left(\frac{t^{-2}}{-2}\right)+C$

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

$=\frac{-1}{18(2+3x^3{)}^2}+C$ , where C is any constant value.

Question:14 Integrate the functions $\frac{1}{x(\log x{)}^m},x>0,m\ne 1$

Answer:

Given function $\frac{1}{x(\log x{)}^m},x>0,m\ne 1$ ,

Assume the $\log x=t$

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

$⟹\int \frac{1}{x(logx{)}^m}dx=\int \frac{dt}{t^m}$

$=\left(\frac{t^{-m+1}}{1-m}\right)+C$

$=\frac{(logx{)}^{1-m}}{(1-m)}+C$ , where C is any constant value.

Question:15 Integrate the functions $\frac{x}{9-4x^2}$

Answer:

Given function $\frac{x}{9-4x^2}$ ,

Assume the $9-4x^2=t$

$\therefore -8xdx=dt$

$⟹\int \frac{x}{9-4x^2}=-\frac{1}{8}\int \frac{1}{t}dt$

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

Now back substituting the value of t ;

$=\frac{-1}{8}\log |9-4x^2|+C$ , where C is any constant value.

Question:16 Integrate the functions $e^{2x+3}$

Answer:

Given function $e^{2x+3}$ ,

Assume the $2x+3=t$

$\therefore 2dx=dt$

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

$=\frac{1}{2}{e}^t+C$

Now back substituting the value of t ;

$=\frac{1}{2}{e}^{2x+3}+C$ , where C is any constant value.

Question:17 Integrate the functions $\frac{x}{e^{x^2}}$

Answer:

Given function $\frac{x}{e^{x^2}}$ ,

Assume the $x^2=t$

$\therefore 2xdx=dt$

$⟹\int \frac{x}{e^{x^2}}dx=\frac{1}{2}\int \frac{1}{e^t}dt$

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

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

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

$=\frac{-1}{2e^{x^2}}+C$ , where C is any constant value.

Question:18 Integrate the functions $\frac{e^{{\tan }^{-1}x}}{1+x^2}$

Answer:

Given,

$\frac{e^{{\tan }^{-1}x}}{1+x^2}$

Let's do the following substitution

$$\begin{gathered} ta{n}^{-1}x=t \\ ⟹\frac{1}{1+x^2}dx=dt \end{gathered}$$

$\therefore \int \frac{e^{{\tan }^{-1}x}}{1+x^2}dx=\int e^{t}dt=e^t+C$

$=e^{ta{n}^{-1}x}+C$

Question:19 Integrate the functions $\frac{e^{2x}-1}{e^{2x}+1}$

Answer:

Given function $\frac{e^{2x}-1}{e^{2x}+1}$ ,

Simplifying it by dividing both numerator and denominator by $e^x$ , we obtain

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

Assume the $e^x+e^{-x}=t$

$\therefore (e^x-e^{-x})dx=dt$

$⟹\int \frac{e^{2x}-1}{e^{2x}+1}dx=\int \frac{e^x-e^{-x}}{e^x+e^{-x}}dx$

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

$=\log |t|+C$

Now, back substituting the value of t,

$=\log |e^x+e^{-x}|+C$ , where C is any constant value.

Question:20 Integrate the functions $\frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}$

Answer:

Given function $\frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}$ ,

Assume the $e^{2x}+e^{-2x}=t$

$\therefore (2e^{2x}-2e^{-2x})dx=dt$

$⟹\int \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}dx=\int \frac{dt}{2t}$

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

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

Now, back substituting the value of t,

$=\frac{1}{2}\log |e^{2x}+e^{-2x}|+C$ , where C is any constant value.

Question:21 Integrate the functions ${\tan }^2(2x-3)$

Answer:

Given function ${\tan }^2(2x-3)$ ,

Assume the $2x-3=t$

$\therefore 2dx=dt$

$⟹\int {\tan }^2(2x-3)dx=\frac{\int {\tan }^2(t)}{2}dt$

$=\frac{1}{2}\int ({\sec }^{2}t-1)dt$ $[\because {\tan }^{2}t+1={\sec }^{2}t]$

$=\frac{1}{2}[\tan t-t]+C$

Now, back substituting the value of t,

$=\frac{1}{2}[\tan (2x-3)-2x+3]+C$

or $\frac{1}{2}\tan (2x-3)-x+C$ , where C is any constant value.

Question:22 Integrate the functions ${\sec }^2(7-4x)$

Answer:

Given function ${\sec }^2(7-4x)$ ,

Assume the $7-4x=t$

$\therefore -4dx=dt$

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

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

Now, back substituted the value of t.

$=-\frac{1}{4}\tan (7-4x)+C$ , where C is any constant value.

Question:23 Integrate the functions $\frac{{\sin }^{-1}x}{\sqrt{1-x^2}}$

Answer:

Given function $\frac{{\sin }^{-1}x}{\sqrt{1-x^2}}$ ,

Assume the ${\sin }^{-1}x=t$

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

$⟹\int \frac{{\sin }^{-1}x}{\sqrt{1-x^2}}dx=\int tdt$

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

Now, back substituted the value of t.

