NCERT Exemplar Class 12 Maths Solutions Chapter 7 Integrals 2021

Q: Are these solutions helpful for competitive examinations?

Indeed, the Class 12 Maths NCERT exemplar solutions chapter 7 covers the syllabus of the competitive exams like NEET and JEE Main to help you ace them.

Q: What are the important topics of this chapter?

Methods of Integration, Integration by Parts and Fundamental Theorem of Calculus are some of the important topics of this chapter. However, rest should not be neglected either.

Q: How many questions are there in this chapter?

The NCERT exemplar solutions for Class 12 Maths chapter 7 consists of 1 exercise with 63 distinct questions for practice.

Q: How many times should one need to read the NCERT books?

Students should read the books enough times months before your exam for better remembering. They can also take help of NCERT exemplar Class 12 Maths solutions chapter 7 pdf download using an online webpage to pdf tool.

NCERT Exemplar Class 12 Maths Solutions Chapter 7 Integrals

Edited By Komal Miglani | Updated on Mar 31, 2025 03:53 AM IST | #CBSE Class 12th

Integrals play a significant role in mathematics. Have you ever thought about how to find the area under a curve, the distance travelled by an object accelerating throughout the distance, or even just the total amount of a quantity accumulated over time? Integrals can be used to find the solution to those types of continuous problems. Within the scope of Class 12 Mathematics, integration is introduced as the anti-derivative, or reverse differentiation, ultimately giving us the ability to retrieve a function from its derivative. In addition, this chapter regarding integration has indefinite integrals, which implies that such a family is represented by an arbitrary constant (∫f(x)dx = F(x) + C).

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LiveCBSE Board Result 2025 LIVE: Class 10th, 12th results soon on cbse.gov.in; marksheets, DigiLocker codeMay 9, 2025 | 10:58 PM IST

Students can follow the steps given below to check the CBSE Class 10, 12 result 2025 through DigiLocker.

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To become proficient in this subject, students should work through the NCERT Solutions for Class 12 Maths, which include clear explanations and very easy-to-follow methods for solving integral problems. The NCERT Solutions encompass a breadth of exercises, such as properties of integrals, applications of integration in physics and engineering, as well as the Fundamental Theorem of Calculus, which connects differentiation and integration. The conceptual understanding and work as preparation for the board exam and competitive examinations, like JEE and NEET.

NCERT Exemplar Class 12 Maths Solutions Chapter 7

Class 12 Maths Chapter 7 exemplar solutions Exercise: 7.3
Page number: 163-169
Total questions: 63

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Question:1

Verify the following:
$\int \frac{2 x - 1}{2 x + 3} d x = x - \log | (2 x + 3)^{2} | + C$

Answer:

$\begin{aligned} To Verify; \\ \int \frac{2 x - 1}{2 x + 3} d x = x - \log | (2 x + 3)^{2} | + C \\ LHS: \int \frac{2 x - 1}{2 x + 3} d x \\ = \int \frac{2 x + 3 - 4}{2 x + 3} d x \\ = \int dx - \int \frac{4}{2 x + 3} dx \end{aligned}$
Let t = 2x + 3
$\Rightarrow dx = \frac{dt}{2} = \int dx - 4 \int \frac{dt}{2 t} = x - 2 \log | t | + C [∵ \int dx = x and \int \frac{1}{x} dx = \log | x |]$

$= x - \log | (2 x + 3)^{2} | + C = RHS [∵ 2 \log | x | = \log | x^{2} |]$
Hence Verified

Question:2

Verify the following:
$\int \frac{2 x + 3}{x^{2} + 3 x} d x = \log | x^{2} + 3 x | + C$

Answer:

To Verify;

$\begin{aligned} \int \frac{2 x + 3}{x^{2} + 3 x} d x = \log | x^{2} + 3 x | + C \\ LHS = \int \frac{2 x + 3}{x^{2} + 3 x} dx Let; t = x^{2} + 3 x \\ \Rightarrow d t = 2 x + 3 \\ = \int \frac{d t}{t} = \log | t | + C \\ [∵ \int \frac{1}{x} d x = \log | x |] \\ \Rightarrow \log | x^{2} + 3 x | + C = RHS \\ [∵ t = x^{2} + 3 x] \end{aligned}$

Question:3

Evaluate the following:
$\int \frac{(x^{2} + 2) d x}{x + 1}$

Answer:

Given; $\int \frac{(x^{2} + 2) d x}{x + 1}$
Let t = x + 1
$\Rightarrow d x = d t = \int \frac{((t - 1)^{2} + 2)}{t} dt$

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

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

$[∵ \int x^{n} dx = \frac{x^{n + 1}}{n + 1} and \int \frac{1}{x} dx = \log | x |]$
$= \frac{t^{2}}{2} - 2 t + 3 \log | t | + C$

$= \frac{t^{2}}{2} - 2 t + \log | t^{3} | + C$

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

Question:4

Evaluate the following:
$\int \frac{e^{6 \log x} - e^{5 \log x}}{e^{4 \log x} - e^{3 \log x}} d x$

Answer:

Given; $\int \frac{e^{6 \log x} - e^{5 \log x}}{e^{4 \log x} - e^{3 \log x}} d x$
As we know $n l o g x = l o g x^{n}$

$\begin{aligned} = \int \frac{e^{\log x^{6}} - e^{\log x^{5}}}{e^{\log x^{4}} - e^{\log x^{2}}} d x \\ = \int \frac{x^{6} - x^{5}}{x^{4} - x^{3}} d x \\ Take x^{3} common out of numerator and denominator to get, \\ = \int \frac{x^{3} (x^{3} - x^{2})}{x^{3} (x - 1)} d x \\ = \int \frac{(x^{3} - x^{2})}{(x - 1)} d x \end{aligned}$

$\begin{aligned} = \int \frac{x^{2} (x - 1)}{(x - 1)} d x \\ = \int x^{2} d x \\ ∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} \\ So, \\ = \frac{x^{3}}{3} + c \end{aligned}$

Question:5

Evaluate the following:
$\int \frac{(1 + \cos x)}{x + \sin x} d x$

Answer:

Given; $\int \frac{(1 + \cos x)}{x + \sin x} d x$
Let t = x + sin x
$\Rightarrow d t = (1 + c o s x) d x = \int \frac{1}{t} d t = \log | t | + C = \log | x + \sin x | + C$

Question:6

Evaluate the following:
$\int \frac{d x}{1 + \cos x}$

Answer:

Given; $\int \frac{d x}{1 + \cos x}$
$= \int \frac{(1 - \cos x)}{(1 + \cos x) (1 - \cos x)} d x = \int \frac{1 - \cos x}{1 - \cos^{2} x} d x$

$= \int \frac{1 - \cos x}{\sin^{2} x} d x [\frac{1}{\sin^{2} x} = {cosec}^{2} x and \frac{\cos x}{\sin^{2} x} = cosec x \cot x]$
$\begin{aligned} = \int ({cosec}^{2} x - cosec x \cot x) d x \\ As we know, \\ \int cosec x \cot x d x = - cosec x + c \\ \int {cosec}^{2} x d x = - \cot x + c \\ = cosec x - \cot x + C \end{aligned}$

Question:7

Evaluate the following:
$\int \tan^{2} x \sec^{4} x d x$

Answer:

Given; $$ \int $ tan\textsuperscript{2} x sec\textsuperscript{4} x dx\\$ $=$ \int $ tan\textsuperscript{2} x sec\textsuperscript{2} x (1+ tan\textsuperscript{2} x) dx\\$ Let; tan x = y
$\\$ \Rightarrow $ sec\textsuperscript{2} x dx = dy\\ \\ =$ \int $ (y\textsuperscript{2}+y\textsuperscript{4} )dy\\$

$= \frac{y^{3}}{3} + \frac{y^{5}}{5} + C = \frac{\tan^{3} x}{3} + \frac{\tan^{5} x}{5} + C$

Question:8

Evaluate the following:
$\int \frac{\sin x + \cos x}{\sqrt{1 + \sin 2 x}} d x$

Answer:

Given; $\int \frac{\sin x + \cos x}{\sqrt{1 + \sin 2 x}} d x$
$= \int c \sin x + \cos x \sqrt{\sin^{2} x + \cos^{2} x + 2 \sin x \cos x}$

$= \int \frac{\sin x + \cos x}{\sqrt{(\sin x + \cos x)^{2}}} d x [∵ \sin 2 x = 2 \sin x \cos x and \sin^{2} x + \cos^{2} x = 1] = \int 1 dx = x + C$

Question:9

Evaluate the following:
$\int \sqrt{1 + \sin x} d x$

Answer:

Given; $\int \sqrt{1 + \sin x} d x$

$= \int \sqrt{\sin^{2} \frac{x}{2} + \cos^{2} \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2}} d x [∵ \sin 2 x = 2 \sin x \cos x and \sin^{2} x + \cos^{2} x = 1]$

$= \int \sqrt{{(\sin \frac{x}{2} + \cos \frac{x}{2})}^{2}} d x = \int (\sin \frac{x}{2} + \cos \frac{x}{2}) d x = 2 \sin \frac{x}{2} - 2 \cos \frac{x}{2} + c$

Question:10

Evaluate the following:
$\int \frac{x}{\sqrt{x} + 1} d x (H i n t : P u t \sqrt{x} = z)$

Answer:

Given;

Let $z = \sqrt{x}$

$\\ \Rightarrow x = z\textsuperscript{2}\\ \\ \Rightarrow dx = 2z dz\\ =\int \frac{z^{2}}{z+1} 2 z d z\\ =2 \int \frac{\mathrm{z}^{3}}{\mathrm{z}+1} \mathrm{~d} z$$
$\\ Let; $t=z+1$\\ i.e. $t=\sqrt{x}+1$ \\ \Rightarrow dt $=\mathrm{dz}$$
$= 2 \int \frac{(t - 1)^{3}}{t} dt = 2 \int \frac{t^{3} - 3 t^{2} + 3 t - 1}{t} dt = 2 \int (t^{2} - 3 t + 3 - \frac{1}{t}) dt$ $[∵ \int x^{n} dx = \frac{x^{n + 1}}{n + 1} and \int \frac{1}{x} dx = \log | x |]$

$= \frac{2 t^{3}}{3} - 3 t^{2} + 3 t - \log | t | + C$

$= \frac{2 (\sqrt{x} + 1)^{3}}{3} - 3 (\sqrt{x} + 1)^{2} + 3 (\sqrt{x} + 1) - \log | \sqrt{x} + 1 | + C$

Question:11

Evaluate the following:
$\int \sqrt{\frac{a + x}{a - x}}$

Answer:

Given, $\int \sqrt{\frac{a + x}{a - x}}$
$= \int \frac{\sqrt{a + x}}{\sqrt{a - x}} \times \frac{\sqrt{a + x}}{\sqrt{a + x}} d x = \int \frac{a + x}{\sqrt{a^{2} - x^{2}}} d x = a \int \frac{1}{\sqrt{a^{2} - x^{2}}} d x - \frac{1}{2} \int \frac{- 2 x}{\sqrt{a^{2} - x^{2}}} d x$

$\begin{aligned} Let t = a^{2} - x^{2} \\ \Rightarrow - 2 x dx = dt \\ = a \sin (\frac{x}{a}) - \frac{1}{2} \int \frac{1}{\sqrt{t}} dt \\ {[∵ \int \frac{d x}{\sqrt{a^{2} - x^{2}}} = \sin^{- 1} \frac{x}{a}]}_{and} [\int \frac{1}{\sqrt{x}} d x = 2 \sqrt{x}] \\ = a \sin (\frac{x}{a}) - \frac{1}{2} \times 2 \sqrt{t} \\ = a \sin (\frac{x}{a}) - \sqrt{a^{2} - x^{2}} + c \end{aligned}$

Question:12

Evaluate the following:
$\int \frac{x^{\frac{1}{2}}}{1 + x^{\frac{3}{4}}} d x$ (Hint : Put x = z⁴)

Answer:

$\begin{aligned} Given \\ \int \frac{x^{\frac{1}{2}}}{1 + x^{\frac{3}{4}}} d x \\ Let x = z^{4} \\ \Rightarrow dx = 4 z^{3} dz \\ = \int \frac{z^{2}}{1 + z^{3}} 4 z^{3} d z \\ = 4 \int \frac{z^{5}}{1 + z^{3}} d z \\ = \frac{4}{3} \int \frac{z^{3}}{1 + z^{3}} 3 z^{2} d z \end{aligned}$
$\begin{aligned} Let t = 1 + z^{3} [\begin{matrix} 1 . e . t = 1 + x^{3 / 4}] \end{matrix} \\ \Rightarrow dt = 3 z^{2} d z \\ = \frac{4}{3} \int \frac{t - 1}{t} d t \end{aligned}$
$= \frac{4}{3} \int dt - \frac{4}{3} \int \frac{1}{t} dt = \frac{4}{3} [t - \log | t |] + C = \frac{4}{3} [(1 + x^{\frac{3}{4}}) - \log | 1 + x^{\frac{3}{4}} |] + C$

Question:13

Evaluate the following:
$\begin{aligned} Given; \\ \int \frac{\sqrt{1 + x^{2}}}{x^{4}} d x \end{aligned}$

Answer:

$\begin{aligned} Given; \\ \int \frac{\sqrt{1 + x^{2}}}{x^{4}} d x \\ Let x = \tan y \\ \Rightarrow dx = \sec^{2} y dx \end{aligned}$
$= \int \frac{\sqrt{1 + \tan^{2} y}}{\tan^{4} y} \sec^{2} y d y$

