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.

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

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.

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

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.

RD Sharma Solutions Class 12 Mathematics Chapter 19 MCQ

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RD Sharma Solutions Class 12 Mathematics Chapter 19 MCQ

RD Sharma Solutions Class 12 Mathematics Chapter 19 MCQ

Updated on 24 Jan 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

RD Sharma Class 12 Solutions Chapter19 MCQ Definite Integrals - Other Exercise

Definite Integrals Excercise: MCQ

Definite Integrals exercise Multiple choice question 1

Answer:
$\frac{\pi }{8}$
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

Answer:
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

Answer:
$\frac{\pi^{2}}{4}$
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

Answer:
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

Answer:
$\frac{\pi }{4}$
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)$
Adding (i) and (ii)
$\begin{aligned} &2 I=\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\cos x}+\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x \\\\ &=\int_{0}^{\frac{\pi}{2}} 1 d x \\\\ &2 I=[x]_{0}^{\frac{\pi}{2}} \\\\ &2 I=\left[\frac{\pi}{2}-0\right] \end{aligned}$
$\begin{aligned} &2 I=\frac{\pi}{2} \\\\ &I=\frac{\pi}{4} \end{aligned}$

Definite Integrals exercise Multiple choice question 6

Answer:
$\log 2$
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

Answer:
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

Answer:
$\log \left(\frac{4}{3}\right)$
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

Answer:
$\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
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

Answer:
$\frac{\pi}{2}-1$
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

Answer:
$\frac{\pi}{\sqrt{a^{2}-b^{2}}}$
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

Answer:
$\frac{\pi }{12}$
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
$=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos \left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}}{\sqrt{\cos \left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}+\sqrt{\sin \left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}} d x$
$=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x,\left[\therefore \sin \left(\frac{\pi}{2}-x\right)=\cos x, \quad \cos \left(\frac{\pi}{2}-x\right)=\sin x\right]$
$I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x \ldots(i i)$
Adding (i) and (ii)
$\begin{aligned} &2 I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x \\\\ &=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 d x \end{aligned}$
$\begin{aligned} &I=\frac{1}{2}[x]_{\frac{\pi}{6}}^{\frac{\pi}{3}} \\\\ &I=\frac{1}{2}\left[\frac{\pi}{3}-\frac{\pi}{6}\right] \end{aligned}$
$\begin{aligned} &I=\frac{1}{2}\left[\frac{2 \pi-2 \pi}{6}\right] \\\\ &I=\frac{1}{2} \times \frac{\pi}{6} \\\\ &=\frac{\pi}{12} \end{aligned}$

Definite Integrals exercise Multiple choice question 13

Answer:
$\frac{\pi}{60}$
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

Answer:
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

Answer:
$\frac{\pi}{12}$
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

Answer:
$\frac{\pi}{12}+\log 2(\sqrt{2})$
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

Answer:
$\frac{\pi }{4}$
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,
$\begin{aligned} &I_{1}=\int_{0}^{\frac{\pi}{2}} \frac{\cos \theta}{\cos \theta+\sin \theta} d \theta \ldots(i i) \\\\ &=\int_{0}^{\frac{\pi}{2}} \frac{\cos \left(\frac{\pi}{2}-\theta\right)}{\cos \left(\frac{\pi}{2}-\theta\right)+\sin \left(\frac{\pi}{2}-\theta\right)} d \theta \end{aligned}$
$\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \frac{\sin \theta}{\sin \theta+\cos \theta} d \theta, \quad\left[\therefore \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right] \\\\ &I_{1}=\int_{0}^{\frac{\pi}{2}} \frac{\sin \theta}{\sin \theta+\cos \theta} d \theta \ldots(i i i) \end{aligned}$
Adding (ii) and (iii)
$\begin{aligned} &I_{1}=\int_{0}^{\frac{\pi}{2}} \frac{\cos \theta+\sin \theta}{\sin \theta+\cos \theta} d \theta \\\\ &=\int_{0}^{\frac{\pi}{2}} 1 d \theta \end{aligned}$
$\begin{aligned} &=\frac{1}{2}[\theta]_{0}^{\frac{\pi}{2}} \\\\ &=\frac{1}{2}\left[\frac{\pi}{2}-0\right] \\\\ &=\frac{\pi}{4} \end{aligned}$
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

