NCERT Solutions for Exercise 7.10 Class 12 Maths Chapter 7 - Integrals

Question 1: Evaluate the integral using substitution.

$\int_0^1\frac{x}{x^2 +1}dx$

Answer:

$\int_0^1\frac{x}{x^2 +1}dx$
let $x^2+1 = t \Rightarrow xdx =dt/2$
when x = 0 then t = 1 and when x =1 then t = 2
$\therefore \int_{o}^{1}\frac{x}{x^2+1}dx=\frac{1}{2}\int_{1}^{2}\frac{dt}{t}$
$\\=\frac{1}{2}[\log\left | t \right |]_{1}^{2}\\ =\frac{1}{2}\log 2$

Question 2: Evaluate the integral using substitution.

$\int^\frac{\pi}{2}_0\sqrt{\sin\phi}\cos^5\phi d\phi$

Answer:

$\int^\frac{\pi}{2}_0\sqrt{\sin\phi}\cos^5\phi d\phi$
let $\sin \phi = t \Rightarrow \cos \phi d\phi = dt$
when $\phi =0,t\rightarrow 0$ and $\phi =\pi/2,t\rightarrow 1$

using the above substitution we can evaluate the integral as

$\therefore \int_{0}^{1} \sqrt{t}(1 - t^2)\,dt$

$= \int_{0}^{1} t^{\frac{1}{2}}(1 + t^4 - 2t^2)\,dt$

$= \int_{0}^{1} t^{\frac{1}{2}}\,dt + \int_{0}^{1} t^{\frac{9}{2}}\,dt - 2\int_{0}^{1} t^{\frac{5}{2}}\,dt$

$= \left[\frac{2t^{3/2}}{3} + \frac{2t^{11/2}}{11} - \frac{4t^{7/2}}{7} \right]_0^1$

$= \frac{64}{231}$

Question 3: Evaluate the integral using substitution.

$\int_0^1 \sin^{-1}\left(\frac{2x}{1+x^2} \right )dx$

Answer:

$\int_0^1 \sin^{-1}\left(\frac{2x}{1+x^2} \right )dx$
let $x = \tan\theta\Rightarrow dx =\sec^2\theta d\theta$
when x = 0 then $\theta= 0$ and when x = 1 then $\theta= \pi/4$

$\\=\int_{0}^{\pi/4}\sin^{-1}(\frac{2\tan\theta}{1+\tan\theta})\sec^2\theta d\theta\\ =\int_{0}^{\pi/4}\sin^{-1}(\sin 2\theta)\sec^2\theta d\theta\\ =\int_{0}^{\pi/4}2\theta \sec^2\theta d\theta\\$
Taking $\theta$ as a first function and $\sec^2\theta$ as a second function, by using by parts method

$\begin{aligned}
&= 2\left[\theta \int \sec^2\theta\, d\theta - \int\left( \frac{d}{d\theta} \theta \cdot \int \sec^2\theta\, d\theta \right) d\theta \right]_0^{\pi/4} \\
&= 2\left[\theta \tan\theta - \int \tan\theta\, d\theta \right]_0^{\pi/4} \\
&= 2\left[\theta \tan\theta + \log\left| \cos\theta \right| \right]_0^{\pi/4} \\
&= 2\left[\frac{\pi}{4} + \log\left(\frac{1}{\sqrt{2}}\right)\right] \\
&= \frac{\pi}{2} - \log 2
\end{aligned}$

Question 4: Evaluate the integral using substitution.

$\int_0^2x\sqrt{x+2}$ . (Put ${x+2} = t^2$ )

Answer:

Let $x+2 = t^2\Rightarrow dx =2t dt$
when x = 0 then t = $\sqrt{2}$ and when x=2 then t = 2

$I=\int_{0}^{2}x\sqrt{x+2}dx$

$= 2\int_{\sqrt{2}}^{2}(t^2 - 2)t^2\, dt$

$= 2\int_{\sqrt{2}}^{2}(t^4 - 2t^2)\, dt$

$= 2\left[ \frac{t^5}{5} - \frac{2}{3}t^3 \right]_{\sqrt{2}}^2$

$= 2\left[ \frac{32}{5} - \frac{16}{3} - \frac{4\sqrt{2}}{5} + \frac{4\sqrt{2}}{3} \right]$

$= \frac{16\sqrt{2}(\sqrt{2} + 1)}{15}$

Question 5: Evaluate the integral using substitution.

$\int_0^{\frac{\pi}{2}}\frac{\sin x}{1 + \cos^2 x}dx$

Answer:

$\int_0^{\frac{\pi}{2}}\frac{\sin x}{1 + \cos^2 x}dx =I$
let $\cos x =t\Rightarrow -\sin x dx = dt$
when x=0 then t = 1 and when x= $\pi/2$ then t = 0

$\\I=\int_{1}^{0}\frac{dt}{1+t^2}\\ =[\tan ^{-1}t]^0_1\\ =\pi/4$

Question 6: Evaluate the integral using substitution.