$= \int \sec y \times \frac{\cos^{4} y}{\sin^{4} y} \times \sec^{2} y d y$

$= \int \frac{\cos y}{\sin^{4} y} d y$

$Let t = \sin y$
$\Rightarrow d t = c o s y d y$
$= \int \frac{dt}{t^{4}} = - \frac{1}{3 t^{3}} + C = - \frac{1}{3 \sin^{3} y} = - \frac{1}{3 \sin^{3} (\sin^{- 1} \frac{x}{\sqrt{x^{2} + 1}})} = - \frac{{(x^{2} + 1)}^{\frac{3}{2}}}{3 x^{3}}$

Question:14

Evaluate the following:
$\int \frac{d x}{\sqrt{16 - 9 x^{2}}}$

Answer:

$\begin{aligned} Given; \\ \int \frac{dx}{\sqrt{16 - 9 x^{2}}} \\ = \int \frac{d x}{\sqrt{4^{2} - (3 x)^{2}}} \\ [∵ \int \frac{d x}{\sqrt{a^{2} - x^{2}}} = \sin^{- 1} \frac{x}{a}] \\ = \frac{1}{3} \sin^{- 1} \frac{3 x}{4} + C \end{aligned}$

Question:15

Evaluate the following:
$\int \frac{dt}{\sqrt{3 t - 2 t^{2}}}$

Answer:

$\begin{aligned} Given \\ \int \frac{dt}{\sqrt{3 t - 2 t^{2}}} \\ = \int \frac{dt}{\sqrt{{(\frac{3}{2 \sqrt{2}})}^{2} - {(\frac{3}{2 \sqrt{2}})}^{2} + 2 \times \sqrt{2} t \times \frac{3}{2 \sqrt{2}} - (\sqrt{2} t)^{2}}} \end{aligned}$
$= \int \frac{d t}{\sqrt{{(\frac{3}{2 \sqrt{2}})}^{2} - {(\sqrt{2} t - \frac{3}{2 \sqrt{2}})}^{2}}} = 2 \sqrt{2} \int \frac{d t}{\sqrt{3^{2} - (4 t - 3)^{2}}}$ $[∵ \int \frac{d x}{\sqrt{a^{2} - x^{2}}} = \sin^{- 1} \frac{x}{a}]$

$= \frac{1}{\sqrt{2}} \sin^{- 1} \frac{4 t - 3}{3} + C$

Question:16

Evaluate the following:
$\int \frac{3 x - 1}{\sqrt{x^{2} + 9}} d x$

Answer:

$Given; \frac{3 x - 1}{\sqrt{x^{2} + 9}} d x$

$= \int \frac{3 x}{\sqrt{x^{2} + 3^{2}}} d x - \int \frac{1}{\sqrt{x^{2} + 3^{2}}} d x$

$[Let; x^{2} + 9 = y \Rightarrow 2 x d x = d y]$

$= \frac{3}{2} \int \frac{d y}{\sqrt{y}} - \log | x + \sqrt{x^{2} + 3^{2}} | + c$

$= 3 \sqrt{x^{2} + 9} - \log | x + \sqrt{x^{2} + 9} | + C$

Question:17

Evaluate the following:
$\int \sqrt{5 - 2 x + x^{2}} d x$

Answer:

$Given; \int \sqrt{5 - 2 x + x^{2}} d x$

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

$[∵ \int \sqrt{x^{2} + a^{2}} = \frac{x}{2} \sqrt{x^{2} + a^{2}} + \frac{a^{2}}{2} \log | x + \sqrt{x^{2} + a^{2}} |$

$= \frac{x - 1}{2} \sqrt{5 - 2 x + x^{2}} + 2 \log | (x + 1) + \sqrt{5 - 2 x + x^{2}} | + C$

Question:18

Evaluate the following:
$\int \frac{x}{x^{4} - 1} d x$

Answer:

$Given; \int \frac{x}{x^{4} - 1} d x$

$[Let; t = x^{2} \Rightarrow d t = 2 x d x]$

$= \int \frac{1}{(t^{2} - 1)} \frac{d t}{2}$
$= \frac{1}{2} \int \frac{1}{(t + 1) (t - 1)} d t = \frac{1}{2} \int \frac{1}{2} [\frac{1}{(t - 1)} - \frac{1}{(t + 1)}] d t$ $[∵ \int \frac{1}{x} d x = \log | x |]$

$= \frac{1}{4} (\log | t - 1 | - \log | t + 1 |) + C$ $[∵ \log a - \log b = \log \frac{a}{b}]$

$= \frac{1}{4} \log | \frac{t - 1}{t + 1} | + c = \frac{1}{4} \log | \frac{x^{2} - 1}{x^{2} + 1} | + c$

Question:19

Evaluate the following:
$\int \frac{x^{2}}{1 - x^{4}} d x put x^{2} = t$

Answer:

Given: $\int \frac{x^{2}}{1 - x^{4}} d x$
$= \int \frac{1}{2} [\frac{2 x^{2}}{(1 + x^{2}) (1 - x^{2})} d x] = \int \frac{1}{2} [\frac{x^{2} + x^{2} - 1 + 1}{(1 + x^{2}) (1 - x^{2})} d x]$

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

$= \int \frac{1}{2} [\frac{(1 + x^{2})}{(1 + x^{2}) (1 - x^{2})} d x - \frac{(1 - x^{2})}{(1 + x^{2}) (1 - x^{2})} d x]$
$= \int \frac{1}{2} [\frac{1}{(1 - x^{2})} d x - \frac{1}{(1 + x^{2})} d x]$
As we know,
$\int \frac{1}{(a^{2} - x^{2})} d x = \frac{1}{2 a} \log \frac{a + x}{a - x}$

$\int \frac{1}{(a^{2} + x^{2})} d x = \frac{1}{a} \tan^{- 1} \frac{x}{a}$

$= \frac{1}{2} \times \frac{1}{2} \log \frac{1 + x}{1 - x} - \frac{1}{2} \tan^{- 1} x + c = \frac{1}{4} \log \frac{1 + x}{1 - x} - \frac{1}{2} \tan^{- 1} x + c$

Question:20

Evaluate the following:
$\int \sqrt{2 a x - x^{2}} d x$

Answer:

$Given; \int \sqrt{2 a x - x^{2}} d x = \int \sqrt{a^{2} - a^{2} + 2 a x - x^{2}} d x$

$= \int \sqrt{a^{2} - (x - a)^{2}} d x$

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

Question:21

Evaluate the following:
$\int \frac{\sin^{- 1} x}{{(1 - x^{2})}^{\frac{3}{2}}} d x$

Answer:

$\begin{aligned} Given; \\ \int \frac{\sin^{- 1} x}{{(1 - x^{2})}^{\frac{3}{2}}} d x \\ \begin{matrix} \begin{array}{c} Let; t = \sin^{- 1} x \\ x = \sin t \\ \Rightarrow d t = \frac{1}{\sqrt{1 - x^{2}}} d x \end{array} \\ \Rightarrow \int \frac{\sin^{- 1} x}{(1 - x^{2}) \sqrt{(1 - x^{2})}} d x \\ = \int \frac{t}{(1 - \sin^{2} t)} d t \\ = \int \frac{t}{\cos^{2} t} d t \end{matrix} \end{aligned}$
$= \int t \sec^{2} t d t$
Apply integration by parts
$[\int f (x) \cdot g (x) d x = f (x) \int g (x) d x - \int \frac{d}{d x} f (x) [\int g (x) d x] d x]$
$= t \int \sec^{2} t dt - \int \frac{d}{dt} (t) [\int \sec^{2} t dt] dt$

$= t tant - \int tant dt = t tant + \int \frac{- \sin t}{\cos t} dt$

$= t tant + \log | \cos t | + C = \frac{x \sin^{- 1} x}{\sqrt{1 - x^{2}}} + \log | \sqrt{1 - x^{2}} | + C$

Question:22

Evaluate the following:
$\int \frac{(\cos 5 x + \cos 4 x)}{1 - 2 \cos 3 x} d x$

Answer:

$Given; \int \frac{(\cos 5 x + \cos 4 x)}{1 - 2 \cos 3 x} dx [\cos a + \cos b = 2 \cos \frac{1}{2} (a + b) \cos \frac{1}{2} (a - b)]$

$= \int \frac{2 \cos \frac{9 x}{2} \cos \frac{x}{2}}{1 - 2 \cos 3 x} d x = \int \frac{2 \cos \frac{9 x}{2} \cos \frac{x}{2} \cos \frac{3 x}{2}}{(1 - 2 \cos 3 x) \cos \frac{3 x}{2}} d x$

$= \int \frac{2 \cos \frac{9 x}{2} \cos \frac{x}{2} \cos \frac{3 x}{2}}{\cos \frac{3 x}{2} - 2 \cos 3 x \cos \frac{3 x}{2}} d x$

$[∵ 2 \cos a \cos b = \cos (a + b) + \cos (a - b)]$

$= \int \frac{2 \cos \frac{9 x}{2} \cos \frac{x}{2} \cos \frac{3 x}{2}}{\cos \frac{3 x}{2} - \cos \frac{9 x}{2} - \cos \frac{3 x}{2}} d x = \int - 2 \cos \frac{3 x}{2} \cos \frac{x}{2} d x$

$= \int - \cos 2 x - \cos x d x = - \frac{\sin 2 x}{2} - \sin x + C$

Question:23

Evaluate the following:
$\int \frac{\sin^{6} x + \cos^{6} x}{\sin^{2} x \cos^{2} x} d x$

Answer:

$Given; \int \frac{\sin^{6} x + \cos^{6} x}{\sin^{2} x \cos^{2} x} dx$

$= \int \frac{(\sin^{2} x + \cos^{2} x) (\sin^{4} x - \sin^{2} x \cos^{2} x + \cos^{4} x)}{\sin^{2} x \cos^{2} x} dx$

$= \int \frac{(\sin^{4} x + \cos^{4} x - \sin^{2} x \cos^{2} x)}{\sin^{2} x \cos^{2} x} d x$

$= \int \frac{\sin^{2} x}{\cos^{2} x} + \frac{\cos^{2} x}{\sin^{2} x} - 1 d x$

$= \int \tan^{2} x + \cot^{2} x - 1 d x$

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

$= \tan x - \cot x - 3 x + C$

Question:24

Evaluate the following:
$\int \frac{\sqrt{x}}{\sqrt{a^{3} - x^{3}}} d x$

Answer:

$\begin{aligned} Given; \\ \int \frac{\sqrt{x}}{\sqrt{a^{3} - x^{3}}} d x \\ [Let; t = \frac{x^{\frac{3}{2}}}{a^{\frac{3}{2}}} \Rightarrow d t = \frac{3 \sqrt{x}}{2 a^{\frac{3}{2}}} d x] \\ = \int \frac{2 a^{\frac{3}{2}} dt}{3 \sqrt{a^{3} - a^{3} t^{2}}} \\ = \int \frac{2 dt}{3 \sqrt{1 - t^{2}}} \\ = \frac{2}{3} \sin^{- 1} t + C = \frac{2}{3} \sin^{- 1} (\frac{x^{\frac{3}{2}}}{a^{\frac{3}{2}}}) + C \end{aligned}$

Question:25

Evaluate the following:
$\int \frac{\cos x - \cos 2 x}{1 - \cos x} d x$

Answer:

Given, $\int \frac{\cos x - \cos 2 x}{1 - \cos x} d x$
$= \int \frac{\cos 2 x - \cos x}{\cos x - 1} d x$

$= \int \frac{2 \cos^{2} x - 1 - \cos x}{\cos x - 1} d x$

$= \int \frac{(2 \cos x + 1) (\cos x - 1)}{\cos x - 1} d x = \int (2 \cos x + 1) d x = 2 \sin x + x + c$

Question:26

Evaluate the following:
$\int \frac{d x}{x \sqrt{x^{4} - 1}} (H int : Put x^{2} = \sec θ)$

Answer:

$\begin{aligned} Given; \\ \int \frac{d x}{x \sqrt{x^{4} - 1}} \\ [Let; x^{2} = \sec θ \Rightarrow 2 xdx = \sec θ \tan θ d θ] \\ = \int \frac{\sec θ \tan θ}{2 \sec θ \sqrt{\sec^{2} θ - 1}} d θ = \int \frac{d θ}{2} \\ = \frac{θ}{2} + c \\ = \frac{\sec^{- 1} x^{2}}{2} + c \end{aligned}$

Question:27

Evaluate the following as limit of sums:
$\int_{0}^{2} (x^{2} + 3) d x$

Answer:

$Given; \int_{0}^{2} (x^{2} + 3) d x$

$We know \int_{a}^{b} f (x) dx = lim_{h \to \infty} h \sum_{r = 0}^{n - 1} f (a + rh)$

$Here a = 0 . b = 2 h = \frac{b - a}{n} = \frac{2 - 0}{n} = \frac{2}{n}$

$\Rightarrow nh = 2$
$= {limh}_{h \to 0} \sum_{r = 0}^{n - 1} f (r h) = lim_{h \to 0} h \sum_{r = 0}^{n - 1} (3 + r^{2} h^{2})$

$= {limh}_{h \to 0} h (3 n + h^{2} (\frac{(n - 1) (n - 1 + 1) (2 n - 2 + 1)}{6})$
$= {limh}_{h \to 0} h (3 n + h^{2} (\frac{(n^{2} - n) (2 n - 1)}{6})$

