RD Sharma Solutions Class 12 Mathematics Chapter 19 MCQ

# RD Sharma Solutions Class 12 Mathematics Chapter 19 MCQ

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 02:20 PM IST

The class 12 RD Sharma chapter 19 exercise MCQ solution deals with the chapter of definite integrals which is a sub part of the indefinite integrals chapter, The Definite integral is defined to be the exact limit and summation that we looked at in the past section to find the net area between a function and the X-axis . The RD Sharma class 12th exercise MCQ gives us the brief practice of the entire chapter and makes us go through all the concepts one by one to revise each and every topic of the chapter thoroughly.

Also Read - RD Sharma Solutions For Class 9 to 12 Maths

## Definite Integrals Excercise: MCQ

Definite Integrals exercise Multiple choice question 1

$\frac{\pi }{8}$
Hint:
Using $\int \sqrt{a^{2}-x^{2}} \; d x$
Explanation:
\begin{aligned} &\int_{0}^{1} \sqrt{x(1-x)} d x \\\\ &=\int_{0}^{1} \sqrt{x-x^{2}} d x \end{aligned}
\begin{aligned} &=\int_{0}^{1} \sqrt{(x)^{2}-x+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}} d x \\\\ &=\int_{0}^{1} \sqrt{-\left(x-\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}} d x \\\\ &=\int_{0}^{1} \sqrt{\left(\frac{1}{2}\right)^{2}-\left(x-\frac{1}{2}\right)^{2}} d x \end{aligned}
We know
$\int \sqrt{a^{2}-x^{2}}=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c$
$I=\left[\frac{x-1 / 2}{2} \sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^{2}}+\frac{1}{8} \sin ^{-1}\left(\frac{x-1 / 2}{2}\right)\right]_{0}^{1}$
$I=\left[\frac{2 x-1}{4} \sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^{2}}+\frac{1}{8} \sin ^{-1}\left(\frac{x-1 / 2}{1 / 2}\right)\right]_{0}^{1}$
$=\left[\frac{2 x-1}{4} \sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^{2}}+\frac{1}{8} \sin ^{-1}(2 x-1)\right]_{0}^{1}$
$=\left(\frac{1}{4} \sqrt{\frac{1}{4}-\left(1-\frac{1}{2}\right)^{2}}+\frac{1}{8} \sin ^{-1}(1)\right)-\left(\frac{-1}{4} \sqrt{\frac{1}{4}-\left(0-\frac{1}{2}\right)^{2}}+\frac{1}{8} \sin ^{-1}(-1)\right)$
\begin{aligned} &=\left(\frac{1}{4} \sqrt{\frac{1}{4}-\frac{1}{4}}+\frac{1}{8} \times \frac{\pi}{2}\right)-\left(\frac{-1}{4} \sqrt{\frac{1}{4}-\frac{1}{4}}+\frac{1}{8} \times\left(-\frac{\pi}{2}\right)\right) \\\\ &=0+\frac{\pi}{16}-0-\left(-\frac{\pi}{16}\right) \\\\ &=\frac{\pi}{8} \end{aligned}

Definite Integrals exercise Multiple choice question 2

2
Hint:
Using $\int x \; d x \text { and } \frac{d}{d x}(\tan x)$
Given:
$\int_{0}^{\pi} \frac{1}{1+\sin x}$
Explanation:
Let
\begin{aligned} &I=\int_{0}^{\pi} \frac{1}{1+\sin x} \\\\ &=\int_{0}^{\pi} \frac{1}{1+\frac{2 \tan {x} / 2}{1+\tan^ {2} x / 2}} d x \end{aligned}
\begin{aligned} &=\int_{0}^{\pi} \frac{1+\tan ^{2 } \frac{x}{2}}{1+\tan ^{2} x / 2+2 \tan x / 2} d x \\\\ &=\int \frac{\sec ^{2 } \frac {x}{2}}{\tan ^{2} x / 2+2 \tan \frac{x}{2}+1} d x \end{aligned}

\begin{aligned} &\text { Put } 2 \tan \frac{x}{2}=t \\\\ &\sec ^{2} x / 2 \cdot 1 / 2 d x=d t \\\\ &\sec ^{2} x / 2 d x=2 d t \end{aligned}
When $x=0$ then $t= 0$
When $x=\pi$ then $t= \infty$
\begin{aligned} &=\int_{0}^{\infty} \frac{2 d t}{t^{2}+2 t+1} \\\\ &=\int_{0}^{\infty} \frac{1}{(t+1)^{2}} d t \end{aligned}
\begin{aligned} &=\int_{0}^{\infty}(t+1)^{-2} d t \\\\ &=\left[\frac{(t+1)^{2}}{-2+1}\right]_{9}^{\infty} \end{aligned}
\begin{aligned} &=-2\left[\frac{1}{(t+1)}\right]_{9}^{\infty} \\\\ &=-2(0-1) \\\\ &=2 \end{aligned}

Definite Integrals exercise Multiple choice question 3

$\frac{\pi^{2}}{4}$
Hint:
$\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x$
Given:
$\int_{0}^{\pi} \frac{x \tan x}{\sec x+\cos x} d x$
Explanation:
Let
$I=\int_{0}^{\pi} \frac{x \tan x}{\sec x+\cos x} d x, \quad\left[\therefore \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]$
$=\int_{0}^{\pi} \frac{-(\pi-x) \tan x}{-\sec x-\cos x} d x$
\begin{aligned} &I=\int_{0}^{\pi} \frac{(\pi-x) \tan x}{\sec x+\cos x} d x \\\\ &I=\int_{0}^{\pi} \frac{\pi \tan x}{\sec x+\cos x} d x-\int_{0}^{\pi} \frac{x \tan x}{\sec x+\cos x} d x \end{aligned}
\begin{aligned} &I=\pi \int_{0}^{\pi} \frac{\tan x}{\sec x+\cos x} d x-I \\\\ &2 I=\pi \int_{0}^{\pi} \frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}+\cos x} d x \\\\ &2 I=\pi \int_{0}^{\pi} \frac{\sin x}{1+\cos ^{2} x} d x \end{aligned}
Put
\begin{aligned} &\cos x=t \\\\ &-\sin x \; d x=d t \end{aligned}
When $x=0$ then $t=1$
When $x=\pi$ then $t=-1$
$=\pi \int_{1}^{-1} \frac{-d t}{1+t^{2}}$
$=\frac{\pi}{2} \int_{-1}^{1} \frac{d t}{1+t^{2}}, \quad\left[\therefore-\int_{b}^{a} f(x) d x=\int_{a}^{b} f(a-x) d x\right]$
\begin{aligned} &=\frac{\pi}{2}\left[\tan ^{-1} t\right]_{-1}^{1} \\\\ &=\frac{\pi}{2}\left[\tan ^{-1}(1)-\tan ^{-1}(-1)\right] \\\\ &=\frac{\pi}{2}\left[\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)\right] \end{aligned}
\begin{aligned} &=\frac{\pi}{2}\left(\frac{\pi}{2}\right) \\\\ &=\frac{\pi^{2}}{4} \end{aligned}

Definite Integrals exercise Multiple choice question 4

4
Given:
$\int_{0}^{2 \pi} \sqrt{1+\sin {x} / 2} d x$
Hint:
Using $\int \sin x \; d x, \int \cos x \; d x$
Explanation:
Let
$I=\int_{0}^{2 \pi} \sqrt{1+\sin {x} / 2} d x$
\begin{aligned} &{\left[\sin 2 \theta=2 \sin \theta \cos \theta, \quad \sin ^{2} \theta+\cos ^{2} \theta=1\right]} \\\\ &=\int_{0}^{2 \pi} \sqrt{\left(\sin {x} / 4+\cos {x} / 4\right)^{2}} d x \\\\ &=\int_{0}^{2 \pi}\left(\sin x / 4+\cos {x} / 4\right) d x \end{aligned}
\begin{aligned} &=\int_{0}^{2 \pi} \sin x / 4 d x+\int_{0}^{2 \pi} \cos ^{x} / 4 d x \\\\ &=\left[-4 \cos \frac{x}{4}\right]_{0}^{2 \pi}+\left[4 \sin \frac{x}{4}\right]_{0}^{2 \pi} \end{aligned}
\begin{aligned} &=-4 \cos [x / 4]_{0}^{2 \pi}+\left[4 \sin \frac{x}{4}\right]_{0}^{2 \pi} \\\\ &=\left[-4 \cos \frac{\pi}{2}+4 \cos 0\right]+\left[4 \sin \frac{\pi}{2}-4 \sin 0\right] \end{aligned}
\begin{aligned} &=0+4+0 \\\\ &=4 \end{aligned}