$\int_0^2\frac{dx}{x + 4 - x^2}$

Answer:

By adjusting, the denominator can also be written as $(\frac{\sqrt{17}}{2})^2-(x-\frac{1}{2})^2 =x+4-x^2$
Now,
$\Rightarrow \int_{0}^{2}\frac{dx}{(\frac{\sqrt{17}}{2})^2-(x-\frac{1}{2})^2}$
let $x-1/2 = t\Rightarrow dx=dt$
when x= 0 then t =-1/2 and when x =2 then t = 3/2

$\Rightarrow \int_{-1/2}^{3/2} \frac{dt}{\left( \frac{\sqrt{17}}{2} \right)^2 - t^2}$
$= \frac{1}{2 \cdot \frac{\sqrt{17}}{2}} \log \frac{\frac{\sqrt{17}}{2} + t}{\frac{\sqrt{17}}{2} - t}$
$= \frac{1}{\sqrt{17}} \left[ \log \frac{\frac{\sqrt{17}}{2} + \frac{3}{2}}{\frac{\sqrt{17}}{2} - \frac{3}{2}} - \log \frac{\frac{\sqrt{17}}{2} - \frac{1}{2}}{\frac{\sqrt{17}}{2} + \frac{1}{2}} \right]$
$= \frac{1}{\sqrt{17}} \left[ \log \left( \frac{\sqrt{17} + 3}{\sqrt{17} - 3} \cdot \frac{\sqrt{17} + 1}{\sqrt{17} - 1} \right) \right]$
$= \frac{1}{\sqrt{17}} \left[ \log \left( \frac{17 + 3 + 4\sqrt{17}}{17 + 3 - 4\sqrt{17}} \right) \right]$
$= \frac{1}{\sqrt{17}} \log \left( \frac{5 + \sqrt{17}}{5 - \sqrt{17}} \right)$
On rationalisation, we get

$=\frac{1}{\sqrt{17}}\log \frac{21+5\sqrt{17}}{4}$

Question 7: Evaluate the integral using substitution.

$\int_{-1}^1\frac{dx}{x^2 +2x + 5}$

Answer:

$\int_{-1}^1\frac{dx}{x^2 +2x + 5}$
it can be written as $x^2+2x+5 = (x+1)^2+2^2$
and put x+1 = t then dx =dt

when x= -1 then t = 0 and when x = 1 then t = 2

$\\\Rightarrow \int_{0}^{2}\frac{dt}{t^2+2^2}\\ =\frac{1}{2}[\tan^{-1}\frac{t}{2}]^2_0\\ =\frac{1}{2}( \pi/4)\\ =\frac{\pi}{8}$

Question 8: Evaluate the integral using substitution.

$\int_1^2\left(\frac{1}{x} - \frac{1}{2x^2} \right )e^{2x}dx$

Answer:

$\int_1^2\left(\frac{1}{x} - \frac{1}{2x^2} \right )e^{2x}dx$
let $2x =t \Rightarrow 2dx =dt$
when x = 1 then t = 2 and when x = 2 then t= 4

$\\=\frac{1}{2}\int_{2}^{4}(\frac{2}{t}-\frac{2}{t^2})e^tdt\\$
let
$\frac{1}{t} = f(t)\Rightarrow f'(t)=-\frac{1}{t^2}$
$\Rightarrow \int_{2}^{4}(\frac{1}{t}-\frac{1}{t^2})e^tdt =\int_{2}^{}4e^t[f(t)+f'(t)]dt$
$\\=[e^tf(t)]^4_2\\ =[e^t.\frac{1}{t}]^4_2\\ =\frac{e^4}{4}-\frac{e^2}{2}\\ =\frac{e^2(e^2-2)}{4}$

Question 9: Choose the correct answer

The value of the integral $\int_{\frac{1}{3}}^1\frac{(x-x^3)^\frac{1}{3}}{x^4}dx$ is

(A) 6

(B) 0

(C) 3

(D) 4

Answer:

$\int_{\frac{1}{3}}^1\frac{(x-x^3)^\frac{1}{3}}{x^4}dx$
$\int_{\frac{1}{3}}^1\frac{(\frac{1}{x^2}-1)^\frac{1}{3}}{x^3}dx\\$
let
$\frac{1}{x^2}-1 = t\Rightarrow \frac{dx}{x^3}=-dt/2$
now, when x = 1/3, t = 8 and when x = 1 , t = 0

$\\=-\frac{1}{2}\int_{8}^{0}t^{1/3}dt\\ =-\frac{1}{2}.\frac{3}{4}[t^4/3]^0_8\\ =-\frac{3}{8}[-2^4]\\ =6$

Hence, The value of integral is 6

Question 10: Choose the correct answer

If $f(x) = \int_0^x t \sin t dt$ , then $f'(x)$ is

(A) $\cos x + x\sin x$

(B) $x\sin x$

(C) $x\cos x$

(D) $\sin x + x\cos x$

Answer:

$f(x) = \int_0^x t \sin t dt$
by using by parts method,
$\\=t\int \sin t dt - \int (\frac{d}{dt}t\int \sin t dt)dt\\ =[t(-\cos t )+\sin t]^x_0$

$f(x) = -x\,\text{cos}\,x + \text{sin}\,x$

So, $f'(x) = -\text{cos}\,x + x\,\text{sin}\,x + \text{cos}\,x = x\,\text{sin}\,x$

Hence, The correct answer is $x\sin x$