$= {limh}_{h \to 0} h (3 n + h^{2} (\frac{2 n^{3} - 3 n^{2} + n}{6}) = lim_{h \to 0} (3 n h + (\frac{2 n^{3} h^{3} - 3 n^{2} h^{3} + n h^{3}}{6})$
$= lim_{h \to 0} (3.2 + (\frac{{2.2}^{3} - {3.2}^{2} . h + 2 h^{2}}{6}) = 6 + \frac{16}{3} = \frac{26}{3}$

Question:28

Evaluate the following as limit of su
$\int_{0}^{2} e^{x} d x$

Answer:

$We know \int_{a}^{b} f (x) d x = lim_{h \to \infty} h \sum_{r = 0}^{n - 1} f (a + r h)$

$Here a = 0, b = 2$

$h = \frac{b - a}{n} = \frac{2 - 0}{n} = \frac{2}{n}$

$\Rightarrow h = 2$
$= {limh}_{h \to 0} \sum_{r = 0}^{n - 1} f (r h) = lim_{h \to 0} h [1 + e^{h} + e^{2 h} + \dots, + e^{(n - 1) h}] = lim_{h \to 0} h [\frac{1 {(e^{h})}^{n} - 1}{e^{h} - 1}]$
$= lim_{h \to 0} h [\frac{e^{n h} - 1}{e^{h} - 1}] = lim_{h \to 0} h [\frac{e^{2} - 1}{e^{h} - 1}] = (e^{2} - 1) lim_{h \to 0} [\frac{h}{e^{h} - 1}] = e^{2} - 1$

Question:29

Evaluate the following:
$\int_{0}^{1} \frac{d x}{e^{x} + e^{- x}}$

Answer:

$Given; \int_{0}^{1} \frac{d x}{e^{x} + e^{- x}}$

$= \int_{0}^{1} \frac{e^{x} d x}{e^{2 x} + 1} = \int_{1}^{e} \frac{d t}{t^{2} + 1} [Let; . t = e^{x} [when x = 0, t = 1 and x = 1, t = e]]$

$\Rightarrow d t = [e^{x} d x] = {[\tan^{- 1} t]}_{1}^{e}$
$= \tan^{- 1} e - \tan^{- 1} 1 = \tan^{- 1} e - \frac{π}{4}$

Question:30

Evaluate the following:
$\int_{0}^{\frac{π}{2}} \frac{\tan x d x}{1 + m^{2} \tan^{2} x}$

Answer:

$Given; \int_{0}^{\frac{π}{2}} \frac{\tan x}{1 + m^{2} \tan^{2} x} dx [Let; t = \tan x \Rightarrow dt = \sec^{2} x dx]$ $[Let; u = t^{2} \Rightarrow du = 2 tdt]$

$\frac{t}{1 + m^{2} t^{2}} \frac{dt}{\sec^{2} x} = \int \frac{t}{(1 + m^{2} t^{2})} \frac{dt}{(1 + t^{2})}$

$= \int \frac{1}{(1 + m^{2} u) (1 + u)} \frac{du}{2}$
By applying partial fraction;
$\frac{1}{(1 + m^{2} u) (1 + u)} = \frac{A}{(1 + m^{2} u)} + \frac{B}{(1 + u)}$

$1 = A (1 + u) + B (1 + m^{2} u)$

$B = \frac{1}{1 - m^{2}}$

$When u = - 1$

$When u = - \frac{1}{m^{2}}$
$A = \frac{m^{2}}{m^{2} - 1}$

$= \frac{1}{2} \int \frac{m^{2}}{(m^{2} - 1) (1 + m^{2} u)} + \frac{1}{(1 - m^{2}) (1 + u)} d u$

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

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

$= \frac{1}{2} \times \frac{m^{2}}{m^{2} - 1} [\log ∣ (1 + m^{2} \tan^{2} x) (1 + \tan^{2} x) ‖ + C$

$= \frac{1}{2} \times \frac{m^{2}}{m^{2} - 1} [\log | (1 + m^{2} \tan^{2} x) \sec^{2} x |] + C$
$= \frac{1}{2} \times \frac{m^{2}}{m^{2} - 1} [\log g (\cos^{2} x + m^{2} \sin^{2} x) \sec^{2} x ‖ + C$

$= \frac{\log | (m^{2} - 1) \sin^{2} x + 1 |}{2 m^{2} - 2} + C$
By applying the given limits 0 to π/2
$= \frac{\log | (m^{2} - 1) \sin^{2} \frac{π}{2} + 1 |}{2 m^{2} - 2} - \frac{\log | (m^{2} - 1) \sin^{2} 0 + 1 |}{2 m^{2} - 2}$

$= \frac{\log | (m^{2}) |}{2 m^{2} - 2}$

Question:31

Evaluate the following:
$\int_{1}^{2} \frac{d x}{\sqrt{(x - 1) (2 - x)}}$

Answer:

$\\ Given \int_{1}^{2} \frac{d x}{\sqrt{(x-1)(2-x)}}$\\ $=>_{1}^{2} \frac{\mathrm{dx}}{\sqrt{(x-1)(2-x)}}$\\ $=\int_{1}^{2} \frac{d x}{\sqrt{-\left(x^{2}-3 x+2\right)}}$$
Using a perfect square method for the denominator
$\Rightarrow x^{2}-3 x+2=x^{2}-3 x+\left(\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}+2$$
$\begin{aligned} = {(x - \frac{3}{2})}^{2} - \frac{1}{4} \\ = \int_{1}^{2} \frac{d x}{\sqrt{{(\frac{1}{2})}^{2} - {(x - \frac{3}{2})}^{2}}} \\ We know \\ \int \frac{dx}{\sqrt{a^{2} - x^{2}}} = \sin^{- 1} (\frac{x}{a}) + C \\ = {[\sin^{- 1} (\frac{(x - \frac{3}{2})}{\frac{1}{2}})]}_{1}^{2} = {[\sin^{- 1} (2 x - 3)]}_{1}^{2} \end{aligned}$
$= \sin^{- 1} (1) - \sin^{- 1} (- 1) We know \sin^{- 1} (- θ) = - \sin θ = \frac{π}{2} + \frac{π}{2} = π$

Question:32

Evaluate the following:
$\int_{0}^{1} \frac{xdx}{\sqrt{1 + x^{2}}}$

Answer:

$\\ Given \int_{0}^{1} \frac{x d x}{\sqrt{1+x^{2}}}$\\ Now put $1+x^{2}=t$\\ $=>2 \mathrm{x} \mathrm{dx}=\mathrm{dt}$\\ At $x=0, t=1$ and at x=1, t=2$

$\Rightarrow \frac{1}{2} \int_{1}^{2} \frac{d t}{\sqrt{t}} = \frac{1}{2} [2 \sqrt{t}]_{1}^{2} = \sqrt{2} - 1 \Rightarrow \int_{0}^{1} \frac{x d x}{\sqrt{1 + x^{2}}} = \sqrt{2} - 1$

Question:33

Evaluate the following:
$\int_{0}^{π} x \sin x \cos^{2} x d x$

Answer:

Using Property

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

$Let I = \int_{0}^{π} x \sin x \cos^{2} x d x$

$\Rightarrow \int_{0}^{π} x \sin x \cos^{2} x d x = \int_{0}^{π} (π - x) \sin (π - x) \cos^{2} (π - x) d x$

$ses (π - x) = - \cos x$

$I = \int_{0}^{π} π \sin x \cos^{2} x d x - \int_{0}^{π} x \sin x \cos^{2} x d x$

$I = \int_{0}^{π} π \sin x \cos^{2} x d x - I$

$2 I = \int_{0}^{π} π \sin x \cos^{2} x d x$

$\begin{aligned} \Rightarrow \int_{0}^{π} π \sin x \cos^{2} x d x = π \int_{0}^{π} \sin x \cos^{2} x d x \\ Now let \cos x = t \\ \Rightarrow - \sin x d x - d t \\ And, at x = 0, t = 1 \\ and at x = π, t = - 1 \end{aligned}$

$\Rightarrow 2 I = - π \int_{1}^{- 1} t^{2} dt = - π {[\frac{t^{2}}{3}]}_{1}^{- 1} = \frac{2 π}{3}$

$\Rightarrow 2 I = \frac{2 π}{3} \Rightarrow I = \frac{π}{3}$

$\Rightarrow \int_{0}^{π} x \sin x \cos^{2} x dx = \frac{π}{3}$

Question:34

Evaluate the following:
$\int_{0}^{\frac{1}{2}} \frac{d x}{(1 + x^{2}) \sqrt{1 - x^{2}}}$ (Hint: let x = sin θ)

Answer:

$Given \int_{0}^{\frac{1}{2}} \frac{d x}{(1 + x^{2}) \sqrt{1 - x^{2}}}$

$\Rightarrow Let x = \sin θ$

$At x = 0, θ = 0 and x = \frac{1}{2}, θ = \frac{π}{6}$
$= \int_{0}^{\frac{1}{2}} \frac{d x}{(1 + x^{2}) \sqrt{1 - x^{2}}} = \int_{0}^{\frac{π}{6}} \frac{\cos θ d θ}{(1 + \sin^{2} θ) \sqrt{1 - \sin^{2} θ}}$

$As 1 - \sin^{2} θ = \cos^{2} θ = \int_{0}^{\frac{π}{6}} \frac{\cos θ d θ}{(1 + \sin^{2} θ) \sqrt{1 - \sin^{2} θ}} = \int_{0}^{\frac{π}{6}} \frac{\cos θ d θ}{(1 + \sin^{2} θ) \sqrt{\cos^{2} θ}} = \int_{0}^{\frac{π}{6}} \frac{\cos θ d θ}{(1 + \sin^{2} θ) \cos θ}$
$\begin{aligned} \Rightarrow \int_{0}^{\frac{π}{6}} \frac{d θ}{(1 + \sin^{2} θ)} \\ \Rightarrow \int_{0}^{\frac{π}{6}} \frac{\sec^{2} θ d θ}{(\sec^{2} θ + \tan^{2} θ)} \\ \Rightarrow As \sec^{2} θ - \tan^{2} θ = 1 \\ \Rightarrow \int_{0}^{\frac{π}{6}} \frac{\sec^{2} θ d θ}{(1 + 2 \tan^{2} θ)} \end{aligned}$
$\begin{aligned} Now put \tan θ = t \\ \Rightarrow \sec^{2} θ d θ = dt \\ At θ = 0, t = 0 \\ at θ = \frac{π}{6}, t = \frac{1}{\sqrt{3}} \\ = \int_{0}^{\frac{1}{\sqrt{2}}} \frac{d t}{(1 + 2 t^{2})} = \frac{1}{2} \int_{0}^{\frac{1}{\sqrt{2}}} \frac{d t}{({(\frac{1}{\sqrt{2}})}^{2} + t^{2})} \\ As \int \frac{d x}{x^{2} + a^{2}} = \frac{1}{2} \tan^{- 1} (\frac{x}{a}) + c \end{aligned}$

$\Rightarrow \int_{0}^{\frac{1}{\sqrt{2}}} \frac{d t}{({(\frac{1}{\sqrt{2}})}^{2} + t^{2})} = \frac{1}{2} {[\frac{1}{\frac{1}{\sqrt{2}}} \tan^{- 1} (\frac{t}{\frac{1}{\sqrt{2}}})]}_{0}^{\frac{1}{\sqrt{2}}} = \frac{\sqrt{2}}{2} \tan^{- 1} (\frac{\sqrt{2}}{\sqrt{3}})$

$\Rightarrow \int_{0}^{\frac{1}{2}} \frac{d x}{(1 + x^{2}) \sqrt{1 - x^{2}}} = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{\sqrt{2}}{\sqrt{3}})$

Question:35

Evaluate the following:
$\int \frac{x^{2} d x}{x^{4} - x^{2} - 12}$

Answer:

$Given: \int \frac{x^{2} d x}{x^{4} - x^{2} - 12} Put x^{2} = t$

$=> \frac{x^{2}}{x^{4} - x^{2} - 12} = \frac{t}{t^{2} - t - 12}$
$\Rightarrow t^{2} - t - 12 = (t + 3) (t - 4)$

$\Rightarrow \frac{t}{(t + 3) (t - 4)} =$

$\frac{A}{(t + 3)} + \frac{B}{(t - 4)} (Concept of partial fraction)$

$\Rightarrow t = t (A + B) + 3 B - 4 A$
On comparing coefficients of ‘t’ we get
$A = \frac{3}{7} & B = \frac{4}{7} = \frac{t}{(t + 3) (t - 4)} = \frac{3}{7 (t + 3)} + \frac{4}{7 (t - 4)}$

$\Rightarrow Now put t = x^{2} back in the above eq.$

$\Rightarrow \frac{x^{2}}{x^{4} - x^{2} - 12} = \frac{3}{7 (x^{2} + 3)} + \frac{4}{7 (x^{2} - 4)}$

$\Rightarrow \frac{x^{2} d x}{x^{4} - x^{2} - 12} = \int (\frac{3}{7 (x^{2} + 3)} + \frac{4}{7 (x^{2} - 4)}) d x = \frac{1}{7} (\int \frac{3 d x}{(x^{2} + 3)} + \int \frac{4 d x}{(x^{2} - 4)})$

$\Rightarrow Now \int \frac{d x}{x^{2} - a^{2}} = \frac{1}{2 a} \ln (\frac{x - a}{x + a}) + c & \int \frac{d x}{x^{2} + a^{2}} = \frac{1}{a} \tan^{- 1} (\frac{x}{a}) + c$

