RD Sharma Class 12 Exercise 18.8 Indefinite Integrals Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 18.8 Indefinite Integrals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 11:58 AM IST

RD Sharma books are considered the gold standard for CBSE. They are informative, easy to understand, and exam-oriented, making them the best choice for students.

RD Sharma Class 12th Exercise 18.8 covers the chapter ‘Indefinite Integrals.’ The questions from this exercise are divided into two parts Level 1and Level 2, based on difficulty and weightage. This exercise contains 52 questions, of which 44 are Level 1, and 8 are Level 2.

RD Sharma Class 12th Exercise 18.8 covers concepts like indefinite integration of trigonometric and logarithmic values. The Level 1 questions are fundamental and easy to solve if students know the basic concepts. In contrast, Level 2 sums are lengthier and require an in-depth understanding of the theorems.

## Indefinite Integrals Excercise:18.8

Indefinite Integrals exercise 18.8 question 1

$\frac{1}{\sqrt{2}}log\left | tan\frac{x}{2} \right |+C$
Hint:
$cos2x=1-2sin^{2}x$
Given:
$\int \! \frac{1}{\sqrt{1-cos2x}}dx$
Explanation:
$\int \! \frac{1}{\sqrt{2sin^{2}\: x}}dx=\frac{1}{\sqrt{2}}\int \frac{1}{sin\: x}dx$
$=\frac{1}{\sqrt{2}}\int cos \: ecxdx$ $\left [ \int cos \: ecx = log\left | tan\frac{x}{2} \right | \right ]$
$=\frac{1}{\sqrt{2}} log\left | tan\frac{x}{2} \right |+C$

Indefinite Integrals exercise 18.8 question 2

$\sqrt{2}log\left | tan\left ( \frac{\pi }{4}+\frac{x}{4} \right ) \right |+C$
Hint:
$cos 2\: x=2cos^{2}\: x-1$
Given:
$\int \frac{1}{\sqrt{1+cos\: x}}dx$
Explanation:
$\int \frac{1}{\sqrt{2cos^{2\: \frac{x}{2}}}}dx$ $\left[\begin{array}{c} 1+\cos 2 x=2 \cos ^{2} x \\ 1+\cos x=2 \cos ^{2} \frac{x}{2} \end{array}\right]$
$=\frac{1}{\sqrt{2}}\int \frac{1}{cos\frac{x}{2}}dx$
$=\frac{1}{\sqrt{2}}\int \! sec\frac{x}{2}dx$
$Let \frac{x}{2}=t$
$dx=2dt$
$=\frac{2}{\sqrt{2}} \int sec\: t\: dt =\sqrt{2} \int sec\: t\: dt$
$=\sqrt{2}\: log\left | tan \left ( \frac{\pi }{4}+\frac{t}{2} \right ) \right |+C$ $\because \int\! sec\: x\: dx= log\: \left | tan \left ( \frac{\pi }{4}+\frac{x}{2} \right ) \right |$
$=\sqrt{2}\: log\! \left | tan \left ( \frac{\pi }{4}+\frac{x}{2} \right ) \right |+C$

### Indefinite Integrals exercise 18.8 question 3Answer:

$log\left | sin\: x \right |+C$
Hint:
$cos\: 2x=2cos^{2}\: x-1=1-2sin^{2}x$
Given:
$\int \sqrt{\frac{1+cos\: 2x}{1-cos\: 2x}}dx$
Explanation:
$\int \sqrt{\frac{2cos^{2}\: x}{2sin^{2}\: x}}dx$ $\left[\begin{array}{c} 1+\cos 2 x=2 \cos ^{2} x \\ 1-\cos 2x=2 \sin ^{2} x \end{array}\right]$
$=\int \sqrt{\left(\frac{\cos x}{\sin x}\right)^{2}} d x$
$=\int cot\; xdx$
$=log\left | sin\; x \right |+C$

Indefinite Integrals exercise 18.8 question 4

$-2log\left | cos\frac{x}{2} \right |+C$
Hint:
$cos \: 2\: x=2cos^{2}\: x-1=1-2sin^{2}\: x$
Given:
$\int \sqrt{\frac{1-cos\: x}{1+cos\: x}}dx$
Explanation:
$\int \sqrt{\frac{2sin^{2}\frac{x}{2}}{2cos^{2}\frac{x}{2}}}dx$ $[cos\: x=2cos^{2}\: \frac{x}{2}-1\: \: or\; \; 1-2sin^{2}\frac{x}{2}]$
$\int tan\frac{x}{2}dx$
$Let \frac{x}{2}=t$
$dx=2dt$
$2\int tan\: tdt=-2log\left | cos\: t \right |+C$
$=-2log\left | cos\: \frac{x}{2} \right |+C$

