NCERT Solutions for Exercise 7.6 Class 12 Maths Chapter 7 - Integrals

Question 2: Integrate the functions $x\sin 3x$

Answer:

Given function is
$f(x)=x\sin 3x$
We will use integration by parts method

$\int x\sin 3xdx=x\int \sin 3xdx-\int (\frac{d(x)}{dx}\cdot \int \sin 3xdx)dx$

$\int x\sin 3xdx=x\cdot \left(\frac{-\cos 3x}{3}\right)-\int (1\cdot \frac{-\cos 3x}{3})dx$

$\int x\sin 3xdx=-\frac{x\cos 3x}{3}+\frac{\sin 3x}{9}+C$
Therefore, the answer is $-\frac{x\cos 3x}{3}+\frac{\sin 3x}{9}+C$

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

Answer:

The given function is
$f(x)=x^2{e}^x$
We will use the integration by parts method

$\int x^2{e}^{x}dx=x^2\int e^{x}dx-\int (\frac{d(x^2)}{dx}\cdot \int e^{x}dx)dx$

$\int x^2{e}^{x}dx=x^2{e}^x-\int (2x{e}^x)dx$
Again use integration by parts in $\int (2x.e^x)dx$

$\int (2x{e}^x)dx=2x\int e^{x}dx-\int (\frac{d(2x)}{dx}\cdot \int e^{x}dx)dx$

$\int 2x{e}^{x}dx=2x{e}^x-\int 2e^{x}dx$

$\int 2x{e}^{x}dx=2x{e}^x-2e^x$
Put this value in our equation
we will get,

$\int x^2{e}^{x}dx=x^2{e}^x-2x{e}^x+2e^x+C$

$\int x^2{e}^{x}dx=e^x(x^2-2x+2)+C$

Therefore, answer is $e^x(x^2-2x+2)+C$

Question 4: Integrate the functions $x\log x$

Answer:

Given function is
$f(x)=x.\log x$
We will use integration by parts method
$\int x\log xdx=\log x\cdot \int xdx-\int (\frac{d(\log x)}{dx}\cdot \int xdx)dx$
$\int x\log xdx=\log x\cdot \frac{x^2}{2}-\int (\frac{1}{x}\cdot \frac{x^2}{2})dx$
$\int x\log xdx=\log x\cdot \frac{x^2}{2}-\int \frac{x}{2}dx$
$\int x\log xdx=\log x\cdot \frac{x^2}{2}-\frac{x^2}{4}+C$

Therefore, the answer is $\frac{x^2}{2}\log x-\frac{x^2}{4}+C$

Question 5: Integrate the functions $x\log 2x$

Answer:

Given function is
$f(x)=x.\log 2x$
We will use integration by parts method

$\int x\log 2xdx=\log 2x\cdot \int xdx-\int (\frac{d(\log 2x)}{dx}\cdot \int xdx)dx$

$\int x\log 2xdx=\log 2x\cdot \frac{x^2}{2}-\int (\frac{2}{2x}\cdot \frac{x^2}{2})dx$

$\int x\log 2xdx=\log 2x\cdot \frac{x^2}{2}-\int \frac{x}{2}dx$

$\int x\log 2xdx=\log 2x\cdot \frac{x^2}{2}-\frac{x^2}{4}+C$
Therefore, the answer is $\log 2x.\frac{x^2}{2}-\frac{x^2}{4}+C$

Question 6: Integrate the functions $x^2\log x$

Answer:

Given function is
$f(x)=x^2.\log x$
We will use integration by parts method

$\int x^2\log xdx=\log x\cdot \int x^{2}dx-\int (\frac{d(\log x)}{dx}\cdot \int x^{2}dx)dx$

$\int x^2\log xdx=\log x\cdot \frac{x^3}{3}-\int (\frac{1}{x}\cdot \frac{x^3}{3})dx$

$\int x^2\log xdx=\log x\cdot \frac{x^3}{3}-\int \frac{x^2}{3}dx$

$\int x^2\log xdx=\log x\cdot \frac{x^3}{3}-\frac{x^3}{9}+C$
Therefore, the answer is $\log x.\frac{x^3}{3}-\frac{x^3}{9}+C$

Question 7: Integrate the functions $x{\sin }^{-1}x$

Answer:

Given function is
$f(x)=x.{\sin }^{-1}x$
We will use integration by parts method

$\int x{\sin }^{-1}xdx={\sin }^{-1}x\int xdx-\int (\frac{d({\sin }^{-1}x)}{dx}\cdot \int xdx)dx$

$\int x{\sin }^{-1}xdx={\sin }^{-1}x\cdot \frac{x^2}{2}-\int (\frac{1}{\sqrt{1-x^2}}\cdot \frac{x^2}{2})dx$
Now, we need to integrate $\int (\frac{1}{\sqrt{1-x^2}}.\frac{x^2}{2})dx$

$\int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\int (\frac{1-x^2}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}})dx$

$\int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\int (\sqrt{1-x^2}-\frac{1}{\sqrt{1-x^2}})dx$

$\int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}(\int \sqrt{1-x^2}dx-\int \frac{1}{\sqrt{1-x^2}}dx)$

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

$\int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{x\sqrt{1-x^2}}{4}-\frac{{\sin }^{-1}x}{4}+C$

Put this value in our equation

Therefore, the answer is $\int x{\sin }^{-1}xdx=\frac{{\sin }^{-1}x}{4}(2x^2-1)-\frac{x\sqrt{1-x^2}}{4}$

Question 8: Integrate the functions $x{\tan }^{-1}x$

Answer:

Given function is
$f(x)=x.{\tan }^{-1}x$
We will use integration by parts method

$\int x{\tan }^{-1}xdx={\tan }^{-1}x\int xdx-\int (\frac{d({\tan }^{-1}x)}{dx}\cdot \int xdx)dx$

$\int x{\tan }^{-1}xdx={\tan }^{-1}x\cdot \frac{x^2}{2}-\int (\frac{1}{1+x^2}\cdot \frac{x^2}{2})dx$

$\int x{\tan }^{-1}xdx={\tan }^{-1}x\cdot \frac{x^2}{2}-\frac{1}{2}\int (\frac{x^2+1}{1+x^2}-\frac{1}{1+x^2})dx$

$\int x{\tan }^{-1}xdx={\tan }^{-1}x\cdot \frac{x^2}{2}-\frac{1}{2}\int (1-\frac{1}{1+x^2})dx$

$\int x{\tan }^{-1}xdx={\tan }^{-1}x\cdot \frac{x^2}{2}-\frac{1}{2}(x-{\tan }^{-1}x)+C$

$\int x{\tan }^{-1}xdx=\frac{{\tan }^{-1}x}{2}(2x^2+1)-\frac{x}{2}+C$
Put this value in our equation
$$\begin{gathered} \int x{\sin }^{-1}xdx={\sin }^{-1}x.\frac{x^2}{2}-\frac{x}{4\sqrt{1-x^2}}-\frac{{\sin }^{-1}x}{4}+C \\ \int x{\sin }^{-1}xdx=\frac{{\sin }^{-1}x}{4}(2x^2-1)-\frac{x}{4\sqrt{1-x^2}} \end{gathered}$$

Therefore, the answer is $\frac{{\tan }^{-1}x}{2}(2x^2+1)-\frac{x}{2}+C$

Question 9: Integrate the functions $x{\cos }^{-1}x$

Answer:

Given function is
$f(x)=x.{\cos }^{-1}x$
We will use integration by parts method

$\int x{\cos }^{-1}xdx={\cos }^{-1}x\int xdx-\int (\frac{d({\cos }^{-1}x)}{dx}\cdot \int xdx)dx$

$\int x{\cos }^{-1}xdx={\cos }^{-1}x\cdot \frac{x^2}{2}-\int (\frac{-1}{\sqrt{1-x^2}}\cdot \frac{x^2}{2})dx$
Now, we need to integrate $\int (\frac{-1}{\sqrt{1-x^2}}.\frac{x^2}{2})dx$

$\int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\int (\frac{1-x^2}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}})dx$

$\int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\int (\sqrt{1-x^2}-\frac{1}{\sqrt{1-x^2}})dx$

$\int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}(\int \sqrt{1-x^2}dx-\int \frac{1}{\sqrt{1-x^2}}dx)$

