NCERT Solutions for Class 12 Maths Chapter 7 Integrals

Question:11 Find the following integrals intergration of $\int \frac{x^3+5x^2-4}{x^2}dx$

Answer:

Given intergral $\int \frac{x^3+5x^2-4}{x^2}dx$ ;

$=\int \frac{x^3}{x^2}dx+\int \frac{5x^2}{x^2}dx-4\int \frac{1}{x^2}dx$

$=\int xdx+5\int 1.dx-4\int x^{-2}dx$

$=\frac{x^2}{2}+5x-4\left(\frac{x^{-1}}{-1}\right)+C$

$=\frac{x^2}{2}+5x+\frac{4}{x}+C$ , where C is any constant value.

Question:12 Find the following integrals $\int \frac{x^3+3x+4}{\sqrt{x}}dx$

Answer:

Given intergral $\int \frac{x^3+3x+4}{\sqrt{x}}dx$ ;

$=\int \frac{x^3}{x^{\frac{1}{2}}}dx+\int \frac{3x}{x^{\frac{1}{2}}}dx+4\int \frac{1}{x^{\frac{1}{2}}}dx$

$=\int x^{\frac{5}{2}}dx+3\int x^{\frac{1}{2}}dx+4\int x^{-\frac{1}{2}}dx$

$=\frac{x^{\frac{7}{2}}}{\frac{7}{2}}+\frac{3\left(x^{\frac{3}{2}}\right)}{\frac{3}{2}}+\frac{4\left(x^{\frac{1}{2}}\right)}{\frac{1}{2}}+C$

$=\frac{2}{7}{x}^{\frac{7}{2}}+2x^{\frac{3}{2}}+8\sqrt{x}+C$, where C is any constant value.

Question:13 Find the following integrals intergration of $\int \frac{x^3-x^2+x-1}{x-1}dx$

Answer:

Given integral $\int \frac{x^3-x^2+x-1}{x-1}dx$

It can be written as

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

Taking $(x-1)$ common out

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

Now, cancelling out the term $(x-1)$ from both the numerator and the denominator.

$=\int (x^2+1)dx$

Splitting the terms inside the brackets

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

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

Question:14 Find the following integrals $\int (1-x)\sqrt{x}dx$

Answer:

Given intergral $\int (1-x)\sqrt{x}dx$ ;

$\int \sqrt{x}dx-\int x\sqrt{x}dx$ or

$\int x^{\frac{1}{2}}dx-\int x^{\frac{3}{2}}dx$

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

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

Question:15 Find the following integrals $\int \sqrt{x}(3x^2+2x+3)dx$

Answer:

Given intergral $\int \sqrt{x}(3x^2+2x+3)dx$ ;

$=\int 3x^2\sqrt{x}dx+\int 2x\sqrt{x}dx+\int 3\sqrt{x}dx$ or $=3\int x^{\frac{5}{2}}dx+2\int x^{\frac{3}{2}}dx+3\int x^{\frac{1}{2}}dx$

$=3\frac{x^{\frac{7}{2}}}{\frac{7}{2}}+2\frac{x^{\frac{5}{2}}}{\frac{5}{2}}+3\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+C$

$=\frac{6}{7}{x}^{\frac{7}{2}}+\frac{4}{5}{x}^{\frac{5}{2}}+2x^{\frac{3}{2}}+C$, where C is any constant value.

Question:16 Find the following integrals $\int (2x-3\cos x+e^x)dx$

Answer:

Given integral $\int (2x-3\cos x+e^x)dx$;

Splitting the integral as the sum of three integrals

$\int 2xdx-3\int \cos xdx+\int e^{x}dx$

$=2\frac{x^2}{2}-3\sin x+e^x+C$

$=x^2-3\sin x+e^x+C$, where C is any constant value.

Question:17 Find the following integrals $\int (2x^2-3\sin x+5\sqrt{x})dx$

Answer:

Given integral $\int (2x^2-3\sin x+5\sqrt{x})dx$ ;

$2\int x^{2}dx-3\int \sin xdx+5\int \sqrt{x}dx$

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

$=\frac{2x^3}{3}+3\cos x+\frac{10}{3}{x}^{\frac{3}{2}}+C$, where C is any constant value.

Question:18 Find the following integrals $\int \sec x(\sec x+\tan x)dx$

Answer:

Given integral $\int \sec x(\sec x+\tan x)dx$ ;

$\int ({\sec }^{2}x+\sec x\tan x)dx$

Using the integral of trigonometric functions

$=\int (se{c}^{2}x)dx+\int \sec x\tan xdx$

$=\tan x+\sec x+C$, where C is any constant value.

Question:19 Find the following integrals intergration of $\int \frac{se{c}^{2}x}{cose{c}^{2}x}dx$

Answer:

Given integral $\int \frac{se{c}^{2}x}{cose{c}^{2}x}dx$ ;

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

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

$=\int {\tan }^{2}xdx$

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

$=\int {\sec }^{2}xdx-\int 1dx$

$=\tan x-x+C$, where C is any constant value.