Indefinite Integrals exercise 18.8 question 5

$2sin\: x-log\left | sec\: x+tan\; x \right |+C$
Hint:
$sec\: x=\frac{1}{cos\: x}$
Given:
$\int \frac{sec\: x}{sec\: 2x }dx$
Explanation:
$\int\left(\frac{1}{\cos x}\right)\left(\frac{1}{\frac{1}{\cos 2 x}}\right) d x$
$=\int\! \frac{cos\: 2x}{cos\: x}dx$
$=\int\! \frac{2cos^{2}\: x-1}{cos\: x}dx$
$=\int\!2\frac{cos^{2}x}{cos\: x}dx-\frac{1}{cos\: x}dx$
$=2\! \int\!cos\: xdx-\int\! sec\: xdx$
$=2 sin\; x-log\left | sec\: x +tan\: x\right |+C$

Indefinite Integrals exercise 18.8 question 6

$log\: log\left | sin\: sin\: x+cos\: cos\: x \right |+C$
Hint:
$cos^{2}\: x-sin^{2}\: x=cos2x$
Given:
$\int\! \frac{cos2x}{(cos\: x+sin\: x)^{2}}dx$
Explanation:
$\int\! \frac{cos^{2}-sin^{2}x}{(cos\: x+sin\: x)^{2}}dx$
$=\int\! \frac{(cos\: x+sin\: x)(cos\: x-sin\: x)}{(cos\: x+sin\: x)^{2}}dx$
$=\int\! \frac{cos\: x-sin\: x}{cos\: x+sin\: x}dx$
$Let \; \; cos\: x+sin\: x=t$
$(-sin\: x+cos\: x)dx=dt$
$=\int \! \frac{1}{t}dt$
$=log\left | t \right |+C$
$=log\left | cos\: x+sin\: x \right |+C$

Indefinite Integrals exercise 18.8 question 7

$xcos(b-a)+sin(b-a)log\left | sin(x-b) \right |+C$
Hint:
$sin(A+B)=sinA\: cosB+cosA\: sinB$
Given:
$\int \! \frac{sin(x-a)}{sin(x-b)}dx$
Explanation:
$\int \! \frac{sin(x-a)}{sin(x-b)}dx$
$=\int\frac{sin(x-a+b-b)}{sin(x-b)}dx$ $[add\; and\; subtract \: b\: in\; (x-a)]$
$=\int \! \frac{sin(x-b+b-a)}{sin(x-b)}dx$
$=\int \! \frac{sin(x-b)cos(b-a)+cos(x-b)sin(b-a)}{sin(x-b))}dx$ $[sin(A+B)=sinA\: cosB+cosA\: sinB]$
$=\int \frac{\sin (x-b) \cos (b-a)}{\sin (x-b)} d x+\int \frac{\cos (x-b) \sin (b-a)}{\sin (x-b)} d x$
$=\int\! cos(b-a)dx+\int \! cot(x-b)sin(b-a)dx$
$=cos(b-a)\int dx+sin(b-a)\int cot(x-b)dx$
$= cos(b-a)x+ sin(b-a)\: log\left | sin(x-b) \right |+C$
$[\int\! cot\: x\: dx= log\left | sin\; x \right |+C]$

### Indefinite Integrals exercise 18.8 question 8

$x\: cos\: cos\: 2a-sin\: sin\: 2a \: log\: log\left | sin\: sin(x+a) \right |+C$
Hint:
$sin(A+B)=sinA\; cosB-cosA\; sinB$
Given:
$\int \! \frac{sin(x-a)}{sin(x+a)}dx$
Explanation:
$\int \! \frac{sin(x-a+a-a)}{sin(x+a)}dx$ $[Add\; and\; subtract\; a\; in\; (x-a) ]$
$=\int \! \frac{sin(x+a-2a)}{sin(x+a)}dx$
$=\int \! \frac{sin(x+a)cos2a-cos(x+a)sin2a}{sin(x+a)}dx$
$=\int \! \frac{sin(x+a)cos2a}{sin(x+a)}dx-\int \! \frac{cos(x+a)sin2a}{sin(x+a)}dx$
$=\int \!cos2a\, dx-\int \! cot(x+a)sin2a\, dx$
$=cos2a\int \! dx-sin2a\int cot(x+a)dx$
$=cos2ax-sin2a\; log\left | sin(x+a) \right |+C$
$=xcos2a-sin2a\; log\left | sin(x+a) \right |+C$