Definite Integrals exercise Multiple choice question 5

$\frac{\pi }{4}$
Given:

$\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x$
Hint:
using $\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x$
Explanation:
Let
$I=\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x \ldots(i)$
$I=\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\cos \left(\frac{\pi}{2}-x\right)}}{\sqrt{\cos \left(\frac{\pi}{2}-x\right)}+\sqrt{\sin \left(\frac{\pi}{2}-x\right)}} d x, \quad\left[\therefore \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]$$=\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x \ldots(i i)$
\begin{aligned} &2 I=\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\cos x}+\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x \\\\ &=\int_{0}^{\frac{\pi}{2}} 1 d x \\\\ &2 I=[x]_{0}^{\frac{\pi}{2}} \\\\ &2 I=\left[\frac{\pi}{2}-0\right] \end{aligned}
\begin{aligned} &2 I=\frac{\pi}{2} \\\\ &I=\frac{\pi}{4} \end{aligned}

Definite Integrals exercise Multiple choice question 6

$\log 2$
Given:

$\int_{0}^{\infty} \frac{1}{1+e^{x}} d x$
Hint:
Explanation:
Let
\begin{aligned} &I=\int_{0}^{\infty} \frac{1}{1+e^{x}} d x \\\\ &=\int_{0}^{\infty} \frac{1}{e^{x}\left(\frac{1}{e^{x}}+1\right)} d x \\\\ &=\int_{0}^{\infty} \frac{e^{-x}}{1+e^{-x}} d x \end{aligned}
Put
\begin{aligned} &1+e^{-x}=t \\\\ &-e^{-x} d x=d t \\\\ &e^{-x} d x=-d t \\\\ &=\int_{0}^{\infty} \frac{-d t}{t} \\\\ &=[-\log |t|]_{0}^{\infty} \end{aligned}
\begin{aligned} &=\left[-\log \left|1+e^{-x}\right|\right]_{0}^{\infty} \\\\ &=\left[-\log \left|1+e^{-\infty}\right|\right]-\left[-\log \left|1+e^{-0}\right|\right] \end{aligned}
\begin{aligned} &=[-\log |1+0|]-[-\log |1+1|],\left[\therefore e^{-\infty}=0, \quad e^{-0}=1\right] \\\\ &=-[0-\log 2] \\\\ &=\log 2 \end{aligned}

Definite Integrals exercise Multiple choice question 7

2
Given:
$\int_{0}^{\frac{\pi^{2}}{4}} \frac{\sin \sqrt{x}}{\sqrt{x}} d x$
Hint:
Using $\int \sin x\; dx$
Explanation:
Let
\begin{aligned} &I=\int_{0}^{\frac{\pi^{2}}{4}} \frac{\sin \sqrt{x}}{\sqrt{x}} d x \\\\ &\text { Put } \sqrt{x}=t \\\\ &\frac{1}{2 \sqrt{x}} d x=d t \\\\ &\frac{1}{\sqrt{x}} d x=2 d t \end{aligned}
\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \sin t \cdot 2 t\; d t \\\\ &=2\left[\frac{-\cos t}{1}\right]_{0}^{\frac{\pi}{2}} \end{aligned}
\begin{aligned} &=2\left[-\cos \frac{\pi}{2}+\cos 0\right] \\\\ &=2[0+1] \\\\ &=2 \end{aligned}

Definite Integrals exercise Multiple choice question 8

$\log \left(\frac{4}{3}\right)$
Hint:
You must know about log properties and using partial fraction.
Given:
$\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{(2+\sin x)+(1+\sin x)} d x$
Explanation:
Let
\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{(2+\sin x)(1+\sin x)} d x \\\\ &I=\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{(1+1+\sin x)(1+\sin x)} d x \end{aligned}
Put
\begin{aligned} &1+\sin x=t \\\\ &\cos x \; d x=d t \end{aligned}
When $x=0$ then $t=1$
When $x=\frac{\pi }{2}$ then $t=2$
\begin{aligned} &=\int_{0}^{2} \frac{d t}{(t+1) t} \\\\ &\frac{1}{(t+1) t}=\frac{A}{t}+\frac{B}{t+1} \end{aligned}
Multiplying by $t(t+1)$
$1=A(t+1)+B(t)$
Putting $t=-1$
$1=A(0)+B(-1)$
$B=-1$
Putting $t=-1$
\begin{aligned} &1=A(1)+B(0) \\\\ &A=1 \end{aligned}
\begin{aligned} &\frac{1}{(t+1) t}=\frac{1}{t}-\frac{1}{t+1} \\\\ &\int_{0}^{2} \frac{d t}{(t+1) t}=\int_{0}^{2} \frac{1}{t}-\frac{1}{t+1} d t \end{aligned}
\begin{aligned} &=\int_{0}^{2} \frac{1}{t} d t-\int_{0}^{2} \frac{1}{t+1} d t \\\\ &=[\log |t|]_{1}^{2}-[\log |t+1|]_{1}^{2} \\\\ &=[\log 2-\log 1]-[\log 3-\log 2] \end{aligned}
\begin{aligned} &=\log 2-0-\log 3+\log 2 \\\\ &=2 \log 2-\log 3 \\\\ &=\log 2^{2}-\log 3 \\\\ &=\log 4-\log 3 \\\\ &=\log \left(\frac{4}{3}\right) \end{aligned}

Definite Integrals exercise Multiple choice question 9

$\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
Given:
$\int_{0}^{\frac{\pi}{2}} \frac{1}{2+\cos x} d x$
Hint:
Using$\int \frac{1}{1+x^{2}} d x$

Explanation:
Let
$I=\int_{0}^{\frac{\pi}{2}} \frac{1}{2+\cos x} d x$
Put
\begin{aligned} &\cos x=\frac{1-\tan ^{2} x / 2}{1+\tan ^{2} x / 2} \\\\ &=\int_{0}^{\frac{\pi}{2}} \frac{1}{2+\frac{1-\tan ^{2} x / 2}{1+\tan ^{2} x / 2}} d x \end{aligned}
$=\int_{0}^{\frac{\pi}{2}} \frac{1+\tan ^{2} x / 2}{2\left(1+\tan ^{2} x / 2\right)+1-\tan ^{2} x / 2} d x$
$=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2 } x/ 2}{3+\tan ^{2} x / 2} d x$
\begin{aligned} &\text { Put } \tan ^{x} /_{2}=t \\\\ &\sec ^{2} x / 2 \cdot 1 / 2 d x=d t \\\\ &\sec ^{2} x /{ }_{2} d x=2 d t \end{aligned}
When $x=0$ then $t= 0$
When $x=\frac{\pi }{2}$ then $t=1$
\begin{aligned} &=\int_{0}^{1} \frac{2 d t}{t^{2}+(\sqrt{3})^{2}} \\\\ &=2\left[\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{t}{\sqrt{3}}\right)\right]_{0}^{1} \\\\ &=\frac{2}{\sqrt{3}}\left[\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)-\tan ^{-1}(0)\right] \end{aligned}
\begin{aligned} &=\frac{2}{\sqrt{3}}\left[\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)-0\right] \\\\ &=\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right) \end{aligned}