$\Rightarrow 7 (\int \frac{3 d x}{(x^{2} + 3)} + \int \frac{4 d x}{(x^{2} - 4)}) = \frac{1}{7} (\frac{3}{\sqrt{3}} \tan^{- 1} (\frac{x}{\sqrt{3}}) + \frac{4}{4} \ln (\frac{x - 2}{x + 2}) + c)$

$\Rightarrow \int \frac{x^{2} d x}{x^{4} - x^{2} - 12} =$

$\frac{\sqrt{3}}{7} \tan^{- 1} (\frac{x}{\sqrt{3}}) + \frac{1}{7} \ln (\frac{x - 2}{x + 2}) + c$

Question:36

Evaluate the following:
$\int \frac{x^{2} d x}{(x^{2} + a^{2}) (x^{2} + b^{2})}$

Answer:

$Given \int \frac{x^{2} d x}{(x^{2} + a^{2}) (x^{2} + b^{2})} Put x^{2} = t$

$\Rightarrow \frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2})} = \frac{t}{(t + a^{2}) (t + b^{2})}$

$\Rightarrow \frac{t}{(t + a^{2}) (t + b^{2})} = \frac{A}{(t + a^{2})} + \frac{B}{(t + b^{2})} (Concept of partial fraction)$

$\Rightarrow t = t (A + B) + a^{2} B + b^{2} A On comparing coefficients of "twe get$

$\Rightarrow A = \frac{a^{2}}{a^{2} - b^{2}} B = \frac{- b^{2}}{a^{2} - b^{2}}$

$\Rightarrow \frac{t}{(t + a^{2}) (t + b^{2})} = \frac{1}{a^{2} - b^{2}} (\frac{a^{2}}{(t + a^{2})} - \frac{b^{2}}{(t + b^{2})})$

$\Rightarrow Now put t = x^{2} back in the above eq.$

$\Rightarrow \frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2})} = \frac{1}{a^{2} - b^{2}} (\frac{a^{2}}{(x^{2} + a^{2})} - \frac{b^{2}}{(x^{2} + b^{2})})$

$\Rightarrow \int \frac{x^{2} d x}{(x^{2} + a^{2}) (x^{2} + b^{2})} = \frac{1}{a^{2} - b^{2}} (\int \frac{a^{2} d x}{(x^{2} + a^{2})} - \int \frac{b^{2} d x}{(x^{2} + b^{2})})$

$\Rightarrow_{Now} \int \frac{d x}{x^{2} + a^{2}} = \frac{1}{a} \tan^{- 1} (\frac{x}{a}) + c$
$= \frac{1}{a^{2} - b^{2}} (\int \frac{a^{2} d x}{(x^{2} + a^{2})} - \int \frac{b^{2} d x}{(x^{2} + b^{2})})$

$\Rightarrow= \frac{1}{a^{2} - b^{2}} (\frac{a^{2}}{a} \tan^{- 1} (\frac{x}{a}) + \frac{b^{2}}{b} \tan^{- 1} (\frac{x}{b}) + c)$

$\Rightarrow \int \frac{x^{2} d x}{(x^{2} + a^{2}) (x^{2} + b^{2})} = \frac{1}{a^{2} - b^{2}} (a \tan^{- 1} (\frac{x}{a}) + b \tan^{- 1} (\frac{x}{b})) + c$

Question:37

Evaluate the following:
$\int_{0}^{π} \frac{x}{1 + \sin x}$

Answer:

$G i v e n \int_{0}^{π} \frac{x d x}{1 + \sin x}$ \$

Let $I = \int_{0}^{π} \frac{xdx}{1 + \sin x}$

$N o w u s i n g P r o p e r t y$ \int_{a}^{b} f(x) d x$

$= \int_{a}^{b} f (a + b - x) d x$

$= \int_{0}^{π} \frac{x d x}{1 + \sin x} = \int_{0}^{π} \frac{(π - x) d x}{1 + \sin (π - x)}$

$\Rightarrow \int_{0}^{π} \frac{x d x}{1 + \sin x} = \int_{0}^{π} \frac{π d x}{1 + \sin x} - \int_{0}^{π} \frac{x d x}{1 + \sin x} \Rightarrow 2 \int_{0}^{π} \frac{x d x}{1 + \sin x} = 2$

$I = π \int_{0}^{π} \frac{d x}{1 + \sin x} \dots (1)$

$\Rightarrow \int_{0}^{π} \frac{d x}{1 + \sin x} = \int_{0}^{π} \frac{1}{1 + \sin x} \times \frac{1 - \sin x}{1 - \sin x} = \int_{0}^{π} \frac{1 - \sin x}{1 - \sin^{2} x} d x$

$\Rightarrow 1 - \sin^{2} x = \cos^{2} x$
$= \int_{0}^{π} \frac{1 - \sin x}{1 - \sin^{2} x} d x = \int_{0}^{π} \frac{1 - \sin x}{\cos^{2} x} d x = \int_{0}^{π} \frac{d x}{\cos^{2} x} - \int_{0}^{π} \frac{\sin x}{\cos^{2} x} d x \dots (2)$

$\Rightarrow \int_{0}^{π} \frac{d x}{\cos^{2} x} = \int_{0}^{π} \sec^{2} x d x = [\tan x]_{0}^{π} = 0$

And $for \int_{0}^{π} \frac{\sin x}{\cos^{2} x} d x Put \cos x = t$
$\Rightarrow - \sin x d x = d t$

$\Rightarrow {At}^{x} = 0 => t = 1 and x = π => t = - 1 = \int_{1}^{- 1} - \frac{d t}{t^{2}} = {[\frac{1}{t}]}_{1}^{- 1} = - 2 \dots$

$2 I = π \int_{0}^{π} \frac{d x}{1 + \sin x} =$

$π (\int_{0}^{π} \frac{d x}{\cos^{2} x} - \int_{0}^{π} \frac{\sin x}{\cos^{2} x} dx) = π (0 - (- 2)) = 2 π$

$I =$ $\int_{0}^{π} \frac{xdx}{1 + \sin x} = π$

Question:38

Evaluate the following:
$\int \frac{2 x - 1}{(x - 1) (x + 2) (x - 3)} d x$

Answer:

Given:
$\int \frac{2 x - 1}{(x - 1) (x + 2) (x - 3)} d x$
Using the concept of partial fractions,
$\Rightarrow \frac{2 x - 1}{(x - 1) (x + 2) (x - 3)} = \frac{A}{(x - 1)} + \frac{B}{(x + 2)} + \frac{C}{(x - 3)}$ \ $\Rightarrow 2 x - 1 = x^{2} (A + B + C) + x (C - 4 B - A) + (3 B - 2 C - 6 A)$ \ Comparing coefficients:\ $\Rightarrow A + B + C = 0 \dots (1)$
$\begin{aligned} \Rightarrow C - 4 B - A = 2 \dots (2) \\ \Rightarrow 3 B - 2 C - 6 A = - 1 \dots (3) \\ \Rightarrow On solving (1), (2) and (3) we get \\ \Rightarrow A = - \frac{1}{6}, B = - \frac{1}{3} and C = \frac{1}{2} \end{aligned}$
$= \frac{2 x - 1}{(x - 1) (x + 2) (x - 3)} = \frac{- \frac{1}{6}}{(x - 1)} + \frac{- \frac{1}{2}}{(x + 2)} + \frac{\frac{1}{2}}{(x - 3)} \Rightarrow \int \frac{2 x - 1}{(x - 1) (x + 2) (x - 3)} d x = \int \frac{1}{2 (x - 3)} d x - \int \frac{1}{3 (x + 2)} d x - \int \frac{1}{6 (x - 1)} d x$
$= \frac{1}{2} \ln (x - 3) - \frac{1}{3} \ln (x + 2) - \frac{1}{6} \ln (x - 1) + c \Rightarrow \int \frac{2 x - 1}{(x - 1) (x + 2) (x - 3)} d x = \frac{1}{2} \ln (x - 3) - \frac{1}{3} \ln (x + 2) - \frac{1}{6} \ln (x - 1) + c$

Question:39

Evaluate the following:
$\int e^{\tan^{- 1} x} (\frac{1 + x + x^{2}}{1 + x^{2}}) d x$

Answer:

Given: $\int e^{\tan^{- 1} x} (\frac{1 + x + x^{2}}{1 + x^{2}}) d x$
$Put \tan^{- 1} x = t$

$l = \frac{d x}{1 + x^{2}} = d t \Rightarrow \int e^{\tan^{- 1} x} (\frac{1 + x + x^{2}}{1 + x^{2}}) d x = \int e^{t} (1 + \tan t + \tan^{2} t) d t$

$\Rightarrow As \sec^{2} θ - \tan^{2} θ = 1$

$\Rightarrow \int e^{t} (1 + \tan t + \tan^{2} t) d t = \int e^{t} (1 + \tan t + \sec^{2} t - 1) d t$
$\Rightarrow \int e^{t} (\tan t + \sec^{2} t) d t$

$Now using the property. \int e^{x} (f (x) + f^{'} (x)) dx = e^{x} f (x)$

$\Rightarrow Now in \int e^{t} (\tan t + \sec^{2} t) d t$
$= f (t) = \tan t \Rightarrow f^{'} (t) = \sec^{2} x$

$\Rightarrow \int e^{t} (\tan t + \sec^{2} t) d t = e^{t} \tan t + C$

$\Rightarrow \int e^{\tan^{- 1} x} (\frac{1 + x + x^{2}}{1 + x^{2}}) d x = e^{\tan^{- 1} x} x + C$

Question:40

Evaluate the following:
$\int \sin^{- 1} \sqrt{\frac{x}{a + x}} d x$ (Hint: Put x = a tan2 θ)

Answer:

Given: $\int \sin^{- 1} \sqrt{\frac{x}{a + x}} d x$
$\Rightarrow Put x = a \tan^{2} θ \Rightarrow dx = 2 a \tan θ \sec^{2} θ d θ \int \sin^{- 1} \sqrt{\frac{x}{a + x}} d x =$ $\int \sin^{- 1} \sqrt{\frac{a \tan^{2} θ}{a + a \tan^{2} θ}} 2 a \tan θ \sec^{2} θ d θ$
$\Rightarrow As \sec^{2} θ - \tan^{2} θ = 1$

$= \int \sin^{- 1} \sqrt{\frac{a \tan^{2} θ}{a (1 + \tan^{2} θ)}} 2 a \tan θ \sec^{2} θ d θ \Rightarrow \int \sin^{- 1} \sqrt{\frac{\tan^{2} θ}{\sec^{2} θ}} 2 a \tan θ \sec^{2} θ d θ$

$= \int \sin^{- 1} \sqrt{\sin^{2} θ} 2 a \tan θ \sec^{2} θ d θ$

$\Rightarrow 2 a \int θ \tan θ \sec^{2} θ d θ$

$\Rightarrow$ $Now put$ $\tan θ = t$

$\Rightarrow \sec^{2} θ d θ = d t$

$\Rightarrow 2 a \int θ \tan θ \sec^{2} θ d θ = 2 a \int t \tan^{- 1} t dt \dots$

$\Rightarrow$ $Now apply integration by part on$ $\int t \tan^{- 1} t dt$

Question:41

Evaluate the following:
$\int_{\frac{π}{3}}^{\frac{π}{2}} \frac{\sqrt{1 + \cos x}}{(1 - \cos x)^{\frac{5}{2}}}$

Answer:

Given: $\int_{\frac{π}{3}}^{\frac{π}{2}} \frac{\sqrt{1 + \cos x}}{(1 - \cos x)^{\frac{5}{2}}}$
Using Trigonometric identities:
$\Rightarrow \cos 2 x = 2 \cos^{2} x - 1 = 1 - 2 \sin^{2} x \frac{\sqrt{1 + \cos x}}{(1 - \cos x)^{\frac{5}{2}}} = \frac{\sqrt{1 + (2 \cos^{2} \frac{x}{2} - 1)}}{{(1 - (1 - 2 \sin^{2} \frac{x}{2}))}^{\frac{5}{2}}} = \frac{\sqrt{2 \cos^{2} \frac{x}{2}}}{{(2 \sin^{2} \frac{x}{2})}^{\frac{5}{2}}}$
$= \frac{\sqrt{2 \cos^{2} \frac{x}{2}}}{{(2 \sin^{2} \frac{x}{2})}^{\frac{5}{2}}} = (\sqrt{2} \cos \frac{x}{2}) / (2^{\frac{5}{2}} \sin^{5} \frac{x}{2}) = \frac{\sqrt{2} \cos \frac{x}{2}}{2^{\frac{5}{2}} \sin^{5} \frac{x}{2}} = \frac{1}{4} \frac{\cos \frac{x}{2}}{\sin^{5} \frac{x}{2}}$
$\Rightarrow \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{\sqrt{1 + \cos x}}{(1 - \cos x)^{\frac{5}{2}}} d x = \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{1}{4} \frac{\cos \frac{x}{2}}{\sin^{5} \frac{x}{2}} d x$

$Put \sin (\frac{x}{2}) = t \cos (\frac{x}{2}) d x = 2 d t At x = \frac{π}{3}$

$\Rightarrow t = \frac{1}{2} and at x = \frac{π}{2}$

$\Rightarrow t = \frac{1}{\sqrt{2}}$
$\Rightarrow \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{1}{4} \frac{\cos \frac{x}{2}}{\sin^{5} \frac{x}{2}} dx = \frac{1}{2} \int_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}} \frac{d t}{t^{5}}$
$\Rightarrow \frac{1}{2} \int_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}} \frac{d t}{t^{5}} = \frac{1}{2} {[\frac{t^{- 4}}{- 4}]}_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}} = \frac{3}{2} \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{\sqrt{1 + \cos x}}{(1 - \cos x)^{\frac{5}{2}}} d x = \frac{3}{2}$