Indefinite Integrals exercise 18.8 question 9

$-log\left | cos\: x-sin\: x \right |+C$
Hint:
$tan\: x=\frac{sin\: x}{cos\: x}$
Given:
$\int \! \frac{1+tan\: x}{1-tan\: x}dx$
Explanation:
$\int \! \frac{1+\frac{sin\: x}{cos\: x}}{1-\frac{sin\: x}{cos\: x}}dx=\int \! \frac{cos\: x+sin\: x}{cos\: x-sin\: x}dx$
Let,
$cos\: x-sin\: x=t$
$(-sin\: x-cos\: x)dx=dt$
$=\int \! \frac{dt}{t}$
$=-log\left | t \right |$
$=-log\left | cos\: x-sin\: x \right |+C$

Indefinite Integrals exercise 18.8 question 10

$(x-a)cos\: a-sin\: a\: log\left | sec(x-a) \right |+C$
Hint:
$cos(A+B)=cosA\: cosB-sinA\: sinB$
Given:
$\int \! \frac{cosx}{cos(x-a)}dx$
Explanation:
$\int \! \frac{cos(x-a+a)}{cos(x-a)}dx$[Add and subtract $a$ in $x$]
$=\int \! \frac{cos(x-a)cos\: a-sin(x-a)sin\: a}{cos(x-a)}dx$
$=\int \! \frac{cos(x-a)cos\: a}{cos(x-a)}dx-\int \! \frac{sin(x-a)sin\: a}{cos(x-a)}dx$
$=\int \! cos\: a\: dx-\int \! tan(x-a)sin\: a\: dx$
$\therefore xcos\: a-sin\: a\: log\left | sec(x-a) \right |+C$

### Indefinite Integrals exercise 18.8 question 11

$log\left | cos(\frac{\pi }{4}-x) \right |+C$
Hint:
$sin^{2}x+cos^{2}x=1\; and\; sin\: 2x=2sin\: x\, cos\: x$
Given:
$\int \! \sqrt{\frac{1-sin2x}{1+sin2x}}dx$
Explanation:
$\int \! \sqrt{\frac{sin^{2}x+cos^{2}x-2sin\: x\: cos\: x}{sin^{2}x+cos^{2}x+2sin\: x\: cos\: x}}dx$
$=\int \! \sqrt{\frac{(sin\: x-cos\: x)^{2}}{(sin\: x+cos\: x)^{2}}}dx$
$=\int \! \sqrt{(\frac{sin\: x-cos\: x}{sin\: x+cos\: x}})^{2}dx$
$=\int \! {\frac{sin\: x-cos\: x}{sin\: x+cos\: x}}dx$
$=\int \!\frac{\frac{sin\: x}{cos\: x}-\frac{cos\: x}{cos\: x}}{\frac{sin\: x}{cos\: x}+\frac{cos\: x}{cos\: x}}dx$
$=\int \!\frac{tan\: x-1}{tan\: x+1}dx$
$=\int \!\frac{tan\: x-tan\frac{\pi }{4}}{1+tan\: x\: tan\frac{\pi }{4}}dx$ $[tan\frac{\pi }{4}=1]$
$=\int \! tan(x-\frac{\pi }{4})dx$
$=-\int \! tan(\frac{\pi }{4}-x)dx$
$=-[-log\left | cos(\frac{\pi }{4}-x) \right |+C]$
$=log\left | cos(\frac{\pi }{4}-x) \right |+C$

Indefinite Integrals exercise 18.8 question 12

$\frac{1}{3}log\left | e^{3x}+1 \right |+C$
Hint:
$\int \! e^{x}dx=e^{x}+C\; \; \; \; or\; \; \; \; \int \! \frac{1}{x}dx=log\left | x \right |+C$
Given:
$\int \! \frac{e^{3x}}{e^{3x}+1}dx$
Explanation:
${e^{3x}+1}=t$
$3e^{3x}dx=dt$
$=\frac{1}{3}\! \int \! \frac{dt}{t}$
$=\frac{1}{3}log\left | t \right |$
$=\frac{1}{3}log\left | e^{3x} \right |+C$

Indefinite Integrals exercise 18.8 question 13

$\frac{1}{3}log\left | 3sec\; x+5 \right |C$
Himt:
$\int \! \frac{1}{x}dx=log\left | x \right |+C$
Given:
$\int \! \frac{sec\: x\: tan\: x}{3sec\: x+5}dx$
Explanation:
Let,
$3sec\: x+5=t$
$3sec\: x \: tan \: xdx=dt$
$=\int \! \frac{dt}{3t}=\frac{1}{3}\int \! \frac{dt}{t}$
$=\frac{1}{3}log \: t+C$
$=\frac{1}{3}log \: \left | 3sec\: x+5 \right |+C$