Answer:
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

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

Definite Integrals exercise Multiple choice question 20

Answer:
$e-1$
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

Answer:
$\frac{1}{2}$
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

Answer:
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

Answer:
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

Answer:
$I=\log \sqrt{3}$
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

Answer:
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

Answer:
$(\log x)^{-1}(x-1) x$
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

Answer:
$10\left(\frac{\pi}{2}\right)^{9}$
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

Answer:
$\frac{1}{2756}$
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

Answer:
$2(\sqrt{2}-1)$
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

Answer:
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

Answer:
$\frac{\pi}{4}$
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
$\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cot x} d x$
$I=\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\frac{\cos x}{\sin x}} d x, \quad\left[\therefore \cot x=\frac{\cos x}{\sin x}\right]$
$=\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\cos x+\sin x} d x \ldots(i), \quad\left[\therefore \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]$
$=\int_{0}^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{2}-x\right)}{\cos \left(\frac{\pi}{2}-x\right)+\sin \left(\frac{\pi}{2}-x\right)} d x$
$=\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\cos x+\sin x} d x \ldots(i i)$
Adding (i) and (ii)
$\begin{aligned} &2 I=\int_{0}^{\frac{\pi}{2}} \frac{\cos x+\sin x}{\cos x+\sin x} d x \\\\ &2 I=\int_{0}^{\frac{\pi}{2}} 1 d x \end{aligned}$
$\begin{aligned} &I=\frac{1}{2}[x]_{0}^{\frac{\pi}{2}} \\\\ &I=\frac{1}{2}\left[\frac{\pi}{2}-0\right] \\\\ &=\frac{\pi}{4} \end{aligned}$

Definite Integrals exercise Multiple choice question 32

Answer:
$\frac{\pi }{4}$
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
$\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x+\cos x} d x \ldots(i) \\\\ &\int_{0}^{\frac{\pi}{2}} f(x) d x=\int_{0}^{\frac{\pi}{2}} f\left(\frac{\pi}{2}-x\right) d x \end{aligned}$
$\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{2}-x\right)}{\sin \left(\frac{\pi}{2}-x\right)+\cos \left(\frac{\pi}{2}-x\right)} d x \\\\ &=\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\cos x+\sin x} d x \ldots(i i) \end{aligned}$
Adding (i) and (ii)
$\begin{aligned} &2 I=\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x+\cos x} d x+\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\cos x+\sin x} d x \\\\ &=\int_{0}^{\frac{\pi}{2}} \frac{1}{\sin x+\cos x}[\cos x+\sin x] d x \end{aligned}$
$\begin{aligned} &2 I=\int_{0}^{\frac{\pi}{2}} 1 d x \\\\ &2 I=[x]_{0}^{\frac{\pi}{2}} \end{aligned}$
$\begin{aligned} &2 I=\left[\frac{\pi}{2}-0\right] \\\\ &=\frac{\pi}{2} \times \frac{1}{2} \\\\ &I=\frac{\pi}{4} \end{aligned}$

Definite Integrals exercise Multiple choice question 33

Answer:
$\frac{\pi }{2}$
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

Answer:
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

Answer:
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
$\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \sin 2 x \log \tan x \; d x \ldots(i) \\\\ &\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x \end{aligned}$
$\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \sin 2\left(\frac{\pi}{2}-x\right) \cdot \log \left[\tan \left(\frac{\pi}{2}-x\right)\right] d x \\\\ &=\int_{0}^{\frac{\pi}{2}} \sin (\pi-2 x) \cdot \log \cot x \; d x \end{aligned}$
$=\int_{0}^{\frac{\pi}{2}} \sin 2 x \cdot \log \cot x\; d x \ldots(i i)$
Adding (i) and (ii)
$2 I=\int_{0}^{\frac{\pi}{2}} \sin 2 x \cdot \log \tan x+\sin 2 x \cdot \log \cot x \; d x$
$\begin{aligned} &2 I=\int_{0}^{\frac{\pi}{2}} \sin 2 x(\log \tan x+\log \cot x) d x \\\\ &2 I=\int_{0}^{\frac{\pi}{2}} \sin 2 x(\log \tan x \cdot \cot x) d x \end{aligned}$
$\begin{aligned} &\log a \cdot b=\log a+\log b \\\\ &2 I=\int_{0}^{\frac{\pi}{2}} \sin 2 x\left(\log \tan x \cdot \frac{1}{\tan x}\right) d x \end{aligned}$
$\begin{aligned} &2 I=\int_{0}^{\frac{\pi}{2}} \sin 2 x(\log 1) d x \\\\ &2 I=0 \\\\ &I=0 \end{aligned}$