Question:42

Evaluate the following:
$\int e^{- 3 x} \cos^{3} x d x$

Answer:

$\begin{aligned} Given: \int e^{- 3 x} \cos^{3} x d x \\ Using trigonometric identity \\ \cos 3 x = 4 \cos^{3} x - 3 \cos x \\ \int e^{- 3 x} \cos^{3} x d x = \frac{1}{4} \int e^{- 3 x} (\cos 3 x + 3 \cos x) d x \\ \frac{1}{4} \int e^{- 3 x} (\cos 3 x + 3 \cos x) d x = \frac{1}{4} \int e^{- 3 x} \cos 3 x d x + \frac{3}{4} \int e^{- 3 x} \cos x d x . . . (1) \end{aligned}$
$\begin{aligned} \Rightarrow Using a generalised formula i.e \\ \int e^{a x} \cos b x d x = \frac{e^{2 x}}{a^{2} + b^{2}} (a \cos b x + b \sin b x) \\ \int e^{- 3 x} \cos 3 x d x = \frac{e^{- 2 x}}{(- 3)^{2} + 3^{2}} ((- 3) \cos 3 x + 3 \sin 3 x) \end{aligned}$
$= \frac{e^{- 2 x}}{(- 3)^{2} + 3^{2}} ((- 3) \cos 3 x + 3 \sin 3 x)$

$= \frac{e^{- 2 x}}{6} (\sin 3 x - \cos 3 x) \dots (2) \int e^{- 3 x} \cos x d x$

$= \frac{e^{- 2 x}}{(- 3)^{2} + 1^{2}} ((- 3) \cos x + \sin x)$

$= \frac{e^{- 2 x}}{10} (\sin x - 3 \cos x) = \frac{e^{- 2 x}}{10} (\sin x - 3 \cos x) \dots (3)$

$\Rightarrow On putting (2) and (3) in (1)$
$\frac{1}{4} \int e^{- 3 x} \cos 3 x d x + \frac{3}{4} \int e^{- 3 x} \cos x d x = \frac{e^{- 2 x}}{4 \times 6} (\sin 3 x - \cos 3 x) + \frac{3 e^{- 2 x}}{4 \times 10} (\sin x - 3 \cos x)$

$\Rightarrow \int e^{- 3 x} \cos^{3} x d x = e^{- 3 x} {\frac{(\sin 3 x - \cos 3 x)}{24} + \frac{3 (\sin x - 3 \cos x)}{40}} + c$

Question:43

Evaluate the following:
$\int \sqrt{\tan x} d x$ (Hint: Put $t a n x = t^{2}$ )

Answer:

Given:
$\int \sqrt{\tan x} d x$
Put $t a n x = t^{2}$
$\begin{aligned} \Rightarrow \sec^{2} x d x = 2 t d t \\ \Rightarrow dx = \frac{2 t dt}{\sec^{2} x} = \frac{2 tdt}{1 + \tan^{2} x} = \frac{2 tdt}{1 + t^{4}} \\ \Rightarrow \int \sqrt{\tan x} d x = \int \sqrt{t^{2}} \frac{2 t dt}{1 + t^{4}} = \int \frac{2 t^{2} dt}{1 + t^{4}} \\ \Rightarrow \int \frac{2 t^{2} d t}{1 + t^{4}} = \int \frac{(2 t^{2} + 1 - 1) d t}{1 + t^{4}} = \int \frac{(t^{2} + 1) + (t^{2} - 1)}{1 + t^{4}} d t \\ \Rightarrow \int \frac{(t^{2} + 1) + (t^{2} - 1)}{1 + t^{4}} d t = \int \frac{(t^{2} + 1)}{1 + t^{4}} d t + \int \frac{(t^{2} - 1)}{1 + t^{4}} d t \end{aligned}$
Taking out $t^{2}$ common in both the numerators
$\Rightarrow \int \frac{(t^{2} + 1)}{1 + t^{4}} d t + \int \frac{(t^{2} - 1)}{1 + t^{4}} d t = \int \frac{t^{2} (1 + (\frac{1}{t^{2}}))}{1 + t^{4}} d t + \int \frac{t^{2} (1 - (\frac{1}{t^{2}}))}{1 + t^{4}} dt \int \frac{t^{2} (1 + (\frac{1}{t^{2}}))}{1 + t^{4}} dt + \int \frac{t^{2} (1 - (\frac{1}{t^{2}}))}{1 + t^{4}} dt$

$= \int \frac{(1 + (\frac{1}{t^{2}}))}{\frac{1}{t^{2}} + t^{2}} dt + \int \frac{(1 - (\frac{1}{t^{2}}))}{\frac{1}{t^{2}} + t^{2}} dt \dots . (1)$
$\Rightarrow Now t^{2} + \frac{1}{t^{2}} = {(t \pm \frac{1}{t})}^{2} \mp 2 \dots (3) = for (a) \int \frac{(1 + (\frac{1}{t^{2}}))}{\frac{1}{t^{2}} + t^{2}} dt taking t - \frac{1}{t} = z$

$\Rightarrow (1 + \frac{1}{t^{2}}) d t = d z$

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

$= \int \frac{(1 + \frac{1}{t^{2}})}{{(t - \frac{1}{t})}^{2} + 2} dt = \int \frac{d z}{z^{2} + 2} = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{z}{\sqrt{2}}) + c \dots (2)$
$for (b) \int \frac{(1 - (\frac{1}{t^{2}}))}{\frac{1}{t^{2}} + t^{2}} dt taking t + \frac{1}{t} = z$

$\Rightarrow (1 - \frac{1}{t^{2}}) dt = dz = \int \frac{(1 - \frac{1}{t^{2}})}{\frac{1}{t^{2}} + t^{2}} dt = \int \frac{(1 - \frac{1}{t^{2}})}{{(t + \frac{1}{t})}^{2} - 2} dt = \int \frac{(1 - \frac{1}{t^{2}})}{{(t + \frac{1}{t})}^{2} - 2} dt = \int \frac{d z}{z^{2} - 2} = \frac{1}{2 \sqrt{2}} \ln | \frac{z - \sqrt{2}}{z + \sqrt{2}} | + c \dots (3)$

Put (2) and (3) in (1)
$= \int \frac{(1 + (\frac{1}{t^{2}}))}{\frac{1}{t^{2}} + t^{2}} dt + \int \frac{(1 - (\frac{1}{t^{2}}))}{\frac{1}{t^{2}} + t^{2}} dt = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{(t - \frac{1}{t})}{\sqrt{2}}) + \frac{1}{2 \sqrt{2}} \ln | \frac{(t + \frac{1}{t}) - \sqrt{2}}{(t + \frac{1}{t}) + \sqrt{2}} | + c$

$\Rightarrow \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{t^{2} - 1}{t \sqrt{2}}) + \frac{1}{2 \sqrt{2}} \ln | \frac{(t^{2} + 1 - t \sqrt{2}}{(t^{2} + 1 + t \sqrt{2}} | + C$
$\Rightarrow Now again putting t = \sqrt{\tan x} to obtain the final result$

$= \int \sqrt{\tan x} d x = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{\tan x - 1}{\sqrt{2 \tan x}}) + \frac{1}{2 \sqrt{2}} \ln | \frac{(\tan x + 1 - \sqrt{2 \tan x}}{\tan x + 1 + \sqrt{2 \tan x}} |$

Question:44

Evaluate the following:
$\int_{0}^{\frac{π}{2}} \frac{d x}{{(a^{2} \cos^{2} x + b^{2} \sin^{2} x)}^{2}}$
(Hint: Divide Numerator and Denominator by $c o s^{4} x$ )

Answer:

Given: $\int_{0}^{\frac{π}{2}} \frac{d x}{{(a^{2} \cos^{2} x + b^{2} \sin^{2} x)}^{2}}$
Dividing Numerator and Denominator by $c o s^{4} x$
$= \int_{0}^{\frac{π}{2}} \frac{(1 / \cos^{4} x) d x}{{((a^{2} \cos^{2} x + b^{2} \sin^{2} x) / \cos^{2} x)}^{2}}$

$\Rightarrow \int_{0}^{\frac{π}{2}} \frac{\sec^{4} x d x}{{(a^{2} + b^{2} \tan^{2} x)}^{2}} \Rightarrow \int_{0}^{\frac{π}{2}} \frac{\sec^{2} x \sec^{2} x d x}{{(a^{2} + b^{2} \tan^{2} x)}^{2}} = \int_{0}^{\frac{π}{2}} \frac{(1 + \tan^{2} x) \sec^{2} x d x}{{(a^{2} + b^{2} \tan^{2} x)}^{2}}$

$\Rightarrow Put \tan x = t$
$\Rightarrow \sec^{2} x d x = d t$

$\Rightarrow A t x = 0, t = 0 and at x = \frac{π}{2}, t = \infty$

$\Rightarrow \int_{0}^{\frac{π}{2}} \frac{(1 + \tan^{2} x) \sec^{2} x d x}{{(a^{2} + b^{2} \tan^{2} x)}^{2}} = \int_{0}^{\infty} \frac{(1 + t^{2}) d t}{{(a^{2} + b^{2} t^{2})}^{2}}$

$\Rightarrow \int_{0}^{\infty} \frac{(1 + t^{2}) d t}{{(a^{2} + b^{2} t^{2})}^{2}} = \frac{1}{b^{2}} \int_{0}^{\infty} \frac{(b^{2} + b^{2} t^{2}) d t}{{(a^{2} + b^{2} t^{2})}^{2}}$
$= \frac{1}{b^{2}} \int_{0}^{\infty} \frac{(b^{2} + b^{2} t^{2}) d t}{{(a^{2} + b^{2} t^{2})}^{2}} = \frac{1}{b^{2}} \int_{0}^{\infty} \frac{(a^{2} + b^{2} t^{2}) + (b^{2} - a^{2})}{{(a^{2} + b^{2} t^{2})}^{2}} d t$

$\Rightarrow \frac{1}{b^{2}} \int_{0}^{\infty} \frac{(a^{2} + b^{2} t^{2}) + (b^{2} - a^{2})}{{(a^{2} + b^{2} t^{2})}^{2}} d t = \frac{1}{b^{2}} \int_{0}^{\infty} \frac{1}{(a^{2} + b^{2} t^{2})} d t + \frac{1}{b^{2}} \int_{0}^{\infty} \frac{(b^{2} - a^{2})}{{(a^{2} + b^{2} t^{2})}^{2}} d t$

$\Rightarrow Let I = \frac{1}{b^{2}} \int_{0}^{\infty} \frac{1}{(a^{2} + b^{2} t^{2})} d t + \frac{1}{b^{2}} \int_{0}^{\infty} \frac{(b^{2} - a^{2})}{{(a^{2} + b^{2} t^{2})}^{2}} d t \dots (1)$
$= Let I_{1} = \int_{0}^{\infty} \frac{1}{(a^{2} + b^{2} t^{2})} d t = \int_{0}^{\infty} \frac{1}{(a^{2} + b^{2} t^{2})} d t$

$= \frac{1}{b^{2}} \int_{0}^{\infty} \frac{1}{((a^{2} / b^{2}) + t^{2})} d t = 1_{1} = \frac{1}{b^{2}} \int_{0}^{\infty} \frac{1}{((a^{2} / b^{2}) + t^{2})} d t$

$= \frac{1}{b^{2}} (\frac{b}{a}) {[\tan^{- 1} (\frac{b t}{a})]}_{0}^{\infty} = \frac{π}{2 a b} \Rightarrow I_{1} = \frac{π}{2 a b} \dots . (2)$

$Let I_{2} = \int_{0}^{\infty} \frac{1}{{(a^{2} + b^{2} t^{2})}^{2}} d t$

$\Rightarrow let b t = a \tan θ$

$\Rightarrow b d t = {asec}^{2} θ d θ = I_{2} = \frac{1}{b} \int_{0}^{\frac{π}{2}} \frac{{asec}^{2} θ d θ}{{(a^{2} + a^{2} \tan^{2} θ)}^{2}} = \frac{1}{b} \int_{0}^{\frac{π}{2}} \frac{a \sec^{2} θ d θ}{a^{4} {(1 + \tan^{2} θ)}^{2}} = \frac{1}{a^{3} b} \int_{0}^{\frac{π}{2}} \frac{\sec^{2} θ d θ}{\sec^{4} θ}$
$= \frac{1}{a^{3} b} \int_{0}^{\frac{π}{2}} \frac{\sec^{2} θ d θ}{\sec^{4} θ} = \frac{1}{a^{3} b} \int_{0}^{\frac{π}{2}} \cos^{2} θ d θ$ $= \frac{1}{2 a^{3} b} \int_{0}^{\frac{π}{2}} (1 + \cos 2 θ) d θ$

$\Rightarrow \frac{1}{2 a^{3} b} \int_{0}^{\frac{π}{2}} (1 + \cos 2 θ) d θ = \frac{1}{2 a^{3} b} {[θ + \frac{\sin 2 θ}{2}]}_{0}^{\frac{π}{2}} = \frac{π}{4 a^{3} b} \dots (3)$