Indefinite Integrals exercise 18.8 question 14

$-log\left | sin\; x+cos\: x \right |+C$
Hint:
$Cot\: x=\frac{cos\: x}{sin\: x}$
Given:
$\int \! \frac{1-cot\: x}{1+cot\: x}dx$
Explanation:
$\int \! \frac{1-\frac{cos\: x}{sin\: x}}{1+\frac{cos\: x}{sin\: x}}dx$ $=\int \! \frac{sin\: x-cos\: x}{sin\: x+cos\: x}dx$
Let
$sin\: x+cos\: x=t$
$(cos\: x-sin\: x)dx=dt$
$=-\int \! \frac{dt}{t}\; \; \; \; =-log\: t+C$
$=-log\left | sin\: x+cos\: x \right |+C$

Indefinite Integrals exercise 18.8 question 15

$log\left | log(tan\: x) \right |+C$
Hint:
$tan\: x=\frac{sin\: x}{cos\: x},\: \: sec\: x=\frac{1}{cos\: x},\: \: cos\: ecx=\frac{1}{sin\: x}$
Given:
$\int \! \frac{sec\: x\: cos\: ecx}{log\left ( tan\: x \right )}dx$
Explanation:
Let
$log(tan\: x)=t$
$\frac{1}{tan\: x}(sec^{2}x)dx=dt$
$\Rightarrow \frac{sec\: x}{\frac{sin\: x}{cos\: x}}\times \frac{1}{cos\: x}dx=dt$
$\Rightarrow sec\: x\: cos\: ecx\: dx=dt$
$=\int \! \frac{dt}{t}\: \: \: =log\: t+C$
$=log\left | log(tan\: x) \right |+C$

Indefinite Integrals exercise 18.8 question 16

$log\left | 3+log\: x \right |+C$
Hint:
$\frac{\mathrm{d} }{\mathrm{d} x}(log\: x)=\frac{1}{x}$
Given:
$\int \! \frac{1}{x(3+log\: x)}dx$
Explanation:
Let
$(3+log\: x)=t$
$\frac{1}{x}dx=dt$
$=\int \! \frac{dt}{t}\: \: \: =log\: t+C$
$=log\left | 3+log\: x \right |+C$

Indefinite Integrals exercise 18.8 question 17

$log\left | e^{x}+x \right |+C$
Hint:
$\frac{\mathrm{d} }{\mathrm{d} x}(e^{x}+x)=e^{x}+1$
Given:
$\int \! \frac{e^{x}+1}{e^{x}+x}dx$
Explanation:
Let
$e^{x}+x=t$
$(e^{x}+1)dx=dt$
$=\int \! \frac{dt}{t}\: \: =log\: t+C$
$=log\left | e^{x} +x\right |+C$

Indefinite Integrals exercise 18.8 question 18

$log\left | log\: x \right |+C$
Hint:
$\frac{\mathrm{d} }{\mathrm{d} x}(log\: x)=\frac{1}{x}$
Given:
$\int \! \frac{1}{x\: log\: x}dx$
Explanation:
Let
$log\: x=t$
$\frac{1}{x}dx=dt$
$=\int \! \frac{dt}{t}$
$=log\: t+C$
$=log\left | log\: x \right |+C$

Indefinite Integrals exercise 18.8 question 19

$\frac{1}{b-a}In\left | a\: cos^{2}\: x+b\: sin^{2}\: x \right |+C$
Hint:
$\int \! \frac{dt}{t}=log\left | t \right |+C$
Given:
$\int \! \frac{sin\: 2x}{a\: cos^{2}\: x+b\: sin^{2}x}dx$........(1)
Explanation:
Let
$a\: cos^{2}\: x+b\: sin^{2}x=t$
$(2a\, cos\, x(-sin\, x)+b(2\, sin\, x)(cos\, x))dx=dt$
$(b-a)[2\, cos\, x\, sin\, x]dx=dt$
$(b-a)sin\, 2x\, dx=dt$
$sin\, 2x\, dx=\frac{dt}{b-a}$
Put in (1)
$\frac{1}{b-a}\int \! \frac{dt}{t}$
$=\frac{1}{b-a}In\left | t \right |+C$
$=\frac{1}{b-a}In\left | a\: cos^{2}\: x+b\: sin^{2} \right |+C$