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

$\int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{x\sqrt{1-x^2}}{4}-\frac{{\cos }^{-1}x}{4}+\frac{{\cos }^{-1}x}{2}+C$
Put this value in our equation
$$\begin{gathered} \int x{\cos }^{-1}xdx={\cos }^{-1}x.\frac{x^2}{2}-(\frac{x\sqrt{1-x^2}}{4}-\frac{{\cos }^{-1}x}{4}+\frac{{\cos }^{-1}x}{2}+C) \\ \int x{\cos }^{-1}xdx=\frac{{\cos }^{-1}x}{4}(2x^2-1)-\frac{x\sqrt{1-x^2}}{4} \end{gathered}$$

Therefore, the answer is $\frac{{\cos }^{-1}x}{4}(2x^2-1)-\frac{x\sqrt{1-x^2}}{4}$

Question 10: Integrate the functions $({\sin }^{-1}x{)}^2$

Answer:

Given function is
$f(x)=({\sin }^{-1}x{)}^2$
we will use integration by parts method

$\int ({\sin }^{-1}x{)}^{2}dx=({\sin }^{-1}x{)}^2\int 1dx-\int (\frac{d(({\sin }^{-1}x{)}^2)}{dx}\cdot \int 1dx)dx$

$\int ({\sin }^{-1}x{)}^{2}dx=({\sin }^{-1}x{)}^2\cdot x-\int ({\sin }^{-1}x\cdot \frac{2x}{\sqrt{1-x^2}})dx$

$\int ({\sin }^{-1}x{)}^{2}dx=({\sin }^{-1}x{)}^2\cdot x+[{\sin }^{-1}x\cdot \int \frac{-2x}{\sqrt{1-x^2}}dx-\int (\frac{d({\sin }^{-1}x)}{dx}\cdot \int \frac{-2x}{\sqrt{1-x^2}}dx)]$

$=({\sin }^{-1}x{)}^2\cdot x+[{\sin }^{-1}x\cdot 2\sqrt{1-x^2}-\int \frac{1}{\sqrt{1-x^2}}\cdot 2\sqrt{1-x^2}dx]$

$=({\sin }^{-1}x{)}^2\cdot x+2{\sin }^{-1}x\sqrt{1-x^2}-2x+C$
Therefore, answer is $({\sin }^{-1}x{)}^2.x+2{\sin }^{-1}x\sqrt{1-x^2}-2x+C$

Question 11: Integrate the functions $\frac{x{\cos }^{-1}}{\sqrt{1-x^2}}$

Answer:

Consider $\int \frac{x{\cos }^{-1}}{\sqrt{1-x^2}}dx=I$

So, we have then: $I=\frac{-1}{2}\int \frac{-2x}{\sqrt{1-x^2}}.{\cos }^{-1}xdx$

After taking ${\cos }^{-1}x$ as a first function and $\left(\frac{-2x}{\sqrt{1-x^2}}\right)$ as second function and integrating by parts, we get

$I=-\frac{1}{2}[{\cos }^{-1}x\int \frac{-2x}{\sqrt{1-x^2}}dx-\int \{(\frac{d}{dx}{\cos }^{-1}x)\int \frac{-2x}{\sqrt{1-x^2}}dx\}dx]$ $=-\frac{1}{2}[{\cos }^{-1}x.2\sqrt{1-x^2}+\int \frac{-1}{\sqrt{1-x^2}}.2\sqrt{1-x^2}dx]$

$=\frac{-1}{2}[2\sqrt{1-x^2}{\cos }^{-1}x-\int 2dx]$

$=\frac{-1}{2}[2\sqrt{1-x^2}{\cos }^{-1}x-2x]+C$

Or, $-(\sqrt{1-x^2}{\cos }^{-1}x+x)+C$

Question 12: Integrate the functions $x{\sec }^{2}x$

Answer:

Consider $x{\sec }^{2}x$

So, we have then: $I=\int x{\sec }^{2}xdx$

After taking $x$ as a first function and ${\sec }^{2}x$ as second function and integrating by parts, we get

$I=x\int {\sec }^{2}xdx-\int \{\left(\frac{d}{dx}x\right)\int {\sec }^{2}xdx\}dx$

$=x\tan x-\int 1.\tan xdx$

$=x\tan x+\log |\cos x|+C$

Question 13: Integrate the functions ${\tan }^{-1}x$

Answer:

Consider ${\tan }^{-1}x$

So, we have then: $I=\int 1.{\tan }^{-1}xdx$

After taking ${\tan }^{-1}x$ as a first function and $1$ as second function and integrating by parts, we get