Question:20 Find the following integrals $\int \frac{2-3\sin x}{{\cos }^{2}x}dx$

Answer:

Given integral $\int \frac{2-3\sin x}{{\cos }^{2}x}dx$ ;

$\int (\frac{2}{{\cos }^{2}x}-\frac{3\sin x}{{\cos }^{2}x})dx$

Using the antiderivative of trigonometric functions

$=2\tan x-3\sec x+C$, where C is any constant value.

Question: 21 Choose the correct answer
The anti derivative of $(\sqrt{x}+1/\sqrt{x})$ equals

$$\begin{gathered} A)\frac{1}{3}{x}^{1/3}+2x^{1/2}+C \\ B)\frac{2}{3}{x}^{2/3}+\frac{1}{2}{x}^2+C \\ C)\frac{2}{3}{x}^{3/2}+2x^{1/2}+C \\ D)\frac{3}{2}{x}^{3/2}+\frac{1}{2}{x}^{1/2}+C \end{gathered}$$

Answer:

Given to find the anti derivative or integral of $(\sqrt{x}+1/\sqrt{x})$ ;

$\int (\sqrt{x}+1/\sqrt{x})dx$

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

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

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

Hence, the correct option is (C).

Question: 22 Choose the correct answer:

If $\frac{d}{dx}f(x)=4x^3-\frac{3}{x^4}$ such that f (2) = 0. Then f (x) is

$$\begin{gathered} A)x^4+\frac{1}{x^3}-\frac{129}{8} \\ B)x^3+\frac{1}{x^4}-\frac{129}{8} \\ C)x^4+\frac{1}{x^3}+\frac{129}{8} \\ D)x^3+\frac{1}{x^4}-\frac{129}{8} \end{gathered}$$

Answer:

Given that the anti derivative of $\frac{d}{dx}f(x)=4x^3-\frac{3}{x^4}$

So, $\frac{d}{dx}f(x)=4x^3-\frac{3}{x^4}$

$f(x)=\int 4x^3-\frac{3}{x^4}dx$

$f(x)=4\int x^3-3\int x^{-4}dx$

$f(x)=4\left(\frac{x^4}{4}\right)-3\left(\frac{x^{-3}}{-3}\right)+C$

$f(x)=x^4+\frac{1}{x^3}+C$

Now, to find the constant C;

Putting the condition, f (2) = 0

$f(2)=2^4+\frac{1}{2^3}+C=0$

$\Rightarrow 16+\frac{1}{8}+C=0$

$\Rightarrow C=\frac{-129}{8}$

$\Rightarrow f(x)=x^4+\frac{1}{x^3}-\frac{129}{8}$

Therefore, the correct answer is A.

Class 12 Integrals NCERT solutions Exercise: 7.2
Page: 240-241, Total Questions: 39

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

Answer:

Given to integrate $\frac{2x}{1+x^2}$ function,

Let us assume $1+x^2=t$

We get, $2xdx=dt$

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

$=\log |t|+C$

$=\log |1+x^2|+C$ now back substituting the value of $t=1+x^2$

As $(1+x^2)$ is positive we can write

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

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

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

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

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

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

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

$=\log |t|+C$

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

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,

$=\cos (\cos x)+C$, where C is any constant value.

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,

$=\frac{-1}{4a}[cos2(ax+b)]+C$, where C is any constant value.

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,

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

$=\frac{2(ax+b{)}^{\frac{3}{2}}}{3a}+C$, where C is any constant value.

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

$\therefore dx=dt$

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

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

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

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

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

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

Substituting the value of $t$ in the above equation,

$=\frac{2}{5}(x+2{)}^{\frac{5}{2}}-\frac{4}{3}(x+2{)}^{\frac{3}{2}}+C$, 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 $1+2x^2=t$

$\therefore 4xdx=dt$

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

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

$=\frac{1}{6}(1+2x^2{)}^{\frac{3}{2}}+C$, 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 $x^3-1=t$

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

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

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

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

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

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

$=\frac{1}{7}(x^3-1{)}^{\frac{7}{3}}+\frac{1}{4}(x^3-1{)}^{\frac{4}{3}}+C$ , 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 $2+3x^3=t$

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

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

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

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

$=\frac{-1}{18(2+3x^3{)}^2}+C$ , 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 $\log x=t$

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

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

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

$=\frac{(logx{)}^{1-m}}{(1-m)}+C$ , 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 $9-4x^2=t$

$\therefore -8xdx=dt$

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

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

Now, 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 $2x+3=t$

$\therefore 2dx=dt$

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

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

Now, 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 $x^2=t$

$\therefore 2xdx=dt$

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

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

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

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

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

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

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

$=e^{ta{n}^{-1}x}+C$, where C is any constant value.

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 the numerator and the 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 $e^x+e^{-x}=t$

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

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

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

$=\log |t|+C$

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

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

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

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

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

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