Indefinite Integrals exercise 18.8 question 20

$\frac{1}{3}log\left | 2+3\, sin\, x \right |+C$
Hint:
$\frac{\mathrm{d} }{\mathrm{d} x}(a+b\, sin\, x)=cos\, x$
Given:
$\int \! \frac{cos\, x}{2+3\, sin\, x}dx$.....(1)
Explanation:
Let
$2+3\, sin\, x=t$
$3\, cos\, x\, dx=dt$
$cos\, x\, dx=\frac{dt}{3}$
Put in (1)
$\int \! \frac{dt}{3t}=\frac{1}{3}\int \! \frac{dt}{t}$
$=\frac{1}{3}log\left | t \right |+C$
$=\frac{1}{3}log\left | 2+3\, sin\, x\right |+C$

Indefinite Integrals exercise 18.8 question 21

$log\left | x+cos\, x \right |+C$
Hint:
$\frac{\mathrm{d} }{\mathrm{d} x}(x+cos\, x)=1-sin\, x$
Given:
$\int \! \frac{1-sin\, x}{x+cos\, x}dx$
Explanation:
Let
$x+cos\; x=t$
$(1-sin\, x)dx=dt$
$\int \! \frac{dt}{t}=log\left | t \right |+C$
$=log\left | x+cos\, x \right |+C$

Indefinite Integrals exercise 18.8 question 22

$-\frac{a}{b}log\left | be^{-x}+c \right |+C$
Hint:
$\int \! \frac{dt}{t}=log\left | t \right |+C$
Given:
$\int\! \! \frac{a}{b+ce^{x}}dx$
Explanation:
$\int\! \! \frac{a}{e^{x}(\frac{b}{e^{x}}+c)}dx$.......(1)
$=\int\! \! \frac{ae^{-x}}{be^{-x}+c}dx$
Let
$be^{-x}+c=t$
$-e^{-x}(b)dx=dt$
$e^{-x}dx=-\frac{1}{b}dt$
Put in (1)
$\int \! \frac{a (-\frac{1}{b})dt}{t}$
$=-\frac{a}{b}\int \! \frac{dt}{t}$
$=-\frac{a}{b}log\left | t \right |+C$
$=-\frac{a}{b}log\left | be^{-x}+c \right |+C$

Indefinite Integrals exercise 18.8 question 23

$-log\left | e^{-x}+1 \right |+C$
Hint:
$\int \! \frac{1}{x}dx=log\left | x \right |+C$
Given:
$\int \! \frac{1}{e^{x}+1}dx$
Explanation:
$\int \! \frac{1}{e^{x}(1+\frac{1}{e^{x}})}dx\; \; \; \; =\int \! \frac{e^{-x}}{e^{-x}+1}dx$....(1)
Let
$e^{-x}+1=t$
$e^{-x}dx=-dt$
From (1)
$\int \! \frac{-dt}{t}=-log\left | t \right |+C$
$=-log\left | e^{-x}+1 \right |+C$

Indefinite Integrals exercise 18.8 question 24

$log\left | log\: sin\: x \right |+C$
Hint:
$cot\: x=\frac{cos\: x}{sin\: x}$
Given:
$\int\! \frac{cot\: x}{log\: sin\: x}dx$ ........(1)
Explanation:
Let
$log\: sin\: x=t$
$\frac{1}{sin\: x}(cos\: x)dx=dt$
$cot\: x\: dx=dt$
Put in (1)
$\int\! \frac{dt}{t}=log\left | t \right |+C$
$=log\left | log\: sin\: x \right |+C$

Indefinite Integrals exercise 18.8 question 25

$\frac{1}{2}log\left | e^{2x}-2 \right |+C$
Hint:
$\int \frac{1}{t}dt=log\left | t \right |+C$
Given:
$\int \frac{e^{2x}}{e^{2x}-2}dx$ .........(1)
Explanation:
Let
$e^{2x}-2=t$
$2e^{2x}dx=dt$
$e^{2x}dx=\frac{dt}{2}$
Put (1) we get
$\int\! \frac{dt}{2t}=\frac{1}{2}\int \! \frac{dt}{t}$
$=\frac{1}{2}log\left | t \right |+C$
$=\frac{1}{2}log\left | e^{2x}-2 \right |+C$

Indefinite Integrals exercise 18.8 question 26

Amswer:
$\frac{1}{2}log\left | 6cos\, x+4sin\, x \right |+C$
Hint:
$\int \! \frac{1}{t}dt=log\left | t \right |+C$
Given:
$\int \! \frac{2cos\, x-3sin\, x}{6cos\, x+4sin\, x}dx$......(1)
Explanation:
Let
$6cos\, x+4sin\, x=t$
$(-6sin\, x+4cos\, x)dx=dt$
$2(2cos\, x-3sin\, x)dx=dt$
Put in (1)
$\int \! \frac{dt}{2t}$
$\frac{1}{2}\int \! \frac{dt}{t}=\frac{1}{2}log\left | t \right |+C$
$=\frac{1}{2}log\left | 6cos\, x+4sin\, x \right |+C$