$I={\tan }^{-1}x\int 1dx-\int \{(\frac{d}{dx}{\tan }^{-1}x)\int 1.dx\}dx$

$={\tan }^{-1}x.x-\int \frac{1}{1+x^2}.xdx$

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

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

$=x{\tan }^{-1}x-\frac{1}{2}\log (1+x^2)+C$

Question 14: Integrate the functions $x(\log x{)}^2$

Answer:

Consider $x(\log x{)}^2$

So, we have then: $I=\int x(\log x{)}^{2}dx$

After taking $(\log x{)}^2$ as a first function and $x$ as second function and integrating by parts, we get

$I=(\log x{)}^2\int xdx-\int \{(\frac{d}{dx}(\log x{)}^2)\int x.dx\}dx$

$=(\log x{)}^2.\frac{x^2}{2}-\int \frac{2\log x}{x}.\frac{x^2}{2}dx$

$=(\log x{)}^2.\frac{x^2}{2}-\int x\log xdx$

$=(\log x{)}^2.\frac{x^2}{2}-(\frac{x^2\log x}{2}-\frac{x^2}{4})+C$

Question 15: Integrate the functions $(x^2+1)\log x$

Answer:

Consider $(x^2+1)\log x$

So, we have then: $I=\int (x^2+1)\log xdx=\int x^2\log xdx+\int \log xdx$

Let us take $I=I_1+I_2$ ....................(1)

Where, $I_1=\int x^2\log xdx$ and $I_2=\int \log xdx$

So, $I_1=\int x^2\log xdx$

After taking $\log x$ as a first function and $x^2$ as second function and integrating by parts, we get

$I=\log x\int x^{2}dx-\int \{(\frac{d}{dx}\log x)\int x^2.dx\}dx$

$=\log x.\frac{x^3}{3}-\int \frac{1}{x}.\frac{x^3}{3}dx$

$=\log x.\frac{x^3}{3}-\frac{x^3}{9}+C_1$ ....................(2)

$I_2=\int \log xdx$

After taking $\log x$ as a first function and $1$ as second function and integrating by parts, we get

$I_2=\log x\int 1.dx-\int \{(\frac{d}{dx}\log x)\int 1.dx\}dx$

$=\log x.x-\int \frac{1}{x}.xdx$

$=x\log x-\int 1dx$

$=x\log x-x+C_2$ ................(3)

Now, using the two equations (2) and (3) in (1) we get,

$I=\frac{x^3}{3}\log x-\frac{x^3}{9}+C_1+x\log x-x+C_2$

$=\frac{x^3}{3}\log x-\frac{x^3}{9}+x\log x-x+(C_1+C_2)$

$=(\frac{x^3}{3}+x)\log x-\frac{x^3}{9}-x+C$

Question 16: Integrate the functions $e^x(\sin x+\cos x)$

Answer:

Let suppose
$I=$ $e^x(\sin x+\cos x)$
$f(x)=\sin x\Rightarrow f^{\prime }(x)=\cos x$
we know that,
$I=\int e^x[f(x)+f^{\prime }(x)]dx=e^x[f(x)]+C$
Thus, the solution of the given integral is given by

$\therefore I=e^x\sin x+C$

Question 17: Integrate the functions $\frac{x{e}^x}{(1+x{)}^2}$

Answer:

$\frac{x{e}^x}{(1+x{)}^2}$
Let suppose
$I=\int \frac{e^x(x)}{(1+x{)}^2}dx$
by rearranging the equation, we get
$\Rightarrow \int e^x[\frac{1}{1+x}-\frac{1}{(1+x{)}^2}]dx$
let
$f(x)=\frac{1}{1+x}\Rightarrow f^{\prime }(x)=-\frac{1}{(1+x{)}^2}$
It is known that $\int e^x[f(x)+f^{\prime }(x)]=e^x[f(x)]+C$
therefore the solution of the given integral is

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

Question 18: Integrate the functions $e^x\left(\frac{1+\sin x}{1+\cos x}\right)$

Answer:

Let
$I=e^x\left(\frac{1+\sin x}{1+\cos x}\right)$
substitute $1={\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}$ and $\sin x=2\sin \frac{x}{2}\cos \frac{x}{2}$

$$\begin{gathered} \Rightarrow e^x(\frac{{\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}}{2{\cos }^2\frac{x}{2}}) \\ =e^x(\frac{1}{2}{\sec }^2\frac{x}{2}+\tan \frac{x}{2}) \end{gathered}$$
let
$f(x)=\tan \frac{x}{2}\Rightarrow f^{\prime }(x)=\frac{1}{2}{\sec }^2\frac{x}{2}$
It is known that $\int e^x[f(x)+f^{\prime }(x)]=e^x[f(x)]+C$
Therefore the solution of the given integral is

$I=e^x\tan \frac{x}{2}+C$

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

Answer:

$e^x(\frac{1}{x}-\frac{1}{x^2})$
It is known that
$\int e^x[f(x)+f^{\prime }(x)]=e^x[f(x)]+C$

let
$f(x)=\frac{1}{x}\Rightarrow f^{\prime }(x)=-\frac{1}{x^2}$
Therefore the required solution of the given above integral is
$I=e^x.\frac{1}{x}+C$

Question 20: Integrate the functions $\frac{(x-3)e^x}{(x-1{)}^3}$

Answer:

$\frac{(x-3)e^x}{(x-1{)}^3}$
It is known that $\int e^x[f(x)+f^{\prime }(x)]=e^x[f(x)]+C$

So, By adjusting the given equation, we get
$\int \frac{(x-3)e^x}{(x-1{)}^3}=\int e^x(\frac{x-1-2}{(x-1{)}^3})=\int e^x(\frac{1}{(x-1{)}^2}-\frac{2}{(x-1{)}^3})dx$

to let
$f(x)=\frac{1}{(x-1{)}^2}\Rightarrow f^{\prime }(x)=-\frac{2}{(x-1{)}^3}$
Therefore the required solution of the given $I=\frac{e^x}{(x-1{)}^2}+C$ integral is

Question 21: Integrate the functions $e^{2x}\sin x$

Answer:

Let
$I=e^{2x}\sin x$
By using integrating by parts, we get

$=\sin x\int e^{2x}dx-\int (\frac{d}{dx}\sin x\cdot \int e^{2x}dx)dx$

$=\frac{\sin x\cdot e^{2x}}{2}-\frac{1}{2}\int e^{2x}\cos xdx$

$=\frac{\sin x\cdot e^{2x}}{2}-\frac{1}{2}[\cos x\int e^{2x}dx-\int (\frac{d}{dx}\cos x\cdot \int e^{2x}dx)dx]$

$=\frac{\sin x\cdot e^{2x}}{2}-\frac{1}{2}[\cos x\cdot \frac{e^{2x}}{2}+\frac{1}{2}\int e^{2x}\sin xdx]$

$=\frac{\sin x\cdot e^{2x}}{2}-\frac{1}{4}\cos x\cdot e^{2x}-\frac{1}{4}I$

$\Rightarrow \frac{5}{4}I=\frac{\sin x\cdot e^{2x}}{2}-\frac{1}{4}\cos x\cdot e^{2x}$

$I=\frac{e^{2x}}{5}(2\sin x-\cos x)+C$

Question 22: Integrate the functions ${\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$

Answer:

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

$\int {\sin }^{-1}\left(\frac{2x}{1+x^2}\right)dx$
let $x=\tan \theta \Rightarrow dx={\sec }^2\theta d\theta$

$$\begin{gathered} =\int {\sin }^{-1}(\frac{2\tan \theta }{1+\tan \theta }){\sec }^2\theta d\theta \\ =\int {\sin }^{-1}(\sin 2\theta ){\sec }^2\theta d\theta \\ =\int 2\theta {\sec }^2\theta d\theta \end{gathered}$$
Taking $\theta$ as a first function and ${\sec }^2\theta$ as a second function, by using by parts method

$=2[\theta \int {\sec }^2\theta d\theta -\int (\frac{d}{d\theta }\theta \cdot \int {\sec }^2\theta d\theta )d\theta ]$

$=2[\theta \tan \theta -\int \tan \theta d\theta ]+C$

$=2[\theta \tan \theta +\log |\cos \theta |]+C$

$=2x{\tan }^{-1}x+2\log ((1+x^2{)}^{-1/2})$

$=2x{\tan }^{-1}x-\log (1+x^2)+C$