$\Rightarrow I = \frac{1}{b^{2}} (I_{1} + (b^{2} - a^{2}) I_{2}) =$

$\frac{1}{b^{2}} (\frac{π}{2 a b} + (b^{2} - a^{2}) \frac{π}{4 a^{3} b})$
$\Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{d x}{{(a^{2} \cos^{2} x + b^{2} \sin^{2} x)}^{2}}$

$= \frac{π}{2 a b^{3}} (1 + (\frac{b^{2}}{a^{2}} - 1) π)$

Question:45

Evaluate the following:
$\int_{0}^{1} x \log (1 + 2 x) d x$

Answer:

Given: $\int_{0}^{1} x \log (1 + 2 x) d x$
$Let 1 + 2 x = t$

$\Rightarrow 2 dx = dt$

$\Rightarrow At x = 0$

$t = 1 and at x = 1$

$\Rightarrow t = 3$

$\Rightarrow \int_{0}^{1} x \ln (1 + 2 x) d x = \frac{1}{4} \int_{1}^{3} (t - 1) \ln t dt . . . . (i)$

$\Rightarrow \int_{1}^{3} (t - 1) \ln t d t = \int_{1}^{3} t \ln t d t - \int_{1}^{3} \ln t d t$

$\Rightarrow Apply Integration by parts = \int t \ln t d t = \ln t \int t d t - \int \frac{d}{d t} (\ln t) (\int t d t) d t$

$= \frac{t^{2}}{2} \ln t - \frac{t^{2}}{4} \dots (2) = \int \ln t d t = \ln t \int d t - \int \frac{d}{d t} (\ln t) (\int d t) d t = t \ln t - t \dots (3)$
$\Rightarrow Put (2) and (3) in (1) = \frac{1}{4} \int_{1}^{3} t \ln t d t - \frac{1}{4} \int_{1}^{3} \ln t d t = \frac{1}{4} {[(\frac{t^{2}}{2} \ln t - \frac{t^{2}}{4}) - (t \ln t - t)]}_{1}^{3} = \frac{3}{8} \ln 3$

$\Rightarrow \int_{0}^{1} x \ln (1 + 2 x) d x = \frac{3}{8} \ln 3$

Question:46

Evaluate the following:
$\int_{0}^{π} x \log \sin x d x$

Answer:

Given: $\int_{0}^{π} x \log \sin x d x$
$Using the property: \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$

$Let I = \int_{0}^{π} x \ln (\sin x) d x = \int_{0}^{π} (π - x) \ln (\sin (π - x)) d x = \int_{0}^{π} π \ln (\sin x) d x \int_{0}^{π} x \ln (\sin x) d x$

$\Rightarrow As \sin (π - x) = \sin x$
$\Rightarrow 2 I = \int_{0}^{π} π \ln (\sin x) d x = π \int_{0}^{π} \ln (\sin x) dx \dots (1)$

$\Rightarrow Now in \int_{0}^{π} \ln (\sin x) d x Using the property$

$\int_{0}^{2 a} f (x) d x = 2 \int_{0}^{a} f (x) d x (for f (2 a - x) = f (x)) = \int_{0}^{π} \ln (\sin x) d x = 2 \int_{0}^{\frac{π}{2}} \ln (\sin x) d x \dots (2)$
$\Rightarrow Let Z = \int_{0}^{\frac{π}{2}} \ln (\sin x) d x$

$\Rightarrow Using the property: \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$

$\Rightarrow z = \int_{0}^{\frac{π}{2}} \ln (\sin (\frac{π}{2} - x) d x = \int_{0}^{\frac{π}{2}} \ln (\cos x) d x \dots (4)$

$\Rightarrow 2 Z = \int_{0}^{\frac{π}{2}} \ln (\sin x) d x + \int_{0}^{\frac{π}{2}} \ln (\cos x) d x = \int_{0}^{\frac{π}{2}} \ln (\sin x \cos x) d x \dots (5)$
$= \int_{0}^{\frac{π}{2}} \ln (\sin x \cos x) d x = \int_{0}^{\frac{π}{2}} \ln (\frac{2 \sin x \cos x}{2}) d x$

$= \int_{0}^{\frac{π}{2}} \ln (\frac{2 \sin x \cos x}{2}) d x = \int_{0}^{\frac{π}{2}} (\ln (\sin 2 x) - \ln 2) d x$

$\Rightarrow \int_{0}^{\frac{π}{2}} (\ln (\sin 2 x)) d x - \int_{0}^{\frac{π}{2}} (\ln 2) d x = \int_{0}^{\frac{π}{2}} \ln (\sin 2 x) d x - \frac{π \ln 2}{2} \dots (6)$

$\Rightarrow Now in \int_{0}^{\frac{π}{2}} \ln (\sin 2 x) d x put 2 x = t$
$\Rightarrow Now in \int_{0}^{\frac{π}{2}} \ln (\sin 2 x) d x put 2 x = t$

$\Rightarrow 2 dx = dt and limits changes from 0 to π$

$\Rightarrow 2 Z = \frac{1}{2} \int_{0}^{π} \ln (\sin t) d t - \frac{π \ln 2}{2}$
$\Rightarrow from equation (2) \frac{1}{2} \int_{0}^{π} \ln (sint) dt again becomes$

$\Rightarrow 2 Z = \frac{2}{2} \int_{0}^{\frac{π}{2}} \ln (\sin t) dt - \frac{π \ln 2}{2}$
$\Rightarrow F r o m e q . (3)$

$\Rightarrow 2 Z = Z - \frac{π \ln 2}{2}$

$\Rightarrow Z = \int_{0}^{\frac{π}{2}} \ln (\sin x) dx = - \frac{π \ln 2}{2} . . . . . . (7)$
$\Rightarrow On putting (7) in (2) and the obtained result in(1)$

$\Rightarrow 2 I = - π^{2} \ln 2 I = \int_{0}^{π} x \ln (\sin x) d x = - \frac{π^{2}}{2} \ln 2$

Question:47

Evaluate the following:
$\int_{- \frac{π}{4}}^{\frac{π}{4}} \log (\sin x + \cos x) d x$

Answer:

Given: $\int_{- \frac{π}{4}}^{\frac{π}{4}} \log (\sin x + \cos x) d x$
$\begin{aligned} \frac{1}{Let} = \int_{- \frac{π}{4}}^{\frac{π}{4}} \ln (\sin x + \cos x) d x \dots (1) \\ Using the property: \\ \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x \\ \int_{- \frac{π}{4}}^{\frac{π}{4}} \ln (\sin x + \cos x) d x = \int_{- \frac{π}{4}}^{\frac{π}{4}} \ln (\sin (- x) + \cos (- x)) d x \\ \Rightarrow As \sin (- x) = \sin x and \cos (- x) = \cos x \end{aligned}$
$\begin{aligned} I = \int_{- \frac{π}{4}}^{\frac{π}{4}} \ln (\cos x - \sin x) d x \dots (2) \\ Adding equation(1) and(2) \\ 2 I = \int_{- \frac{π}{4}}^{\frac{π}{4}} \ln (\sin x + \cos x) dx + \int_{- \frac{π}{4}}^{\frac{π}{4}} \ln (\cos x - \sin x) dx \\ 2 I = \int_{- \frac{π}{4}}^{\frac{π}{4}} \ln (\cos^{2} x - \sin^{2} x) d x = \int_{- \frac{π}{4}}^{\frac{π}{4}} \ln (\cos 2 x) d x \end{aligned}$
$\Rightarrow Put 2 x = t$

$\Rightarrow 2 x dx = dt 2 I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \ln (\cos t) dt$

$\begin{aligned} \Rightarrow As \cos (- x) = \cos x \\ Using property: \\ \int_{- a}^{a} f (x) d x = 2 \int_{0}^{a} f (x) d x (f o r f (- x) = f (x)) \\ \Rightarrow 2 I = 2 \int_{0}^{\frac{π}{2}} \ln (\cos t) dt \\ Using the property: \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x \end{aligned}$
$\begin{aligned} 2 I = \int_{0}^{\frac{π}{2}} \ln (\cos (\frac{π}{2} - t)) dt = \int_{0}^{\frac{π}{2}} \ln (sint) dt \\ \Rightarrow Now From previous question eq (7) we obtained \\ \int_{0}^{\frac{π}{2}} \ln (sint) dt = - \frac{π \ln 2}{2} = 2 I \\ I = \int_{- \frac{π}{4}}^{\frac{π}{4}} \ln (\sin x + \cos x) d x = - \frac{π \ln 2}{4} \end{aligned}$

Question:48

$\int \frac{\cos 2 x - \cos 2 θ}{\cos x - \cos θ} d x$ is equal to

$A . 2 (s i n x + x c o s θ) + C B . 2 (s i n x - x c o s θ) + C C . 2 (s i n x + 2 x c o s θ) + C D . 2 (s i n x - 2 x c o s θ) + C$

Answer:

$\begin{aligned} Using Trigonometric identity \cos 2 x = 2 \cos^{2} x - 1 \\ \Rightarrow \int \frac{\cos 2 x - \cos 2 θ}{\cos x - \cos θ} d x = \int \frac{(2 \cos^{2} x - 1) - (2 \cos^{2} θ - 1)}{\cos x - \cos θ} d x \end{aligned}$
$= 2 \int \frac{\cos^{2} x - \cos^{2} θ}{\cos x - \cos θ} d x = 2 \int (\frac{(\cos x + \cos θ) (\cos x - \cos θ)}{\cos x - \cos θ})$

$= 2 {(\cos x + \cos θ) d x = 2 \int \cos x d x + 2 \int \cos θ d x$
$= 2 \int \cos x d x + 2 \int \cos θ d x = 2 (\sin x + x \cos θ) + c$

Question:49

$\int \frac{d x}{\sin (x - a) \sin (x - b)}$ is equal to

$\\ A. \sin (b-a) \log \left|\frac{\sin (x-b)}{\sin (x-a)}\right|+C$\\\\ B. $\operatorname{cosec}(b-a) \log \left|\frac{\sin (x-a)}{\sin (x-b)}\right|+C$\\\\ C. $\operatorname{cosec}(b-a) \log \left|\frac{\sin (x-b)}{\sin (x-a)}\right|+C$\\\\ D. $\sin (b-a) \log \left|\frac{\sin (x-a)}{\sin (x-b)}\right|+C$\\$

Answer:

$\begin{aligned} Given: \int \frac{d x}{\sin (x - a) \sin (x - b)} \\ Multiply Nr and Dr by \sin (b - a) \\ \Rightarrow \frac{1}{\sin (b - a)} \int \frac{\sin (b - a) d x}{\sin (x - a) \sin (x - b)} \end{aligned}$
$\Rightarrow \sin (b - a) = \sin ((x - a) - (x - b))$

$\Rightarrow Also \sin (A - B) = \sin A \cos B - \cos A \sin B$

$\Rightarrow \frac{\sin (b - a)}{\sin (x - a) \sin (x - b)} = \frac{\sin ((x - a) - (x - b))}{\sin (x - a) \sin (x - b)} = \frac{\sin (x - a) \cos (x - b) - \cos (x - a) \sin (x - b)}{\sin (x - a) \sin (x - b)}$

$\Rightarrow \frac{\sin (x - a) \cos (x - b) - \cos (x - a) \sin (x - b)}{\sin (x - a) \sin (x - b)} = \frac{\cos (x - b)}{\sin (x - b)} - \frac{\cos (x - a)}{\sin (x - a)}$
$= \frac{\cos (x - b)}{\sin (x - b)} - \frac{\cos (x - a)}{\sin (x - a)}$

$= \cot (x - b) - \cot (x - a) \int \frac{d x}{\sin (x - a) \sin (x - b)}$

$= \frac{1}{\sin (b - a)} (\int \cot (x - b) d x - \int \cot (x - a) d x)$

$\Rightarrow Now \int \cot x d x = \ln | \sin x | + c$

$\Rightarrow \frac{1}{\sin (b - a)} (\int \cot (x - b) d x - \int \cot (x - a) d x)$
$= \frac{1}{\sin (b - a)} (\ln | \sin (x - b) | - \ln | \sin (x - a) |)$ $\Rightarrow \int \frac{dx}{\sin (x - a) \sin (x - b)} = \frac{1}{\sin (b - a)} \ln (\frac{\sin (x - b)}{\sin (x - a)}) = cosec (b - a) \ln (\frac{\sin (x - b)}{\sin (x - a)})$

Question:50

$\int \tan^{- 1} \sqrt{x} d x$ is equal to
$\\A.(x+1) \tan ^{-1} \sqrt{x}-\sqrt{x}+C$\\ B. $x \tan ^{-1} \sqrt{x}-\sqrt{x}+C$\\ C. $\sqrt{x}-x \tan ^{-1} \sqrt{x}+C$\\ D. $\sqrt{x}-(x+1) \tan ^{-1} \sqrt{x}+C$$

Answer:

Given: $\int \tan^{- 1} \sqrt{x} d x$
$Put x = t^{2} \Rightarrow d x = 2 d t$

$\Rightarrow \int \tan^{- 1} \sqrt{x} d x = \int 2 \tan^{- 1} \sqrt{t^{2}} d t$

$\Rightarrow 2 \int \tan^{- 1} tdt \dots (1)$
$\Rightarrow Now apply integration by part on \int t \tan^{- 1} t d t$

$\Rightarrow \int t \tan^{- 1} t d t = \tan^{- 1} t \int t d t - \int (\frac{d}{d t} \tan^{- 1} t) (\int t d t) d t$

$\Rightarrow \frac{t^{2}}{2} \tan^{- 1} t - \frac{1}{2} \int \frac{t^{2}}{1 + t^{2}} d t$