Indefinite Integrals exercise 18.8 question 27

$\frac{1}{2}log\left | x^{2}+sin\, 2x+2x \right |+C$
Hint:
$\int\! \frac{1}{t}dt=log\left | t \right |+C$
Given:
$\int\! \frac{cos\, 2x+x+1}{x^{2}+sin\, 2x+2x}dx$.......(1)
Explanation:
Let
$x^{2}+sin\, 2x+2x=t$
$(2x+2cos\, 2x+2)dx=dt$
$2(x+cos\, 2x+1)dx=dt$
From (1)
$\frac{1}{2}\int \! \frac{dt}{t}=\frac{1}{2}log\left | t \right |+C$
$=\frac{1}{2}log\left | x^{2}+sin\, 2x+2x \right |+C$

Indefinite Integrals exercise 18.8 question 28

$\frac{1}{sin(b-a)}[log(\frac{sec(x+b)}{sec(x+a)})]+C$
Hint:
$M\! ultipl\! y\; and\; divide\; by\; sin(b-a)$
Given:
$\int \! \frac{1}{cos(x+a)cos(x+b)}dx$......(1)
Explanation:
$M\! ultipl\! y\; and\; divide\; by\; sin(b-a)$
From (1)
$=\frac{1}{sin(b-a)}\int \! \frac{sin(b-a)}{cos(x+a)cos(x+b)}dx$
$=\frac{1}{sin(b-a)}\int \! \frac{sin[(x+b)-(x+a)]}{cos(x+a)cos(x+b)}dx$ Add and subtract $x$ in $b-a$
$=\frac{1}{sin(b-a)}\int \! \frac{sin(x+b)cos(x+a)-sin(x+a)cos(x+b)}{cos(x+a)cos(x+b)}dx$
$=\frac{1}{sin(b-a)}[\int \! \frac{sin(x+b)}{cos(x+b)}dx-\int \! \frac{sin(x+a)}{cos(x+a)}dx]$
$=\frac{1}{sin(b-a)}[\int \! tan(x+b)dx-\int \! tan(x-a)dx]$
$=\frac{1}{sin(b-a)}[log\left | sec(x+b) \right |-log\left | sec(x+a) \right |]+C$
$=\frac{1}{sin(b-a)}[log\left | \frac{sec(x+b)}{sec(x+a)} \right |]+C$ $[log\, a-log\, b=log\frac{a}{b}]$

Indefinite Integrals exercise 18.8 question 29

$log\left | 2sin\, x+cos\, x \right |+C$
Hint:
$\int \! \frac{dt}{t}=log\left | t \right |+C$
Given:
$\int \! \frac{-sin\, x+2cos\, x}{2sin\, x+cos\, x}dx$....(1)
Explanation:
Let
$2sin\, x+cos\, x=t$
$(2cos\, x-sin\, x)dx=dt$
From (1)
$\int \! \frac{dt}{t}=log\left | t \right |+C$
$=log\left | 2sin\, x+cos\, x \right |+C$

Indefinite Integrals exercise 18.8 question 30

$\frac{1}{3}log\left | cos\, 3x \right |+C$
Hint:
$U\! se\; cos\, A-cos\, B \: and\: sin\, A-sin\, B f\! ormula$
Given:
$\int \! \frac{cos4x-cos2x}{sin4x-sin2x}dx$
Explanation:
$\int \frac{-2 \sin \left(\frac{4 x-2 x}{2}\right) \sin \left(\frac{4 x+2 x}{2}\right)}{2 \cos \left(\frac{4 x+2 x}{2}\right) \sin \left(\frac{4 x-2 x}{2}\right)} d x \quad\left[\begin{array}{l} \cos A-\cos B=-2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right) \\ \sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right) \end{array}\right]$
$=-\int \! \frac{sin3x}{cos3x}dx$
$=-\int \! \tan\, 3xdx$
$=-[\frac{-log\left | cos\, 3x \right |}{3}]+C$
$=\frac{1}{3}log\left | cos\, 3x \right |+C$