Definite Integrals exercise Multiple choice question 36

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

Answer:
$\pi \log 2$
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

Answer:
$\int_{a}^{2 a} f(x) d x=\int_{0}^{a} f(x) d x+\int_{0}^{2 a}f(2 a-x) d x$
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

Answer:
$I=\frac{a+b}{2} \int_{a}^{b} f(x) d x$
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
$\begin{aligned} &I=\int_{a}^{b} x \cdot f(x) d x \ldots(i) \\\\ &\int_{a}^{b} f(x) d x=\int_{a}^{b}(a+b-x) d x \end{aligned}$
$\begin{aligned} &I=\int_{a}^{b}(a+b-x) \cdot f(a+b-x) d x \\\\ &I=\int_{a}^{b}(a+b-x) \cdot f(x) d x \end{aligned}$
$I=\int_{a}^{b}(a+b) \cdot f(x) d x-\int_{a}^{b} x f(x) d x \ldots(i i)$
Adding equation (i) and (ii)
$I+I=\int_{a}^{b} x f(x) d x+\int_{a}^{b}(a+b) \cdot f(x) d x-\int_{a}^{b} x f(x) d x$
$\begin{aligned} &2 I=\int_{a}^{b}(a+b) f(x) d x \\\\ &I=\frac{1}{2} \int_{a}^{b}(a+b) f(x) d x \\\\ &I=\frac{a+b}{2} \int_{a}^{b}(a+b) f(x) d x \end{aligned}$

Definite Integrals exercise Multiple choice question 40

Answer:
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)
Adding Eq(i) and Eq(ii)
$\begin{aligned} &2 I=0 \\\\ &I=0 \end{aligned}$

Definite Integrals exercise Multiple choice question 41

Answer:
$I=0$
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
$\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \log \left[\frac{4+3 \sin x}{4+3 \cos x}\right] d x \ldots(i) \\\\ &\int_{0}^{a} f(x) d x=\int_{0}^{a}(a-x) d x \end{aligned}$
$\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \log \left[\frac{4+3 \sin \left(\frac{\pi}{2}-x\right)}{4+3 \cos \left(\frac{\pi}{2}-x\right)}\right] d x \\\\ &I=\int_{0}^{\frac{\pi}{2}} \log \left[\frac{4+3 \cos x}{4+3 \sin x}\right] d x \ldots(i i) \end{aligned}$
Adding (i) and (ii)
$\begin{aligned} &2 I=\int_{0}^{\frac{\pi}{2}}\left(\log \left[\frac{4+3 \cos x}{4+3 \sin x}\right]+\log \left[\frac{4+3 \sin x}{4+3 \cos x}\right]\right) d x \\\\ &2 I=\int_{0}^{\frac{\pi}{2}}\left(\log \left[\frac{4+3 \cos x}{4+3 \sin x}\right] \cdot \log \left[\frac{4+3 \sin x}{4+3 \cos x}\right]\right) d x \end{aligned}$
$\begin{aligned} &2 I=\int_{0}^{\frac{\pi}{2}} 1 d x \\\\ &2 I=0 \\\\ &I=0 \end{aligned}$

Definite Integrals exercise Multiple choice question 42

Answer:
$\pi$
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

Answer:
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

Answer:
$\int_{a}^{b} f(x+c) d x$
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

Answer:
$\frac{a}{2} \int_{0}^{a} f(x) d x$
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

Answer:
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

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