Definite Integrals exercise Multiple choice question 10

$\frac{\pi}{2}-1$
Given:
$\int_{0}^{1} \sqrt{\frac{1-x}{1+x}} d x$
Hint:
Explanation:
Let
\begin{aligned} &I=\int_{0}^{1} \sqrt{\frac{1-x}{1+x}} d x \\\\ &=\int_{0}^{1} \sqrt{\frac{1-x}{1+x} \times \frac{1-x}{1-x}} d x \end{aligned}
\begin{aligned} &=\int_{0}^{1} \sqrt{\frac{(1-x)^{2}}{1-x^{2}}} d x \\\\ &=\int_{0}^{1} \frac{1-x}{\sqrt{1-x^{2}}} d x \end{aligned}
\begin{aligned} &=\int_{0}^{1} \frac{1}{\sqrt{1-x^{2}}} d x-\int_{0}^{1} x\left(1-x^{2}\right)^{-1 / 2} d x \\\\ &=\int_{0}^{1} \frac{1}{\sqrt{1-x^{2}}} d x+\int_{0}^{1}\left(1-x^{2}\right)^{-1 / 2}(-2 x) d x \end{aligned}
\begin{aligned} &=\left[\sin ^{-1} x\right]_{0}^{1}+\frac{1}{2}\left[\frac{\left(1-x^{2}\right)^{1 / 2}}{1 / 2}\right]_{0}^{1} \\\\ &=\left[\sin ^{-1}(1)-\sin ^{-1}(0)\right]+\frac{1}{2}\left[0-\frac{1}{1 / 2}\right] \end{aligned}
\begin{aligned} &=\left(\frac{\pi}{2}-0\right)+\frac{1}{2}(-2) \\\\ &=\frac{\pi}{2}-1 \end{aligned}

Definite Integrals exercise Multiple choice question 11

$\frac{\pi}{\sqrt{a^{2}-b^{2}}}$
Given:
$\int_{0}^{\pi} \frac{1}{a+b \cos x} d x$
Hint:
Using$\int \frac{1}{1+x^{2}} d x$
Explanation:

Let
\begin{aligned} &I=\int_{0}^{\pi} \frac{1}{a+b \cos x} d x \\\\ &\cos x=\frac{1-\tan ^{2} x / 2}{1+\tan ^{2} x / 2} \end{aligned}
\begin{aligned} &=\int_{0}^{\pi} \frac{1}{a+b\left(\frac{1-\tan ^{2} x / 2}{1+\tan ^{2} x / 2}\right)} d x \\\\ &=\int_{0}^{\pi} \frac{1+\tan ^{2} x / 2}{2 a+a \tan ^{2} x / 2+b-b \tan ^{2} x / 2} d x \end{aligned}
\begin{aligned} &=\int_{0}^{\pi} \frac{\sec ^{2 } \frac{x}{2}}{(a-b) \tan ^{2} x / 2+(a+b)} d x \\\\ &\text { Put } \tan {x} /_{2}=t \\\\ &\sec ^{2} x / 2 \cdot{ }^{1} /{ }_{2} d x=d t \\\\ &\sec ^{2} x /{2} d x=2 d t \end{aligned}
$=\int_{0}^{\infty} \frac{2 d t}{(a-b) t^{2}+(a+b)}$
$=\frac{2}{(a-b)} \int_{0}^{\infty} \frac{d t}{t^{2}+\left(\sqrt{\frac{a+b}{a-b}}\right)^{2}}$
$=\frac{2}{a-b}\left\{\tan ^{-1}\left[t \sqrt{\frac{a-b}{a+b}}\right]\right\}_{0}^{\infty} \times \frac{1}{\sqrt{\frac{a+b}{a-b}}}$
\begin{aligned} &=\frac{2}{a-b} \frac{\sqrt{a-b}}{\sqrt{a+b}}\left(\tan ^{-1} \infty-\tan ^{-1} 0\right) \\\\ &=\frac{2}{\sqrt{(a+b)(a-b)}} \cdot\left(\frac{\pi}{2}\right) \\\\ &=\frac{\pi}{\sqrt{a^{2}-b^{2}}} \end{aligned}

Definite Integrals exercise Multiple choice question 12

$\frac{\pi }{12}$
Given:
$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\sqrt{\cot x}} d x$
Hint:
Using$\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x$

Explanation:
Let
\begin{aligned} &I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\sqrt{\cot x}} d x \\\\ &I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\sqrt{\frac{\cos x}{\sin x}}} d x, \quad\left[\therefore \cot x=\frac{\cos x}{\sin x}\right] \end{aligned}
$=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x \ldots(i)$
We have a property that
$\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x$
Using this, we get
$=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos \left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}}{\sqrt{\cos \left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}+\sqrt{\sin \left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}} d x$
$=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x,\left[\therefore \sin \left(\frac{\pi}{2}-x\right)=\cos x, \quad \cos \left(\frac{\pi}{2}-x\right)=\sin x\right]$
$I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x \ldots(i i)$
\begin{aligned} &2 I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x \\\\ &=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 d x \end{aligned}
\begin{aligned} &I=\frac{1}{2}[x]_{\frac{\pi}{6}}^{\frac{\pi}{3}} \\\\ &I=\frac{1}{2}\left[\frac{\pi}{3}-\frac{\pi}{6}\right] \end{aligned}
\begin{aligned} &I=\frac{1}{2}\left[\frac{2 \pi-2 \pi}{6}\right] \\\\ &I=\frac{1}{2} \times \frac{\pi}{6} \\\\ &=\frac{\pi}{12} \end{aligned}

Definite Integrals exercise Multiple choice question 13

$\frac{\pi}{60}$
Given:
$\int_{0}^{\infty} \frac{x^{2}}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)\left(x^{2}+c^{2}\right)} d x=\frac{\pi}{2(a+b)(b+c)(c+a)}$
Hint:
Using given condition find a,b,c.

Explanation:
Given that
$\int_{0}^{\infty} \frac{x^{2}}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)\left(x^{2}+c^{2}\right)} d x=\frac{\pi}{2(a+b)(b+c)(c+a)} \ldots(i)$
We have to evaluate
$\int_{0}^{\infty} \frac{d x}{\left(x^{2}+4\right)\left(x^{2}+9\right)}$
Let
$I=\int_{0}^{\infty} \frac{d x}{\left(x^{2}+4\right)\left(x^{2}+9\right)}$
Multiply and divide by $x^{2}$
$I=\int_{0}^{\infty} \frac{x^{2} d x}{\left(x^{2}+4\right)\left(x^{2}+9\right)\left(x^{2}+0\right)}$
On comparing with equ (i)
We get
\begin{aligned} &a^{2}=4 ; b^{2}=9 ; c^{2}=0 \\\\ &a=2 ; b=3 ; c=0 \end{aligned}
To given
$\int_{0}^{\infty} \frac{x^{2}}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)\left(x^{2}+c^{2}\right)} d x=\frac{\pi}{2(a+b)(b+c)(c+a)}$
$\int_{0}^{\infty} \frac{x^{2}}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)\left(x^{2}+c^{2}\right)} d x=\frac{\pi}{2(2+3)(3+0)(0+2)}$
$=\frac{\pi}{2 \times 5 \times 3 \times 2}$
$=\frac{\pi}{60}$

Definite Integrals exercise Multiple choice question 14

1
Given:
$\int_{1}^{\theta} \frac{\log x}{x} d x$
Hint:
Using by parts.
Explanation:
Let
\begin{aligned} I &=\int_{1}^{e} \log x d x \\\\ &=\left[\log x \int 1 d x-\int \frac{d}{d x}(\log x)-\int d x\right]_{1}^{e} \end{aligned}
\begin{aligned} &=\left[x \log x-\int \frac{1}{x} \cdot x d x\right]_{1}^{e} \\\\ &=[x \log x-x]_{1}^{e} \end{aligned}

\begin{aligned} &=[x(\log x-1)]_{1}^{e} \\\\ & \\\\ &=1[0-1]=1 \end{aligned}

Definite Integrals exercise Multiple choice question 15

$\frac{\pi}{12}$
Given:
$\int_{1}^{\sqrt{3}} \frac{1}{1+x^{2}} d x$
Hint:
Use formula $\int \frac{1}{1+x^{2}} d x=\tan ^{-1} x$

Explanation:
\begin{aligned} I &=\int_{1}^{\sqrt{3}} \frac{1}{1+x^{2}} d x \\\\ &=\left[\tan ^{-1} x\right]_{1}^{\sqrt{3}} \\\\ &=\tan ^{-1} \sqrt{3}-\tan ^{-1} 1 \end{aligned}
\begin{aligned} &=\frac{\pi}{3}-\frac{\pi}{4} \\\\ &=\frac{4 \pi-3 \pi}{12} \\\\ &=\frac{\pi}{12} \end{aligned}

Definite Integrals exercise Multiple choice question 16

$\frac{\pi}{12}+\log 2(\sqrt{2})$
Given:
$\int_{0}^{3} \frac{3 x+1}{x^{2}+9} d x$
Hint:
Using$\int \frac{d x}{x} \text { and } \int \frac{1}{1+x^{2}} d x$