Indefinite Integrals exercise 18.8 question 31

$log\left | log(sec\, x+tan\, x) \right |+C$
Hint:
$\int \! \frac{dt}{t}=log\left | t \right |+C$
Given:
$\int \! \frac{sec\, x}{log(sec\, x+tan\, x)}dx$......(1)
Explanation:
Let
$log(sec\, x+tan\, x)=t$
$\frac{1}{sec\, x+tan\, x}(sec\, x\: tan\, x+sec^{2}x)dx=dt$
$\Rightarrow \frac{\frac{sin\, x}{cos^{2}x}+\frac{1}{cos^{2}x}}{\frac{sin\, x}{cos\, x}+\frac{1}{cos\, x}}dx=dt$
$\Rightarrow \frac{\frac{1+sin\, x}{cos^{2}x}}{\frac{1+sin\, x}{cos\, x}}dx=dt$
$=sec\, xdx=dt$
Put in (1)
$\int \! \frac{dt}{t}=log\left | t \right |+C$
$=log\left | log(sec\, x+tan\, x) \right |+C$

Indefinite Integrals exercise 18.8 question 32

$log\left | log \: tan\frac{x}{2} \right |+C$
Hint:
$sin\, 2x=2sin\, x\, cos\, x$
Given:
$\int \! \frac{cos\, ecx}{log\, tan\frac{x}{2}}dx$.....(1)
Explanation:
Let
$log\, tan\frac{x}{2}=t$
$\frac{1}{tan\frac{x}{2}}sec^{2}\frac{x}{2}(\frac{1}{2})dx=dt$
$\frac{1}{2cos\frac{x}{2}sin\frac{x}{2}}dx=dt$
$\frac{1}{sin2\times \frac{x}{2}}dx=dt$
$cos\, ecx\, dx=dt$
Put in (1)
$\int \! \frac{dt}{t}=log\left | t \right |+C$
$=log\left | log \: tan\frac{x}{2} \right |+C$

Indefinite Integrals exercise 18.8 question 33

$=log\left | log(log\, x) \right |+C$
Hint:
$log(log\, x)=t$
$\frac{1}{log\, x(x)}dx=dt$
Given:
$\int \! \frac{1}{x\, log\, x\, log(log\, x)}dx$....(1)
Explanation:
Let
$log\, (log\, x)=t$
$\frac{1}{log\, x\times x}dx=dt$
$\frac{dx}{x\, log\, x}=dt$
From (1) we get
$\int \! \frac{dt}{t}=log\left | t \right |+C$
$=log\left | log(log\, x) \right |+C$

Indefinite Integrals exercise 18.8 question 34

$-log\left | 1+cot\, x \right |+C$
Hint:
$\frac{\mathrm{d} }{\mathrm{d} x}cot\, x=-cos\, ec^{2}x$
Given:
$\int \! \frac{cos\, ec^{2}x}{1+cot\, x}dx$.....(1)
Explanation:
Let
$1+cot\, x=t$
$-cos\, ec^{2}xdx=dt$
From (1)
$-\int \! \frac{dt}{t}=-log\left | t \right |+C$
$=-log\left | 1+cot\, x \right |+C$

Indefinite Integrals exercise 18.8 question 35

$log\left | 10^{x}+x^{10} \right |+C$
Hint:
$\frac{\mathrm{d} }{\mathrm{d} x}(10^{x}+x^{10})=10x^{9}+10^{x}log_{e}10$
Given:
$\int \! \frac{10x^{9}+10^{x}log_{e}10}{10^{x}+x^{10}}dx$.......(1)
Explanation:
Let
$10^{x}+x^{10}=t$
$(10x^{9}+10^{x}log_{e}10)dx=dt$
From (1) we get
$\int \! \frac{dt}{t}=log\left | t \right |+C$
$=log\left | 10^{x}+x^{10} \right |+C$

Indefinite Integrals exercise 18.8 question 36

$log\left | x+cos^{2}x \right |+C$
Hint:
Put denominator=t
Given:
$\int \! \frac{1-sin\, 2x}{x+cos^{2}x}dx$.......(1)
Explanation:
Let
$x+cos^{2}x=t$
$[1+2cos\, x(-sin\, x)]dx=dt$
$(1-2sin\, x\, cos\, x)dx=dt$
$(1-2sin\, 2x)dx=dt$
Put in (1) we get
$\int \! \frac{dt}{t}=log\left | t \right |+C$
$=log\left | x+cos^{2}x \right |+C$

Indefinite Integrals exercise 18.8 question 37

$log\left | x+log\, sec\, x \right |+C$
Hint:
Put denominator = t
Given:
$\int \! \frac{1+tan\, x}{x+log\, sec\, x}dx$......(1)
Explanation:
Let
$x+log\, sec\, x=t$
$[1+\frac{1}{sec\, x}(sec\, x\, tan\, x)]dx=dt$
$(1+tan\, x)dx=dt$
Put in (1)
$\int \frac{dt}{t}=log\left | t \right |+C$
$=log\left | x+log\, sec\, x \right |+C$