$\Rightarrow Now \int \frac{t^{2}}{1 + t^{2}} d t = \int \frac{t^{2} + 1 - 1}{1 + t^{2}} d t = \int \frac{t^{2} + 1}{1 + t^{2}} d t - \int \frac{1}{1 + t^{2}} d t$
$\Rightarrow \int \frac{t^{2} + 1}{1 + t^{2}} dt - \int \frac{1}{1 + t^{2}} dt = \int dt - \int \frac{1}{1 + t^{2}} dt = t - \tan^{- 1} t \dots$

$\Rightarrow Put (3) in (2) and the resulting equation in (1)$

$\Rightarrow 2 \int t \tan^{- 1} t d t = 2 (\frac{t^{2}}{2} \tan^{- 1} t - \frac{1}{2} (t - \tan^{- 1} t))$

$\Rightarrow 2 \int t \tan^{- 1} t dt = t^{2} \tan^{- 1} t - t + \tan^{- 1} t$
$\Rightarrow t^{2} \tan^{- 1} t - t + \tan^{- 1} t = \tan^{- 1} t (t^{2} + 1) - t$

$\Rightarrow \tan^{- 1} t (t^{2} + 1) - t = \tan^{- 1} \sqrt{x} (x + 1) - \sqrt{x}$

$\Rightarrow \int \tan^{- 1} \sqrt{x} d x = \tan^{- 1} \sqrt{x} (x + 1) - \sqrt{x} + C$

Question:51

$\int e^{x} {(\frac{1 - x}{1 + x^{2}})}^{2} d x$ is equal to
$\\A. \frac{e^{x}}{1+x^{2}}+C \\ B. \frac{-\mathrm{e}^{\mathrm{x}}}{1+\mathrm{x}^{2}}+\mathrm{C}$\\ C.$\frac{\mathrm{e}^{\mathrm{x}}}{\left(1+\mathrm{x}^{2}\right)^{2}}+\mathrm{C}$\\ D. $\frac{-\mathrm{e}^{\mathrm{x}}}{\left(1+\mathrm{x}^{2}\right)^{2}}+\mathrm{C}$$

Answer:

Given: $\int e^{x} {(\frac{1 - x}{1 + x^{2}})}^{2} d x$
$= \int e^{x} {(\frac{1 - x}{1 + x^{2}})}^{2} d x = \int e^{x} (\frac{1 + x^{2} - 2 x}{{(1 + x^{2})}^{2}}) d x$

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

$= \int e^{x} {(\frac{1}{(1 + x^{2})}) + (\frac{- 2 x}{{(1 + x^{2})}^{2}})} d x$
$\begin{aligned} Now using the property: \\ \int e^{x} (f (x) + f^{'} (x)) d x = e^{x} f (x) \\ \Rightarrow Now in \int e^{x} {(\frac{1}{(1 + x^{2})}) + (\frac{- 2 x}{{(1 + x^{2})}^{2}})} d x \\ f (x) = \frac{1}{(1 + x^{2})} \\ \Rightarrow f^{'} (x) = \frac{- 2 x}{{(1 + x^{2})}^{2}} \end{aligned}$
$= \int e^{x} {(\frac{1}{(1 + x^{2})}) + (\frac{- 2 x}{{(1 + x^{2})}^{2}})} d x = \frac{e^{x}}{1 + x^{2}} + c \int e^{x} {(\frac{1 - x}{1 + x^{2}})}^{2} d x = \frac{e^{x}}{1 + x^{2}} + c$

Question:52

$\int \frac{x^{9}}{{(4 x^{2} + 1)}^{6}} d x$ is equal to
$\begin{aligned} A . & \frac{1}{5 x} {(4 + \frac{1}{x^{2}})}^{- 5} + C \\ B . & \frac{1}{5} {(4 + \frac{1}{x^{2}})}^{- 5} + C \\ C . & \frac{1}{10 x} (1 + 4)^{- 5} + C \\ D . & \frac{1}{10} {(\frac{1}{x^{2}} + 4)}^{- 5} + C \end{aligned}$

Answer:

Given: $\int \frac{x^{9}}{{(4 x^{2} + 1)}^{6}} d x$
Taking $x^{2}$ out from the denominator
$\int (\frac{x^{9}}{x^{12} {(4 + \frac{1}{x^{2}})}^{6}}) d x$

$\Rightarrow \int (\frac{1}{x^{3} {(4 + \frac{1}{x^{2}})}^{6}}) dx = \int (\frac{\frac{1}{x^{3}}}{{(4 + \frac{1}{x^{2}})}^{6}}) dx$

$\Rightarrow Now put 4 + \frac{1}{x^{2}} = t$
$\Rightarrow - \frac{2}{x^{3}} d x = d t$

$\Rightarrow \int (\frac{\frac{1}{x^{3}}}{{(4 + \frac{1}{x^{2}})}^{6}}) d x = \int - \frac{1}{2 t^{6}} d t$

$\Rightarrow - \frac{1}{2} t^{- 6} d t = \frac{- t^{- 5}}{- 5 \times 2} + C = \frac{1}{10} {(4 + \frac{1}{x^{2}})}^{- 5} + C$

Question:53

If $\int \frac{d x}{(x + 2) (x^{2} + 1)} = a \log | 1 + x^{2} | + b \tan^{- 1} x + \frac{1}{5} \log | x + 2 | + c$ then
$\\ A.a=\frac{-1}{10}, b=\frac{-2}{5}$\\ B. $a=\frac{1}{10}, b=-\frac{2}{5}$\\ C. $\mathrm{a}=\frac{-1}{10}, \mathrm{~b}=\frac{2}{5}$\\ D. $a=\frac{1}{10}, b=\frac{2}{5}$\\$

Answer:

Given: $\int \frac{d x}{(x + 2) (x^{2} + 1)} = a \log | 1 + x^{2} | + b \tan^{- 1} x + \frac{1}{5} \log | x + 2 | + c$
Using concept of partial fractions
$= \frac{1}{(x + 2) (x^{2} + 1)} = \frac{A}{(x + 2)} + \frac{B x + C}{(x^{2} + 1)} \Rightarrow A (x^{2} + 1) + (B x + C) (x + 2) = 1$
$\Rightarrow x^{2} (A + B) + x (C + 2 B) + (A + 2 C) = 1$

$\Rightarrow A + B = 0 \dots (1)$

$\Rightarrow C + 2 B = 0 \dots (2)$

$\Rightarrow A + 2 C = 1. . . (3)$
On solving the above three equations we get
$\Rightarrow A = \frac{1}{5}, B = - \frac{1}{5} and C = \frac{2}{5}$

$\Rightarrow \frac{1}{(x + 2) (x^{2} + 1)} = \frac{\frac{1}{5}}{(x + 2)} + \frac{- \frac{1}{5} x + \frac{2}{5}}{(x^{2} + 1)}$ $\Rightarrow \frac{\frac{1}{5}}{(x + 2)} + \frac{- \frac{1}{5} x + \frac{2}{5}}{(x^{2} + 1)} = \frac{1}{5 (x + 2)} - \frac{x}{5 (x^{2} + 1)} + \frac{2}{5 (x^{2} + 1)}$
$\int \frac{d x}{(x + 2) (x^{2} + 1)}$

$= \int (\frac{1}{5 (x + 2)} - \frac{x}{5 (x^{2} + 1)} + \frac{2}{5 (x^{2} + 1)}) d x$

$= \frac{1}{5} \ln | x + 2 | - \frac{1}{10} \ln | x^{2} + 1 | + \frac{2}{5} \tan^{- 1} x + c \dots (2)$
On comparing (1) and (2) we get,
$\Rightarrow a = - \frac{1}{10} and b = \frac{2}{5}$

Question:54

$\int \frac{x^{3}}{x + 1}$ is equal to

$\\A. x+\frac{x^{2}}{2}+\frac{x^{3}}{3}-\log |1-x|+C$\\ B. $x+\frac{x^{2}}{2}-\frac{x^{3}}{3}-\log |1-x|+C$\\ C. $x-\frac{x^{2}}{2}-\frac{x^{3}}{3}-\log |1+x|+C$\\ D. $x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\log |1+x|+C$\\$

Answer:

Given: $\int \frac{x^{3}}{x + 1}$
$\frac{x^{3}}{x + 1} = \frac{x^{3} + 1 - 1}{x + 1}$

$\Rightarrow \frac{x^{2} + 1}{x + 1} - \frac{1}{x + 1} = \frac{(x + 1) (x^{2} - x + 1)}{x + 1} - \frac{1}{x + 1}$
$\Rightarrow \int \frac{x^{3}}{x + 1} d x = \int ((x^{2} - x + 1) - \frac{1}{x + 1}) d x$

$\Rightarrow \int (x^{2} - x + 1) d x - \int \frac{1}{x + 1} d x = \frac{x^{3}}{3} - \frac{x^{2}}{2} + x - \ln | 1 + x | + c$

Question:55

$\int \frac{x + \sin x}{1 + \cos x} d x$ is equal to

$\\A. \log |1+\cos x|+c$\\ B. $\log |\mathrm{x}+\sin \mathrm{x}|+\mathrm{C}$\\ C. $x-\tan \frac{x}{2}+C$\\ D. $x \cdot \tan \frac{x}{2}+C$\\$

Answer:

$D. x \cdot \tan \frac{x}{2}+C$\\$

Given: $\int \frac{x + \sin x}{1 + \cos x} d x$
As we know
$\sin 2 x = \frac{2 \tan x}{1 + \tan^{2} x}, 1 + \tan^{2} x = \sec^{2} x and \cos 2 x = \frac{1 + \tan^{2} x}{1 - \tan^{2} x}$
$\Rightarrow \frac{x + \sin x}{1 + \cos x}$ $= \frac{x + \frac{2 \tan (\frac{x}{2})}{1 + \tan^{2} (\frac{x}{2})}}{1 + \frac{1 + \tan^{2} (\frac{x}{2})}{1 - \tan^{2} (\frac{x}{2})}}$
$\Rightarrow \frac{x + x \tan^{2} (\frac{x}{2}) + 2 \tan (\frac{x}{2})}{2}$

$\Rightarrow \int \frac{x + \sin x}{1 + \cos x} d x = \int \frac{x + x \tan^{2} (\frac{x}{2}) + 2 \tan (\frac{x}{2})}{2} d x \Rightarrow let \frac{x}{2} = t \Rightarrow \frac{d x}{2} = d t$
$\Rightarrow \int \left(2 t+2 t \tan ^{2} t+2 \tan t\right) d t=2 \int \left(t+t \tan ^{2 }t+\tan t \right) d t \\ \Rightarrow 2 \int \mathrm{tdt}+2 \int \mathrm{t} \left( \sec ^{2} \mathrm{t}-1 \right) \mathrm{dt}+2 \int \left(\tan \mathrm{t} \mathrm{dt}\right) \\ \Rightarrow 2 \int operatorname{tdt}+2 \int t \sec ^{2} t d t-2 \int operatorname{tdt}+2 \int \tan t d t\\ \Rightarrow 2 \int \mathrm{t} \sec ^{2} \mathrm{t} \mathrm{dt}+2 \int \text{ tant } \mathrm{dt} \ldots .(1)$ $\\\text{Applying Integration by parts on } \int \mathrm{t} \sec ^{2} \mathrm{t} dt \\ \Rightarrow \int t \sec ^{2} t d t=t \int \sec ^{2} t d t-\int\left(\frac{d}{d t} t\right)\left(\int \sec ^{2} t d t\right) d t $$$ $\\ \begin{aligned} &\Rightarrow \int \mathrm{t} \sec ^{2} \mathrm{t} \mathrm{dt}=\mathrm{t} \tan \mathrm{t}-\int \mathrm{tan} \mathrm{t} \mathrm{dt} \ldots(2)\\ \end{aligned}$

$\begin{aligned} \Rightarrow Put (2) in (1) \\ \Rightarrow 2 (t tant - \int \tan t dt) + 2 \int tant dt = 2 t tant + c = xtan (\frac{x}{2}) + c \\ \Rightarrow \int \frac{x + \sin x}{1 + \cos x} d x = xtan (\frac{x}{2}) + c \end{aligned}$

Question:56

If $\frac{x^{3} d x}{\sqrt{1 + x^{2}}} = a {(1 + x^{2})}^{\frac{3}{2}} + b \sqrt{1 + x^{2}} + C$ , then
$\\ A.a=\frac{1}{3}, b=1$\\ B. $a=\frac{-1}{3}, b=1$\\ C.$a=\frac{-1}{3}, b=-1$\\ D. $a=\frac{1}{3}, b=-1$\\$

Answer:

Given: $\frac{x^{3} d x}{\sqrt{1 + x^{2}}} = a {(1 + x^{2})}^{\frac{3}{2}} + b \sqrt{1 + x^{2}} + C$ ...........(1)
$\begin{aligned} Put \\ 1 + x^{2} = t \\ \Rightarrow 2 x d x = d t \\ \Rightarrow \int \frac{x^{3} d x}{\sqrt{1 + x^{2}}} = \int \frac{x^{2} \cdot x d x}{\sqrt{1 + x^{2}}} \end{aligned}$
$\Rightarrow= \frac{1}{2} \int \frac{(t - 1) dt}{\sqrt{t}}$

$\Rightarrow \frac{1}{2} \int \frac{(t - 1) d t}{\sqrt{t}}$

$\Rightarrow= \frac{1}{2} \int \frac{t dt}{\sqrt{t}} - \frac{1}{2} \int \frac{dt}{\sqrt{t}}$