Explanation:
Let
\begin{aligned} &I=\int_{0}^{3} \frac{3 x+1}{x^{2}+9} d x \\\\ &=\int_{0}^{3} \frac{3 x}{x^{2}+9} d x+\int_{0}^{3} \frac{1}{x^{2}+9} d x \end{aligned}
\begin{aligned} &= \int_{0}^{3} \frac{3 x}{x^{2}+9} d x+\int_{0}^{3} \frac{1}{x^{2}+9} d x \\\\ &=\left[\frac{3}{2} \log \left|x^{2}+9\right|+\frac{1}{3} \tan ^{-1} \frac{x}{3}\right]_{0}^{3} \end{aligned}
\begin{aligned} &=\frac{3}{2} \log 18+\frac{1}{3} \tan ^{-1} 1-\frac{3}{2} \log 9+\frac{1}{3} \tan ^{-1} 0 \\\\ &=\frac{3}{2}(\log 18-\log 9)+\frac{1}{3} \tan ^{-1}\left(\tan \frac{\pi}{4}\right)+\frac{1}{3} \tan ^{-1}(\tan 0) \end{aligned}
\begin{aligned} &=\frac{3}{2}\left(\log \frac{18}{9}\right)+\frac{1}{3} \times \frac{\pi}{4}+\frac{1}{3} \times 0, \quad\left[\therefore \log m-\log n=\log \frac{m}{n}\right] \\\\ &=\frac{3}{2} \log 2+\frac{\pi}{12}+0 \end{aligned}
\begin{aligned} &=\log (2)^{3 / 2}+\frac{\pi}{12} \\\\ &=\frac{\pi}{12}+\log 2(\sqrt{2}) \end{aligned}

Definite Integrals exercise Multiple choice question 17

$\frac{\pi }{4}$
Given:
$\int_{0}^{\infty} \frac{x}{(1+x)\left(1+x^{2}\right)} d x$
Hint:
You must know derivation of sinx, cosx and tanx.

Explanation:
Let
\begin{aligned} &I=\int_{0}^{\infty} \frac{x}{(1+x)\left(1+x^{2}\right)} d x \\\\ &=\int_{0}^{\frac{\pi}{2}} \frac{\tan \theta}{(1+\tan \theta)\left(1+\tan ^{2} \theta\right)} \cdot \sec ^{2} \theta d \theta \end{aligned}
\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \frac{\tan \theta}{(1+\tan \theta)\left(\sec ^{2} \theta\right)} \cdot \sec ^{2} \theta d \theta \\\\ &=\int_{0}^{\frac{\pi}{2}} \frac{\tan \theta+1-1}{1+\tan \theta} d \theta \end{aligned}
\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} 1 d \theta-\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\tan \theta} d \theta \\\\ &=\int_{0}^{\frac{\pi}{2}} 1 d \theta-\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\frac{\sin \theta}{\cos \theta}} d \theta, \quad\left[\therefore \tan \theta=\frac{\sin \theta}{\cos \theta}\right] \end{aligned}
\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} 1 d \theta-\int_{0}^{\frac{\pi}{2}} \frac{\cos \theta}{\cos \theta+\sin \theta} d \theta \\\\\ &=[\theta]_{0}^{\frac{\pi}{2}}-I_{1}, \quad \text { where } I_{1}=\int_{0}^{\frac{\pi}{2}} \frac{\cos \theta}{\cos \theta+\sin \theta} d \theta \end{aligned}
\begin{aligned} &=\left(\frac{\pi}{2}-0\right)-I_{1} \\\\ &=\frac{\pi}{2}-I_{1} \ldots(i) \end{aligned}
Now,
\begin{aligned} &I_{1}=\int_{0}^{\frac{\pi}{2}} \frac{\cos \theta}{\cos \theta+\sin \theta} d \theta \ldots(i i) \\\\ &=\int_{0}^{\frac{\pi}{2}} \frac{\cos \left(\frac{\pi}{2}-\theta\right)}{\cos \left(\frac{\pi}{2}-\theta\right)+\sin \left(\frac{\pi}{2}-\theta\right)} d \theta \end{aligned}
\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \frac{\sin \theta}{\sin \theta+\cos \theta} d \theta, \quad\left[\therefore \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right] \\\\ &I_{1}=\int_{0}^{\frac{\pi}{2}} \frac{\sin \theta}{\sin \theta+\cos \theta} d \theta \ldots(i i i) \end{aligned}
\begin{aligned} &I_{1}=\int_{0}^{\frac{\pi}{2}} \frac{\cos \theta+\sin \theta}{\sin \theta+\cos \theta} d \theta \\\\ &=\int_{0}^{\frac{\pi}{2}} 1 d \theta \end{aligned}
\begin{aligned} &=\frac{1}{2}[\theta]_{0}^{\frac{\pi}{2}} \\\\ &=\frac{1}{2}\left[\frac{\pi}{2}-0\right] \\\\ &=\frac{\pi}{4} \end{aligned}
Put in (i)
\begin{aligned} &I=\frac{\pi}{2}-\frac{\pi}{4} \\\\ &I=\frac{2 \pi-\pi}{4} \\\\ &I=\frac{\pi}{4} \end{aligned}

Definite Integrals exercise Multiple choice question 18

2
Given:
$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin |x| d x$
Hint:
You must know about $\left | x \right |$ function and $\int \sin x\; dx$

Explanation:
Let
$I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin |x| d x$
We know that
\begin{aligned} &|x|=\left\{\begin{aligned} x, & x \geq 0 \\ -x, & x<0 \end{aligned}\right. \\\\ &I=\int_{-\frac{\pi}{2}}^{0} \sin (-x) d x+\int_{0}^{\frac{\pi}{2}} \sin x \; d x \end{aligned}
\begin{aligned} &I=-\int_{-\frac{\pi}{2}}^{0} \sin x \: d x+\int_{0}^{\frac{\pi}{2}} \sin x \; d x \\\\ &=-[-\cos x]_{-\frac{\pi}{2}}^{0}+[-\cos x]_{0}^{\frac{\pi}{2}} \end{aligned}
\begin{aligned} &=[\cos x]_{-\frac{\pi}{2}}^{0}-[\cos x]_{0}^{\frac{\pi}{2}} \\\\ &=\cos (0)-\cos \left(-\frac{\pi}{2}\right)-\cos \frac{\pi}{2}+\cos (0) \\\\ &=1-0-0+1 \\\\ &=2 \end{aligned}

Definite Integrals exercise Multiple choice question 19

$\frac{\pi}{4}$
Given:
$\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\tan x} d x$
Hint:
using$\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x$

Explanation:
Let
\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\tan x} d x \\\\ &I=\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\frac{\sin x}{\cos x}} d x \end{aligned}
\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\cos x+\sin x} d x \ldots(i) \\\\ &I=\int_{0}^{\frac{\pi}{2}} \frac{\cos \left(\frac{\pi}{2}-2\right)}{\cos \left(\frac{\pi}{2}-2\right)+\sin \left(\frac{\pi}{2}-2\right)} d x \end{aligned}
$I=\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x+\cos x} d x \ldots(i i)$
\begin{aligned} &2 I=\int_{0}^{\frac{\pi}{2}} \frac{\sin x+\cos x}{\sin x+\cos x} d x \\\\ &2 I=\int_{0}^{\frac{\pi}{2}} 1 d x \end{aligned}
\begin{aligned} &I=\frac{1}{2}[x]_{0}^{\frac{\pi}{2}} \\\\ &I=\frac{1}{2}\left[\frac{\pi}{2}-0\right] \\\\ &=\frac{\pi}{4} \end{aligned}

Definite Integrals exercise Multiple choice question 20

$e-1$
Given:
$\int_{0}^{\frac{\pi}{2}} \cos x \cdot e^{\sin x} d x$
Hint:
You must know about $\frac{d}{d x}(\sin x)$ and $\int e^{t} d t$

Explanation:
Let
$I=\int_{0}^{\frac{\pi}{2}} \cos x \cdot e^{\sin x} d x$
Put
\begin{aligned} &\sin x=t \\\\ &\cos x \; d x=d t \end{aligned}
When$x=0$ then $t= 0$
When $x=\frac{\pi }{2}$ then $t=1$
\begin{aligned} &=\int_{0}^{1} e^{t} d x \\\\ &=\left[e^{t}\right]_{0}^{1} \\\\ &=e^{1}-e^{0} \\\\ &=e-1 \end{aligned}

Definite Integrals exercise Multiple choice question 21

$\frac{1}{2}$
Given:
$\int_{0}^{a} \frac{1}{1+4 x^{2}} d x=\frac{\pi}{8}$
Hint:
To solve this equation, we will do 4x in whole system.