$=\frac{({\sin }^{-1}x{)}^2}{2}+C$ , where C is any constant value.

Question:24 Integrate the functions $\frac{2\cos x-3\sin x}{6\cos x+4\sin x}$

Answer:

Given function $\frac{2\cos x-3\sin x}{6\cos x+4\sin x}$ ,

or simplified as $\frac{2\cos x-3\sin x}{2(3\cos x+2\sin x)}$

Assume the $3\cos x+2\sin x=t$

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

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

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

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

Now, back substituted the value of t.

$=\frac{1}{2}\log |3\cos x+2\sin x|+C$ , where C is any constant value.

Question:25 Integrate the functions $\frac{1}{{\cos }^{2}x(1-\tan x{)}^2}$

Answer:

Given function $\frac{1}{{\cos }^{2}x(1-\tan x{)}^2}$ ,

or simplified as $\frac{1}{{\cos }^{2}x(1-\tan x{)}^2}=\frac{{\sec }^{2}x}{(1-\tan x{)}^2}$

Assume the $(1-\tan x)=t$

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

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

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

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

Now, back substituted the value of t.

$=\frac{1}{1-\tan x}+C$

where C is any constant value.

Question:26 Integrate the functions $\frac{\cos \sqrt{x}}{\sqrt{x}}$

Answer:

Given function $\frac{\cos \sqrt{x}}{\sqrt{x}}$ ,

Assume the $\sqrt{x}=t$

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

$⟹\int \frac{\cos \sqrt{x}}{\sqrt{x}}dx=2\int \cos tdt$

$=2\sin t+C$

Now, back substituted the value of t.

$=2\sin \sqrt{x}+C$ , where C is any constant value.

Question:27 Integrate the functions $\sqrt{\sin 2x}\cos 2x$

Answer:

Given function $\sqrt{\sin 2x}\cos 2x$ ,

Assume the $\sin 2x=t$

$\therefore 2\cos 2xdx=dt$

$⟹\int \sqrt{\sin 2x}\cos 2xdx=\frac{1}{2}\int \sqrt{t}dt$

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

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

Now, back substituted the value of t.

$=\frac{1}{3}(\sin 2x{)}^{\frac{3}{2}}+C$ , where C is any constant value.

Question:28 Integrate the functions $\frac{\cos x}{\sqrt{1+\sin x}}$

Answer:

Given function $\frac{\cos x}{\sqrt{1+\sin x}}$ ,

Assume the $1+\sin x=t$

$\therefore \cos xdx=dt$

$⟹\int \frac{\cos x}{\sqrt{1+\sin x}}dx=\int \frac{dt}{\sqrt{t}}$

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

$=2\sqrt{t}+C$

Now, back substituted the value of t.

$=2\sqrt{1+\sin x}+C$ , where C is any constant value.

Question:29 Integrate the functions $\cot xlog\sin x$

Answer:

Given function $\cot xlog\sin x$ ,

Assume the $\log \sin x=t$

$\therefore \frac{1}{\sin x}.\cos xdx=dt$

$\cot xdx=dt$

$⟹\int \cot x\log \sin xdx=\int tdt$

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

Now, back substituted the value of t.

$=\frac{1}{2}(\log \sin x{)}^2+C$ , where C is any constant value.

Question:30 Integrate the functions $\frac{\sin x}{1+\cos x}$

Answer:

Given function $\frac{\sin x}{1+\cos x}$ ,

Assume the $1+\cos x=t$

$\therefore -\sin xdx=dt$

$⟹\int \frac{\sin x}{1+\cos x}dx=\int -\frac{dt}{t}$

$=-\log |t|+C$

Now, back substituted the value of t.

$=-\log |1+\cos x|+C$ , where C is any constant value.

Question:31 Integrate the functions $\frac{\sin x}{(1+\cos x{)}^2}$

Answer:

Given function $\frac{\sin x}{(1+\cos x{)}^2}$ ,

Assume the $1+\cos x=t$

$\therefore -\sin xdx=dt$

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

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

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

Now, back substituted the value of t.

$=\frac{1}{1+\cos x}+C$ , where C is any constant value.

Question:32 Integrate the functions $\frac{1}{1+\cot x}$

Answer:

Given function $\frac{1}{1+\cot x}$

Assume that $I=\int \frac{1}{1+\cot x}dx$

Now solving the assumed integral;

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

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

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

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

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

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

Now, to solve further we will assume $\sin x+\cos x=t$

Or, $(\cos x-\sin x)dx=dt$

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

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

Now, back substituting the value of t,

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

Question:33 Integrate the functions $\frac{1}{1-\tan x}$

Answer:

Given function $\frac{1}{1-\tan x}$

Assume that $I=\int \frac{1}{1-\tan x}dx$

Now solving the assumed integral;

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

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