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

$\Rightarrow (\frac{1}{3} t^{\frac{3}{2}} - \sqrt{t} + c) = \frac{1}{3} {(1 + x^{2})}^{\frac{3}{2}} - 1 \sqrt{1 + x^{2}} + c Comparing (1) and (3)$

$\Rightarrow a = \frac{1}{3} and b = - 1$

Question:57

$\int_{\frac{- π}{4}}^{\frac{π}{4}} \frac{d x}{1 + \cos 2 x}$ is equal to
A. 1
B. 2
C. 3
D. 4

Answer:

Given: $\int_{\frac{- π}{4}}^{\frac{π}{4}} \frac{d x}{1 + \cos 2 x}$
Using trigonometric identities:
$\cos^{2} x + \sin^{2} x = 1 and \cos 2 x = \cos^{2} x - \sin^{2} x$

$\frac{1}{1 + \cos 2 x} = \frac{1}{(\cos^{2} x + \sin^{2} x + \cos^{2} x - \sin^{2} x)} = \frac{1}{2 \cos^{2} x}$
$\int_{- \frac{π}{4}}^{\frac{π}{4}} (\frac{d x}{1 + \cos 2 x}) = \int_{- \frac{π}{4}}^{\frac{π}{4}} (\frac{d x}{2 \cos^{2} x}) = \frac{1}{2} \int_{- \frac{π}{4}}^{\frac{π}{4}} \sec^{2} x d x$

$= \frac{1}{2} \int_{- \frac{π}{4}}^{\frac{π}{4}} \sec^{2} x d x = \frac{1}{2} [\tan x]_{- \frac{π}{4}}^{\frac{π}{4}} = \frac{1 + 1}{2} = 1$

Question:58

$\int_{0}^{\frac{π}{2}} \sqrt{1 - \sin 2 x} d x$ is equal to
A. $2 \sqrt{2}$
B. $2 (\sqrt{2} + 1)$
C. $2$
D. $2 (\sqrt{2} - 1)$

Answer:

D)
As

$\sin 2 x = 2 \sin x \cos x and \sin^{2} x + \cos^{2} x = 1$

$\Rightarrow \int_{0}^{\frac{π}{2}} \sqrt{1 - \sin 2 x} d x = \int_{0}^{\frac{π}{2}} \sqrt{\sin^{2} x + \cos^{2} x - 2 \sin x \cos x} d x$

$\Rightarrow \int_{0}^{\frac{π}{2}} \sqrt{(\cos x - \sin x)^{2}} d x = \int_{0}^{\frac{π}{2}} | (\cos x - \sin x) | d x$
$From 0 < x < \frac{π}{4}, \cos x > \sin x and from \frac{π}{4} < x < \frac{π}{2}, \cos x < \sin x$

$\Rightarrow \int_{0}^{\frac{π}{2}} | (\cos x - \sin x) | d x = \int_{0}^{\frac{π}{4}} (\cos x - \sin x) d x + \int_{\frac{π}{4}}^{\frac{π}{2}} (\sin x - \cos x) d x$
$= 2 [\sin x + \cos x]_{0}^{π / 4}$
On solving the Above Integral we get $2 (\sqrt{2} - 1)$

Question:59

Fill in the blanks in each of the following
$\int_{0}^{\frac{π}{2}} \cos x e^{\sin x} d x$ is equal to ___________.

Answer:

$e - 1$
Given: $\int_{0}^{\frac{π}{2}} \cos x e^{\sin x} d x$
$Put \sin x = t \Rightarrow \cos x d x = d t$

$\Rightarrow At x = 0$

$\Rightarrow t = 0 and x = \frac{π}{2}$

$\Rightarrow t = 1$

$\Rightarrow \int_{0}^{1} e^{t} d t = {[e^{t}]}_{0}^{1} = e - 1$

Question:60

Fill in the blanks in each of the following
$\int \frac{x + 3}{(x + 4)^{2}} e^{x} d x =$ ___________.

Answer:

$\begin{aligned} \frac{e^{x}}{x + 4} + c \\ Given \\ \int \frac{x + 3}{(x + 4)^{2}} e^{x} d x \\ = \int \frac{x + 3}{(x + 4)^{2}} e^{x} d x = \int \frac{(x + 3) + 1 - 1}{(x + 4)^{2}} e^{x} d x = \int \frac{(x + 4) - 1}{(x + 4)^{2}} e^{x} d x \\ = \int \frac{(x + 4) - 1}{(x + 4)^{2}} e^{x} d x = \int e^{x} (\frac{(x + 4)}{(x + 4)^{2}} - \frac{1}{(x + 4)^{2}}) d x \end{aligned}$
$\begin{aligned} \int e^{x} (\frac{(x + 4)}{(x + 4)^{2}} - \frac{1}{(x + 4)^{2}}) d x = \int e^{x} (\frac{1}{(x + 4)} - \frac{1}{(x + 4)^{2}}) d x \\ Now using the property: \\ \int e^{x} (f (x) + f^{'} (x)) d x = e^{x} f (x) \\ \Rightarrow Now in \int e^{x} (\frac{1}{(x + 4)} - \frac{1}{(x + 4)^{2}}) dx \end{aligned}$
$\Rightarrow f (x) = \frac{1}{x + 4}$

$\Rightarrow f^{'} (x) = - \frac{1}{(x + 4)^{2}}$

$\Rightarrow \int e^{x} (\frac{1}{(x + 4)} - \frac{1}{(x + 4)^{2}}) d x = \frac{e^{x}}{x + 4} + c$

$\Rightarrow \int \frac{x + 3}{(x + 4)^{2}} e^{x} d x = \frac{e^{x}}{x + 4} + c$

Question:61

Fill in the blanks in each of the following
If $\int_{0}^{a} \frac{1}{1 + 4 x^{2}} d x = \frac{π}{8}$ , then a = ____________.

Answer:

$a = \frac{1}{2}$

$Given: \int_{0}^{a} \frac{1}{1 + 4 x^{2}} dx = \frac{π}{8}$

$\frac{1}{1 + 4 x^{2}} = \frac{\frac{1}{4}}{\frac{1}{4} + x^{2}} = \frac{\frac{1}{4}}{{(\frac{1}{2})}^{2} + x^{2}}$

$\int_{0}^{a} \frac{1}{1 + 4 x^{2}} dx = \int_{0}^{a} \frac{\frac{1}{4}}{{(\frac{1}{2})}^{2} + x^{2}} dx$
$Now \int \frac{d x}{x^{2} + a^{2}} = \frac{1}{a} \tan^{- 1} (\frac{x}{a}) + c$

$\int_{0}^{a} \frac{\frac{1}{4}}{{(\frac{1}{2})}^{2} + x^{2}} d x = \frac{\frac{1}{4}}{\frac{1}{2}} \tan^{- 1} (\frac{x}{\frac{1}{2}})$

$= {[\frac{1}{2} \tan^{- 1} (2 x)]}_{0}^{a} = \frac{π}{8}$

$\frac{1}{2} \tan^{- 1} (2 a) = \frac{π}{8}$

$2 a = \tan (\frac{π}{4}) = 1$

$a = \frac{1}{2}$

Question:62

Fill in the blanks in each of the following
$\int \frac{\sin x}{3 + 4 \cos^{2} x} d x =$ __________.

Answer:

$\begin{aligned} - \frac{1}{2 \sqrt{3}} \tan^{- 1} (\frac{2 \cos x}{\sqrt{3}}) + c \\ Given \\ \int \frac{\sin x}{3 + 4 \cos^{2} x} d x \\ \Rightarrow Now let \cos x = t \end{aligned}$
$\Rightarrow - \sin xdx = dt$

$\Rightarrow \int \frac{\sin x}{3 + 4 \cos^{2} x} dx = \int \frac{- dt}{3 + 4 t^{2}} = \frac{1}{4} \int \frac{- dt}{\frac{3}{4} + t^{2}}$

$\Rightarrow \frac{1}{4} \int \frac{- dt}{{(\frac{\sqrt{3}}{2})}^{2} + t^{2}}$
$\Rightarrow Now \int \frac{d x}{x^{2} + a^{2}} = \frac{1}{a} \tan^{- 1} (\frac{x}{a}) + c = \frac{1}{4} \int \frac{- d t}{{(\frac{2}{2})}^{2} + t^{2}} = \frac{- \frac{1}{4}}{\frac{\sqrt{3}}{2}} \tan^{- 1} (\frac{2 t}{\sqrt{3}}) + c$

$\Rightarrow \int \frac{\sin x}{3 + 4 \cos^{2} x} d x = - \frac{1}{2 \sqrt{3}} \tan^{- 1} (\frac{2 \cos x}{\sqrt{3}}) + c$

Question:63

Fill in the blanks in each of the following
The value of $\int_{- π}^{π} \sin^{3} x \cos^{2} x d x$ is ____________.

Answer:

Using the property: $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$
$Let I = \int_{- π}^{π} \sin^{3} x \cos^{2} x d x$

$\Rightarrow \int_{- π}^{π} \sin^{3} x \cos^{2} x d x = \int_{- π}^{π} \sin^{3} (π + (- π) - x) \cos^{2} (π + (- π) - x) d x$

$\Rightarrow As \sin (- x) = - \sin x and \cos (- x) = \cos x$
$= \int_{- π}^{π} \sin^{3} (π + (- π) - x) \cos^{2} (π + (- π) - x) d x = \int_{- π}^{π} \sin^{3} (- x) \cos^{2} (- x) dx$

$\Rightarrow \int_{- π}^{π} \sin^{3} (- x) \cos^{2} (- x) dx = - \int_{- π}^{π} \sin^{3} x \cos^{2} x dx = - I$

$\Rightarrow I = - I \Rightarrow 2 I = 0$
$\Rightarrow I = \int_{- π}^{π} \sin^{3} x \cos^{2} x d x = 0$

NCERT Exemplar Class 12 Maths Solutions

NCERT Exemplar Class 12 Mathematics Chapter 1: Relations and Functions

NCERT Exemplar Class 12 Mathematics Chapter 2: Inverse Trigonometric Functions

NCERT Exemplar Class 12 Mathematics Chapter 3: Matrices

NCERT Exemplar Class 12 Mathematics Chapter 4: determinants

NCERT Exemplar Class 12 Mathematics Chapter 5: Continuity and Differentiability

NCERT Exemplar Class 12 Mathematics Chapter 6: Applications of derivatives

NCERT Exemplar for Class 12 Mathematics Chapter 7: Integrals

NCERT Exemplar Class 12 Mathematics Chapter 8: Applications of Integrals

NCERT Exemplar Class 12 Mathematics Chapter 9: Differential equations

NCERT Exemplar Class 12 Mathematics Chapter 10: Vector Algebra

NCERT Exemplar Class 12 Mathematics Chapter 11: Three Dimensional Geometry

NCERT Exemplar Class 12 Mathematics Chapter 12: Linear Programming

NCERT Exemplar Class 12 Mathematics Chapter 13: Probability

Importance of NCERT Exemplar Class 12 Maths Solutions Chapter 7

Class 12 Maths NCERT exemplar solutions Chapter 7 Integrals touches upon an exhaustive explanation of how we do integration by using properties.

These solutions provide a wide range of problems that will help students during their exams.
These solutions are easy to understand as they are well explained.
These also cover some major properties and rules, like King's property and Leibnitz's rule.

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NCERT Solutions for Class 12 Maths: Chapter Wise

NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions

NCERT solutions for class 12 maths chapter 2 Inverse Trigonometric Functions

NCERT solutions for class 12 maths chapter 3 Matrices

NCERT solutions for class 12 maths chapter 4 Determinants

NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability

NCERT solutions for class 12 maths chapter 6 Application of Derivatives

NCERT solutions for class 12 maths chapter 7 Integrals

NCERT solutions for class 12 maths chapter 8 Application of Integrals

NCERT solutions for class 12 maths chapter 9 Differential Equations

NCERT solutions for class 12 maths chapter 10 Vector Algebra

NCERT solutions for class 12 maths chapter 11 Three Dimensional Geometry

NCERT solutions for class 12 maths chapter 12 Linear Programming

NCERT solutions for class 12 maths chapter 13 Probability

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Indeed, the Class 12 Maths NCERT exemplar solutions chapter 7 covers the syllabus of the competitive exams like NEET and JEE Main to help you ace them.

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Methods of Integration, Integration by Parts and Fundamental Theorem of Calculus are some of the important topics of this chapter. However, rest should not be neglected either.

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The NCERT exemplar solutions for Class 12 Maths chapter 7 consists of 1 exercise with 63 distinct questions for practice.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Can I change my board from CBSE to CHSE Odisha in Class 12th?

Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.

Read Complete Answer

Manasvini Gupta 30 January,2025

quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

Read Complete Answer

Jefferson 25 September,2024

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

Re-evaluate Your Study Strategies:
- Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
- Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
- Practice Regularly: Consistent practice is key to mastering chemistry.
Consider Professional Help:
- Tutoring: A tutor can provide personalized guidance and support.
- Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
Explore Alternative Options:
- Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
- Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
Focus on NEET 2025 Preparation:
- Stay Dedicated: Continue your NEET preparation with renewed determination.
- Utilize Resources: Make use of study materials, online courses, and mock tests.
Seek Support:
- Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
- Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.

Read Complete Answer

Kanishka kaushikii 17 September,2024

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

Read Complete Answer

Yuvan 30 August,2024

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Read Complete Answer

Ashvadha 12 July,2024

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Central Board of Secondary Education Class 12th Examination

JEE Test Series

Applications Open

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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