Explanation:
Let
\begin{aligned} &I=\int_{0}^{a} \frac{1}{1+4 x^{2}} d x \\\\ &I=\int_{0}^{a} \frac{1}{1+(2 x)^{2}} d x \end{aligned}
\begin{aligned} &=\left[\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1}\right)\right]_{0}^{a} \\\\ &=\frac{\pi}{8} \\\\ &\frac{1}{2}\left[\tan ^{-1}(2 a)-\tan ^{-1}(0)\right]=\frac{\pi}{8} \end{aligned}
\begin{aligned} &\frac{1}{2}\left[\tan ^{-1}(2 a)\right]=\frac{\pi}{8} \\\\ &\tan ^{-1}(2 a)=\frac{\pi}{4} \end{aligned}
\begin{aligned} &2 a=\tan \frac{\pi}{4} \\\\ &2 a=1 \\\\ &a=\frac{1}{2} \end{aligned}

Definite Integrals exercise Multiple choice question 22

0
Given:
$\int_{0}^{1} f(x) d x=1 ; \int_{0}^{1} x f(x) d x=a ; \int_{0}^{1} 2 x^{2} f(x) d x=a^{2}$
Hint:
To solve this equation, we have to integrate differentially.

Explanation:
\begin{aligned} &\int_{0}^{1} f(x) d x=1 \ldots(i) \\\\ &\int_{0}^{1} x f(x) d x=a \ldots(i i) \\\\ &\int_{0}^{1} 2 x^{2} f(x) d x=a^{2} \ldots(i i i) \end{aligned}
\begin{aligned} &I=\int_{0}^{1}(a-x)^{2} f(x) d x \\\\ &I=\int_{0}^{1}\left(a^{2}+x^{2}-2 a x\right) f(x) d x \end{aligned}
\begin{aligned} &I=\int_{0}^{1} a^{2} f(x) d x+\int_{0}^{1} x^{2} f(x) d x-\int_{0}^{1} 2 a x f(x) d x \\\\ &I=a^{2} \int_{0}^{1} f(x) d x+\int_{0}^{1} x^{2} f(x) d x-2 a \int_{0}^{1} x f(x) d x \end{aligned}
\begin{aligned} &I=a^{2}+a^{2}-2(a)(a) \\\\ &I=2 a^{2}-2 a^{2} \\\\ &I=0 \end{aligned}

Definite Integrals exercise Multiple choice question 23

0
Given:
$\int_{-\pi}^{\pi} \sin ^{3} x \cos ^{2} x \; d x$
Hint:
This equation is solve by $\int_{-a}^{a} f(x) d x$

Explanation:
\begin{aligned} &\int_{-a}^{a} f(x) d x \\\\ &f(-x)=\sin ^{3}(-x) \cdot \cos ^{2}(-x) \end{aligned}
\begin{aligned} &=-(\sin x)^{3} \cdot \cos ^{2} x \\\\ &=-\sin ^{3} x \cdot \cos ^{2} x \\\\ &=-f(x) \end{aligned}
f(x) is odd function.
$\int_{-\pi}^{\pi} \sin ^{3} x \cos ^{2} x \; d x=0$

Definite Integrals exercise Multiple choice question 24

$I=\log \sqrt{3}$
Given:
$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{\sin 2 x} d x$
Hint:
To solve this equation we convert sin into cosec
Explanation:
Let
\begin{aligned} &I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{\sin 2 x} d x \\\\ &=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \operatorname{cosec} 2 x \; d x \end{aligned}
\begin{aligned} &I=-\frac{1}{2}[\log |\operatorname{cosec} 2 x+\cot 2 x|]_{\frac{\pi}{6}}^{\frac{\pi}{3}} \\\\ &I=-\frac{1}{2}\left[\log \left(\operatorname{cosec} \frac{2 \pi}{3}+\cot \frac{2 \pi}{3}\right)\right]-\left[\log \left(\operatorname{cosec} \frac{\pi}{3}+\cot \frac{\pi}{3}\right)\right] \end{aligned}
\begin{aligned} &I=-\frac{1}{2}\left[\log \left|\frac{2}{\sqrt{3}}-\frac{1}{\sqrt{3}}\right|\right]-\left[\log \left|\frac{2}{\sqrt{3}}+\frac{1}{\sqrt{3}}\right|\right] \\\\ &I=-\frac{1}{2}\left[\log \frac{1}{\sqrt{3}}-\log \sqrt{3}\right] \end{aligned}
\begin{aligned} &I=-\frac{1}{2}[-\log \sqrt{3}-\log \sqrt{3}] \\\\ &I=-\frac{1}{2} \times-2 \log \sqrt{3} \\\\ &I=\log \sqrt{3} \end{aligned}

Definite Integrals exercise Multiple choice question 25

2
Given:
$\int_{-1}^{1}|1-x| d x$
Hint:
To solve this equation we use $\int \frac{x^{n}}{n+1}$ formula.

Explanation:
Let
\begin{aligned} &\int_{-1}^{1}|1-x| d x \\\\ &=\left[x-\frac{x^{2}}{2}\right]_{-1}^{1} \end{aligned}
\begin{aligned} &=\left[1-\frac{1}{2}\right]-\left[-1-\frac{1}{2}\right] \\\\ &=1-\frac{1}{2}+1+\frac{1}{2} \\\\ &=2 \end{aligned}

Definite Integrals exercise Multiple choice question 26

$(\log x)^{-1}(x-1) x$
Given:
$\int_{x^{2}}^{x^{3}} \frac{1}{\log _{e} t} d x$
Hint:
This equation will solve by$f'(x)$formula.

Solution:
\begin{aligned} &f^{\prime}(x)=\frac{1}{\log x^{3}} \cdot 3 x^{2}-\frac{1}{\log x^{2}} \cdot(2 x) \\\\ &f^{\prime}(x)=\frac{3 x^{2}}{3 \log x^{3}}-\frac{2 x}{2 \log x^{2}} \end{aligned}
\begin{aligned} f^{\prime}(x) &=\frac{x^{2}}{\log x}-\frac{x}{\log x} \\\\ &=\frac{x^{2}-x}{\log x} \end{aligned}
\begin{aligned} &=\frac{x(x-1)}{\log x} \\\\ &=(\log x)^{-1}(x-1) x \end{aligned}

Definite Integrals exercise Multiple choice question 27

$10\left(\frac{\pi}{2}\right)^{9}$
Hint:
To solve this equation we convert sin into cos.
Given:
$I_{10}=\int_{0}^{\frac{\pi}{2}} x^{10} \sin x \; d x$
Solution:
\begin{aligned} &I_{10}=\int_{0}^{\frac{\pi}{2}} x^{10} \sin x\; d x \\\\ &=-x^{10} \cdot \cos x-\int_{0}^{\frac{\pi}{2}} 10 x^{9} \cdot \cos x \; d x \end{aligned}
\begin{aligned} &=0-10 \int_{0}^{\frac{\pi}{2}} x^{9} \cdot \cos x \; d x \\\\ &=10\left[x^{9} \sin x-9 \int_{0}^{\frac{\pi}{2}} x^{8} \sin x \; d x\right]_{0}^{\frac{\pi}{2}} \end{aligned}
\begin{aligned} &=10\left(\frac{\pi}{2}\right)^{9}-90 \int_{0}^{\frac{\pi}{2}} x^{8} \sin x \; d x \\\\ &=10\left(\frac{\pi}{2}\right)^{9}-90 I_{8} \\\\ &=10\left(\frac{\pi}{2}\right)^{9} \end{aligned}

Definite Integrals exercise Multiple choice question 28

$\frac{1}{2756}$
Hint:
Integrate then use limits
Given:
$\int_{0}^{1} \frac{x}{(1-x)^{54}} d x$
Solution:
\begin{aligned} &I=\int_{0}^{1} \frac{x}{(1-x)^{54}} d x \\\\ &I=\int_{0}^{1} \frac{x}{(x-1)^{54}} \quad\left[\mathrm{Q}(x-y)^{2}=(y-x)^{2}\right] \end{aligned}
\begin{aligned} &=\int_{0}^{1} \frac{x+1-1}{(x-1)^{54}} d x \\\\ &=\int_{0}^{1} \frac{(x-1)}{(x-1)^{54}} d x+\int_{0}^{1} \frac{1}{(x-1)^{54}} d x \end{aligned}
\begin{aligned} &=\int_{0}^{1}(x-1)^{-53} d x+\int_{0}^{1}(x-1)^{-54} d x \\\\ &=\left[-\frac{(x-1)^{-52}}{52}\right]_{0}^{1}+\left[-\frac{(x-1)^{-53}}{53}\right]_{0}^{1} \end{aligned}
\begin{aligned} &=\left[\frac{(x-1)^{-52}}{52}\right]_{0}^{1}+\left[\frac{(x-1)^{-53}}{53}\right]_{0}^{1} \\\\ &=\frac{(-1)^{52}}{52}+\frac{(-1)^{53}}{53} \end{aligned}
\begin{aligned} &=\frac{1}{52}-\frac{1}{53}=\frac{1}{52 \times 53} \\\\ &=\frac{1}{2756} \end{aligned}

Definite Integrals exercise Multiple choice question 29

$2(\sqrt{2}-1)$
Hint:
First integrate and then use limits
Given:
$\int_{0}^{\frac{\pi}{2}} \sqrt{1-\sin 2 x} \; d x$
Solution:
\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \sqrt{1-\sin 2 x} d x \\\\ &I=\int_{0}^{\frac{\pi}{2}} \sqrt{|\sin x-\cos x|^{2}} d x \end{aligned}
\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}}|\sin x-\cos x| d x \\\\ &=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(\sin x-\cos x)dx-\int_{0}^{\frac{\pi}{4}}(\sin x-\cos x)dx \end{aligned}
\begin{aligned} &I=[-\cos x-\sin x]_{\frac{\pi}{4}}^{\frac{\pi}{2}}-|-\cos x-\sin x|_{0}^{\frac{\pi}{4}} \\\\ &I=-0-1+\sqrt{2}-(-\sqrt{2}+1)=2 \sqrt{2}-2 \\\\ &I=2(\sqrt{2}-1) \end{aligned}

Definite Integrals exercise Multiple choice question 30

4
Hint:
To solve this we will split $1-x^{2}$ in differential form.
Given:
$\int_{-2}^{2}\left|1-x^{2}\right| d x$
Solution:
\begin{aligned} &\int_{-2}^{2}\left|1-x^{2}\right| d x \\\\ &1-x^{2}=0 \\\\ &x^{2}=1 \\\\ &x=\pm 1 \end{aligned}
\begin{aligned} &=\int_{-2}^{-1}\left(1-x^{2}\right) d x+\int_{-1}^{1}\left(1-x^{2}\right) d x+\int_{1}^{2}\left(1-x^{2}\right) d x \\\\ &=\left[\frac{x^{3}}{3}-x\right]_{-2}^{-1}+\left[x-\frac{x^{3}}{3}\right]_{-1}^{1}+\left[\frac{x^{3}}{3}-x\right]_{1}^{2} \end{aligned}
\begin{aligned} &=\left[-\frac{1}{3}+1-\left(-\frac{8}{3}\right)+2\right]+\left[1-\frac{1}{3}-(-1)+\frac{1}{3}\right]+\left[\frac{8}{3}-2-\left(\frac{1}{3}-1\right)\right] \\\\ &=\frac{2}{3}-\left(-\frac{2}{3}\right)+\left[\frac{2}{3}-\left(-\frac{2}{3}\right)\right]+\left[\frac{2}{3}+\frac{2}{3}\right] \end{aligned}
\begin{aligned} &=\frac{4}{3}+\frac{4}{3}+\frac{4}{3}+\frac{12}{3} \\\\ &=4 \end{aligned}

Definite Integrals exercise Multiple choice question 31

$\frac{\pi}{4}$
Hint:
To solve this equation we should simplify cot x.
Given:
$\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cot x} d x$
Solution:
Let
$\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cot x} d x$
$I=\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\frac{\cos x}{\sin x}} d x, \quad\left[\therefore \cot x=\frac{\cos x}{\sin x}\right]$
$=\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\cos x+\sin x} d x \ldots(i), \quad\left[\therefore \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]$
$=\int_{0}^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{2}-x\right)}{\cos \left(\frac{\pi}{2}-x\right)+\sin \left(\frac{\pi}{2}-x\right)} d x$
$=\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\cos x+\sin x} d x \ldots(i i)$
\begin{aligned} &2 I=\int_{0}^{\frac{\pi}{2}} \frac{\cos x+\sin x}{\cos x+\sin x} d x \\\\ &2 I=\int_{0}^{\frac{\pi}{2}} 1 d x \end{aligned}
\begin{aligned} &I=\frac{1}{2}[x]_{0}^{\frac{\pi}{2}} \\\\ &I=\frac{1}{2}\left[\frac{\pi}{2}-0\right] \\\\ &=\frac{\pi}{4} \end{aligned}

Definite Integrals exercise Multiple choice question 32

$\frac{\pi }{4}$
Hint:
To solve this equation we use $\int_{b}^{a} f(x) d x$ formula.
Given:
$\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x+\cos x} d x$
Solution:
Let
\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x+\cos x} d x \ldots(i) \\\\ &\int_{0}^{\frac{\pi}{2}} f(x) d x=\int_{0}^{\frac{\pi}{2}} f\left(\frac{\pi}{2}-x\right) d x \end{aligned}
\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{2}-x\right)}{\sin \left(\frac{\pi}{2}-x\right)+\cos \left(\frac{\pi}{2}-x\right)} d x \\\\ &=\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\cos x+\sin x} d x \ldots(i i) \end{aligned}
\begin{aligned} &2 I=\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x+\cos x} d x+\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\cos x+\sin x} d x \\\\ &=\int_{0}^{\frac{\pi}{2}} \frac{1}{\sin x+\cos x}[\cos x+\sin x] d x \end{aligned}
\begin{aligned} &2 I=\int_{0}^{\frac{\pi}{2}} 1 d x \\\\ &2 I=[x]_{0}^{\frac{\pi}{2}} \end{aligned}
\begin{aligned} &2 I=\left[\frac{\pi}{2}-0\right] \\\\ &=\frac{\pi}{2} \times \frac{1}{2} \\\\ &I=\frac{\pi}{4} \end{aligned}

Definite Integrals exercise Multiple choice question 33

$\frac{\pi }{2}$
Hint:
To solve this equation we use $\int_{a}^{b} \frac{d}{d x} f(x) d x$ formula.
Given:
$\int_{0}^{1} \frac{d}{d x}\left\{\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\right\} d x$
Solution:
Let
\begin{aligned} I &=\int_{0}^{1} \frac{d}{d x}\left\{\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\right\} d x \\\\ &=\int_{0}^{1} \frac{d}{d x}\left(2 \tan ^{-1} x\right) d x \end{aligned}
\begin{aligned} &=2 \int_{0}^{1} \frac{d}{d x}\left(\tan ^{-1} x\right) d x \\\\ &=2 \int_{0}^{1} \frac{1}{1+x^{2}} d x \end{aligned}
\begin{aligned} &=2\left[\tan ^{-1} x\right]_{0}^{1} \\\\ &=2\left[\tan ^{-1} 1-\tan ^{-1} 0\right] \\\\ &=2\left[\frac{\pi}{4}\right]-0 \\\\ &=\frac{\pi}{2} \end{aligned}

Definite Integrals exercise Multiple choice question 34

1
Hint:
To solve this equation we suppose sin x as cos x.
Given:
$\int_{0}^{\frac{\pi}{2}} x \sin x \; d x$
Solution:
Let
$I=\int_{0}^{\frac{\pi}{2}} x \sin x \; d x$
\begin{aligned} &=\left[x \int \sin x \; d x+\int \cos x \; d x\right]_{0}^{\pi / 2} \\\\ &=[-x \cos x+\sin x]_{0}^{\pi / 2} \end{aligned}
\begin{aligned} &=-\frac{\pi}{2} \cos \frac{\pi}{2}+\sin \frac{\pi}{2}-0 \\\\ &=-\frac{\pi}{2} \cos \frac{\pi}{2}+1 \\\\ &=0+1=1 \end{aligned}

Definite Integrals exercise Multiple choice question 35

0
Hint:
To solve this equation we use $\int_{a}^{b} f(x) d x$ formula.
Given:
$\int_{0}^{\frac{\pi}{2}} \sin 2 x \log \tan x \; d x$
Solution:
Let
\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \sin 2 x \log \tan x \; d x \ldots(i) \\\\ &\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x \end{aligned}
\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \sin 2\left(\frac{\pi}{2}-x\right) \cdot \log \left[\tan \left(\frac{\pi}{2}-x\right)\right] d x \\\\ &=\int_{0}^{\frac{\pi}{2}} \sin (\pi-2 x) \cdot \log \cot x \; d x \end{aligned}
$=\int_{0}^{\frac{\pi}{2}} \sin 2 x \cdot \log \cot x\; d x \ldots(i i)$
$2 I=\int_{0}^{\frac{\pi}{2}} \sin 2 x \cdot \log \tan x+\sin 2 x \cdot \log \cot x \; d x$
\begin{aligned} &2 I=\int_{0}^{\frac{\pi}{2}} \sin 2 x(\log \tan x+\log \cot x) d x \\\\ &2 I=\int_{0}^{\frac{\pi}{2}} \sin 2 x(\log \tan x \cdot \cot x) d x \end{aligned}
\begin{aligned} &\log a \cdot b=\log a+\log b \\\\ &2 I=\int_{0}^{\frac{\pi}{2}} \sin 2 x\left(\log \tan x \cdot \frac{1}{\tan x}\right) d x \end{aligned}
\begin{aligned} &2 I=\int_{0}^{\frac{\pi}{2}} \sin 2 x(\log 1) d x \\\\ &2 I=0 \\\\ &I=0 \end{aligned}

Definite Integrals exercise Multiple choice question 36

$\frac{\pi }{4}$
Hint:
To solve this equation we convert cos in term of tan.
Given:
$\int_{0}^{\pi} \frac{1}{5+3 \cos x} d x$
Solution:
Let
\begin{aligned} &I=\int_{0}^{\pi} \frac{1}{5+3 \cos x} d x \\\\ &\therefore \cos 2 \theta=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta} \end{aligned}
$\cos \theta=\frac{1-\tan ^{2} \theta / 2}{1+\tan ^{2} \theta / 2}$
$I=\int_{0}^{\pi} \frac{1}{5+3\left(\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)} d x$
\begin{aligned} &I=\int_{0}^{\pi} \frac{\sec ^{2} \frac{x}{2}}{8+2 \tan ^{2} \frac{x}{2}} d x \\\\ &\text { Let } \tan \frac{x}{2}=t, \sec ^{2} \frac{x}{2} \cdot \frac{1}{2} d x=d t \end{aligned}
\begin{aligned} &=\frac{2}{2} \int_{0}^{\infty} \frac{d t}{4+t^{2}} \\\\ &=\int_{0}^{\infty} \frac{1}{(2)^{2}+t^{2}} d t \end{aligned}
\begin{aligned} &=\left[\frac{1}{2} \tan ^{-1}\left(\frac{t}{2}\right)\right]_{0}^{\infty} \\\\ &=\frac{1}{2}\left(\frac{\pi}{2}\right)=\frac{\pi}{4} \end{aligned}

Definite Integrals exercise Multiple choice question 37

$\pi \log 2$
Hint:
To solve this equation we suppose x as tan x.
Given:
$\int_{0}^{\infty} \log \left(x+\frac{1}{x}\right)\left(\frac{1}{1+x^{2}}\right) d x$
Solution:
Let
\begin{aligned} &I=\int_{0}^{\infty} \log \left(x+\frac{1}{x}\right)\left(\frac{1}{1+x^{2}}\right) d x \\\\ &x=\tan x \end{aligned}
\begin{aligned} &d x=\sec ^{2} \theta d \theta \\\\ &\text { When } x=0, \tan \theta=0, \theta=0 \\\\ &\text { When } x=\infty, \tan \theta=\infty, \theta=\frac{\pi}{2} \end{aligned}
\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \log (\tan \theta+\cot \theta) \cdot \frac{\sec ^{2} \theta d \theta}{1+\tan ^{2} \theta} \\\\ &=\int_{0}^{\frac{\pi}{2}} \log \left(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\right) d \theta \end{aligned}
\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cos \theta} d \theta \\\\ &=\int_{0}^{\frac{\pi}{2}} \log 1-\log (\sin \theta \cos \theta) d \theta \end{aligned}
\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \log \sin \theta d \theta-\int_{0}^{\frac{\pi}{2}} \log \cos \theta d \theta \\\\ &=-\left(\frac{\pi}{2} \log 2\right)-\left(-\frac{\pi}{2} \log 2\right) \end{aligned}
$I=0$

Definite Integrals exercise Multiple choice question 38

$\int_{a}^{2 a} f(x) d x=\int_{0}^{a} f(x) d x+\int_{0}^{2 a}f(2 a-x) d x$
Hint:
Given:
$\int_{a}^{2 a} f(x) d x$
Solution:
$\int_{a}^{2 a} f(x) d x$
\begin{aligned} &f(2 a-x)=f(x) \\\\ &\int_{0}^{a} 2 f(x) d x \end{aligned}
\begin{aligned} &=2 \int_{0}^{a} f(x) d x \\\\ &\int_{a}^{2 a} f(x) d x=\int_{0}^{a} f(x) d x+\int_{0}^{2 a}f(2 a-x) d x \end{aligned}

Definite Integrals exercise Multiple choice question 39

$I=\frac{a+b}{2} \int_{a}^{b} f(x) d x$
Hint:
To solve this we will split (a+b-x)
Given:
$\int(a+b-x)=f(x)$
Solution:
$\int(a+b-x)=f(x)$
Let
\begin{aligned} &I=\int_{a}^{b} x \cdot f(x) d x \ldots(i) \\\\ &\int_{a}^{b} f(x) d x=\int_{a}^{b}(a+b-x) d x \end{aligned}
\begin{aligned} &I=\int_{a}^{b}(a+b-x) \cdot f(a+b-x) d x \\\\ &I=\int_{a}^{b}(a+b-x) \cdot f(x) d x \end{aligned}
$I=\int_{a}^{b}(a+b) \cdot f(x) d x-\int_{a}^{b} x f(x) d x \ldots(i i)$
$I+I=\int_{a}^{b} x f(x) d x+\int_{a}^{b}(a+b) \cdot f(x) d x-\int_{a}^{b} x f(x) d x$
\begin{aligned} &2 I=\int_{a}^{b}(a+b) f(x) d x \\\\ &I=\frac{1}{2} \int_{a}^{b}(a+b) f(x) d x \\\\ &I=\frac{a+b}{2} \int_{a}^{b}(a+b) f(x) d x \end{aligned}

Definite Integrals exercise Multiple choice question 40

0
Hint:
Given:
$\int_{0}^{1} \tan ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}\right) d x$
Explanation:
\begin{aligned} I &=\int_{0}^{1} \tan ^{-1}\left(\frac{2 x-1}{1-x-x^{2}}\right) d x \\\\ &=\int_{0}^{1} \tan ^{-1}\left[\frac{x-(1-x)}{1+x(1-x)}\right] d x \end{aligned}
$=\int_{0}^{1} \tan ^{-1} x-\tan ^{-1}(1-x) d x$ .........Eq(i)
Using properties,
\begin{aligned} &\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x \\\\ &I=\int_{0}^{1} \tan ^{-1}(1-x)-\tan ^{-1}(1-(1-x)) d x \end{aligned}
$=\int_{0}^{1} \tan ^{-1}(1-x)-\tan ^{-1} x \; d x$ ..........Eq(ii)
\begin{aligned} &2 I=0 \\\\ &I=0 \end{aligned}

Definite Integrals exercise Multiple choice question 41

$I=0$
Hint:
To solve this equation we use $\int_{0}^{a} f(x) d x$ formula.
Given:
$\int_{0}^{\frac{\pi}{2}} \log \left[\frac{4+3 \sin x}{4+3 \cos x}\right] d x$
Solution:
Let
\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \log \left[\frac{4+3 \sin x}{4+3 \cos x}\right] d x \ldots(i) \\\\ &\int_{0}^{a} f(x) d x=\int_{0}^{a}(a-x) d x \end{aligned}
\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \log \left[\frac{4+3 \sin \left(\frac{\pi}{2}-x\right)}{4+3 \cos \left(\frac{\pi}{2}-x\right)}\right] d x \\\\ &I=\int_{0}^{\frac{\pi}{2}} \log \left[\frac{4+3 \cos x}{4+3 \sin x}\right] d x \ldots(i i) \end{aligned}
\begin{aligned} &2 I=\int_{0}^{\frac{\pi}{2}}\left(\log \left[\frac{4+3 \cos x}{4+3 \sin x}\right]+\log \left[\frac{4+3 \sin x}{4+3 \cos x}\right]\right) d x \\\\ &2 I=\int_{0}^{\frac{\pi}{2}}\left(\log \left[\frac{4+3 \cos x}{4+3 \sin x}\right] \cdot \log \left[\frac{4+3 \sin x}{4+3 \cos x}\right]\right) d x \end{aligned}
\begin{aligned} &2 I=\int_{0}^{\frac{\pi}{2}} 1 d x \\\\ &2 I=0 \\\\ &I=0 \end{aligned}

Definite Integrals exercise Multiple choice question 42

$\pi$
Hint:
We use $\int_{-a}^{a} f(x) d x=0$ function.
Given:
$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(x^{3}+x \cos x+\tan ^{5} x+1\right) d x$
Explanation:
\begin{aligned} &\int_{-a}^{a} f(x) d x=0\\\\ &f(x) \text { odd function }\\\\ &f(-x)=-f(x)\\\\ &(-x)^{3}=-x^{3} \ldots(i) \end{aligned}
\begin{aligned} &x \cos x \\\\ &(-x) \cos (-x)=-x \cos x \ldots(i i) \\\\ &\tan ^{5} x \\\\ &\tan ^{5}(-x)=-\tan ^{5} x \ldots(i i i) \end{aligned}
\begin{aligned} &=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^{3} d x+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x \cos x d x+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \tan ^{5} x d x+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 d x \\\\ &=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 d x \end{aligned}
\begin{aligned} &=[x]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \\\\ &=\frac{\pi}{2}-\left(-\frac{\pi}{2}\right) \\\\ &=\frac{2 \pi}{2} \\\\ &=\pi \end{aligned}

Definite Integrals exercise Multiple choice question 43

1
Hint:
To solve this equation.
Given:
$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{1+\cos 2 x} d x$
Explanation:
\begin{aligned} &\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{1+\cos 2 x} d x \\\\ &=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2 \cos ^{2} x} d x \\\\ &1+\cos 2 x=2 \cos ^{2} x \end{aligned}
\begin{aligned} &=\frac{1}{2} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{\cos ^{2} x} d x \\\\ &=\frac{1}{2} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec ^{2} x d x \\\\ &\frac{1}{\cos x}=\sec x \end{aligned}
\begin{aligned} &=\frac{1}{2}[\tan x]_{\frac{\pi}{4}}^{\frac{\pi}{4}} \\\\ &=\frac{1}{2}\left[\tan \frac{\pi}{4}-\tan \left(-\frac{\pi}{4}\right)\right] \\\\ &=\frac{1}{2}[1-(-1)] \end{aligned}
\begin{aligned} &=\frac{1}{2}[1+1] \\\\ &=\frac{2}{2} \\\\ &=1 \end{aligned}

Definite Integrals exercise Multiple choice question 44

$\int_{a}^{b} f(x+c) d x$
Hint:
To solve this $\int_{a}^{b} f(x) d x$ formula is used.
Given:
$\int_{a+c}^{b+c} f(x) d x$
Explanation:
$\int_{a+c}^{b+c} f(x) d x$
\begin{aligned} &=f[x]_{a+c}^{b+c} \\ \\ &=f(b+c)-f(a+c) \end{aligned}
Now, solving option (b),
\begin{aligned} &\int_{a}^{b} f(x+c) d x \\\\ &=[f(x+c)]_{a}^{b} \\\\ &=f(b+c)-f(a+c) \end{aligned}
Hence, option (b) is correct.

Definite Integrals exercise Multiple choice question 45

$\frac{a}{2} \int_{0}^{a} f(x) d x$
Hint:
To solve this equation we use $\int_{0}^{a} f(x) d x$ formula.
Given:
$f(x)=f(a-x), \quad g(x)+g(a-x)=a$
Explanation:
\begin{aligned} &I=\int_{0}^{a} g(x) f(x) d x \ldots(i) \\\\ &\text { Put } x=(a-x) \\\\ &=\int_{0}^{a} g(a-x) f(a-x) d x \ldots(i i) \end{aligned}
\begin{aligned} &=\int_{0}^{a} f(x)(a-g(x)) d x \\\\ &I=a \int_{0}^{a} f(x) d x-\int_{0}^{a} f(x) \cdot g(x) d x \end{aligned}
$\begin{gathered} I=a \int_{0}^{a} f(x) d x-I \\\\ 2 I=a \int_{0}^{a} f(x) d x \\\\ I=\frac{a}{2} \int_{0}^{a} f(x) d x \end{gathered}$

Definite Integrals exercise Multiple choice question 46

1
Hint:
To solve this question we have to use substitution method.
Given:
$\int_{0}^{1} \tan \left(\sin ^{-1} x\right) d x$
Explanation:
\begin{aligned} &\sin ^{-1} x=t \\\\ &x=\sin t \\\\ &d x=\cos t d t \\\\ &\text { When } x=0, \sin ^{-1} 0=0 \\\\ &\text { When } x=1, \sin ^{-1} 1=\frac{\pi}{2} \end{aligned}
\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \tan t \cdot \cos t \; d t \\\\ &\frac{\sin t}{\cos t}=\tan t \\\\ &\sin t=\cos t \cdot \cos t \end{aligned}
\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \sin t d t \\\\ &I=[-\cos t]_{0}^{\frac{\pi}{2}} \\\\ &I=0-(-1) \\\\ &=1 \end{aligned}

The RD Sharma class 12 solution of Definite integrals exercise MCQ covers the chapter 'Definite Integrals.' There are about 46 MCQs in this exercise which gives a brief insight of the chapter. The RD Sharma class 12th exercise MCQ covers all the essential concepts like,

• Limit sum of definite integrals

• Fundamentals of a theorem in definite integral

• Basic properties

• Evaluation method of integrals

The beneficial factors of using the RD Sharma class 12 solution chapter 19 exercise MCQ are :-

• RD Sharma class 12th exercise MCQ is available for download on the careers360 website, you can visit the site for free PDFs and study materials.

• Any student who has previously used the RD Sharma class 12 chapter 19 exercise MCQ solution for practicing will always raise the study material of RD Sharma as it has always helped them to score better and to understand maths easily.

• The RD Sharma class 12th exercise MCQ contains questions that are prepared by maths experts with some helpful tips to solve the questions easily.

• Any student can refer to the RD Sharma class 12th exercise MCQ solution for solving questions that they find trouble in as the solution provides solved question examples for the betterment of the students.

• Every student should make use of the RD Sharma class 12th exercise MCQ to stay ahead of the class as most of the Teacher's take reference of the solution to provide lectures and prepare question papers.

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RD Sharma Chapter-wise Solutions

1. Mention some definite rephrasing rules?

The integral of a sum is said to be  the sum of the integrals whereas the integral of a difference is said to be the difference of the integrals.Similarly, we can say that  the integral of the product of a constant and a function is equal to the constant multiplied by the integral of the function.

2. What is a definite integral used for?

definite integrals can be used to determine the mass of an object if its density function is known

3. Can a definite integral be negative?

expressed more compactly,  The Definite integral is the sum of the areas above minus the sum of the areas below, therefore area is always non negative.

4. Is Definite integral important for exams?

Yes,  The Definite integral chapter is important for exams as most of the questions asked for higher marks are from these chapters

5. Can the RD Sharma solutions be used for self practice?

Yes, you can definitely use the RD Sharma solutions for self practice and then you can evaluate the performance according to the  answers given in the book.

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