Indefinite Integrals exercise 18.8 question 38

$\frac{1}{b^{2}}log(a^{2}+b^{2}sin^{2}x)+C$
Hint:
Put denominator = t
Given:
$\int \! \frac{sin\, 2x}{a^{2}+b^{2}sin^{2}x}dx$......(1)
Explanation:
Let
$a^{2}+b^{2}sin^{2}x=t$
$b^{2}(2sin\, xcos\, x)dx=dt$
$b^{2}sin\, 2xdx=dt$
$sin\, 2xdx=\frac{dt}{b^{2}}$
Put in (1) we get
$\int \! \frac{dt}{b^{2}(t)}$
$=\frac{1}{b^{2}}\int \! \frac{dt}{t}$
$=\frac{1}{b^{2}}log\left | t \right |+C$
$=\frac{1}{b^{2}}log(a^{2}+b^{2}sin^{2}x)+C$

Indefinite Integrals exercise 18.8 question 39

$log\left | x+log\, x \right |+C$
Hint:
Put
$x+log\, x=t$
$(x+\frac{1}{x})dx=dt$
$\frac{x+1}{x}dx=dt$
Given:
$\int \! \frac {x+1}{x(x+log\, x)}dx$....(1)
Explanation:
Let
$x+log\, x=t$
$(x+\frac{1}{x})dx=dt$
$(\frac{x+1}{x})dx=dt$
Put in (1) we get
$\int \! \frac{dt}{t}$
$=log\left | t \right |+C$
$=log\left | x+log\, x \right |+C$

Indefinite Integrals exercise 18.8 question 40

$\frac{1}{3}log\left | 2+3sin^{-1}x \right |+C$
Hint:
$\frac{\mathrm{d} }{\mathrm{d} x} sin^{-1}x=\frac{1}{\sqrt{1-x^{2}}}$
Given:
$\int \! \frac{1}{\sqrt{1-x^{2}}(2+3sin^{-1}x)}dx$......(1)
Explanation:
Let
$2+3sin^{-1}x=t$
$\frac{3}{\sqrt{1-x^{2}}}dx=dt$
$\frac{dx}{\sqrt{1-x^{2}}}=\frac{dt}{3}$
Put in (1) we get
$\frac{1}{3}\int \! \frac{dt}{t}=\frac{1}{3}log\left | t \right |+C$
$=\frac{1}{3}log\left | 2+3sin^{-1}x \right |+C$

Indefinite Integrals exercise 18.8 question 41

$log\left | tan\, x+2 \right |+C$
Hint:
$\frac{\mathrm{d} }{\mathrm{d} x}(tan\, x)=sec^{2}x$
Given:
$\int \! \frac{sec^{2}x}{tan\, x+2}dx$....(1)
Explanation:
Let
$tan\, x+2=t$
$sec^{2}xdx=dt$
Put in (1)
$\int \! \frac{dt}{t}=log\left | t \right |+C$
$=log\left | tan\, x+2 \right |+C$

Indefinite Integrals exercise 18.8 question 42

$log\left | sin2x+tan\, x-5 \right |+C$
Hint:
Put denominator=t
Given:
$\int \frac{2cos2x+sec^{2}x}{sin2x+tan\, x-5}dx$.....(1)
Explanation:
Let
$sin2x+tan\, x-5=t$
$(2cos2x+sec^{2}x)dx=dt$
Put in (1) we get
$\int \! \frac{dt}{t}=log\left | t \right |+C$
$=log\left | sin2x+tan\, x-5 \right |+C$

Indefinite Integrals exercise 18.8 question 43

$\frac{1}{3}log\left | sin3x \right |-\frac{1}{5}log\left | sin5x \right |+C$
Hint:
$sin(A-B)=sinAcosB-cosAsinB$
Given:
$\int \! \frac{sin2x}{sin5x\, sin3x}dx$
Explanation:
$\int \! \frac{sin(5x-3x)}{sin5x\, sin3x}dx$
$=\int \! \frac{sin5x\, cos3x-cos5x\, sin3x}{sin5x\, sin3x}dx$
$=\int \! \frac{cos3x}{sin3x}dx-\int \! \frac{cos5x}{sin5x}dx$
$=\int \! \cot3xdx-\int \! cot5xdx$
$=\frac{1}{3}log\left | sin3x \right |-\frac{1}{5}log\left | sin5x \right |+C$

Indefinite Integrals exercise 18.8 question 44

$log\left | x+log\, sin\, x \right |+C$
Hint:
Put denominator = t
Given:
$\int \! \frac{1+cot\, x}{x+log\, sin\, x}dx$........(1)
Explanation:
Let
$x+log\, sin\, x=t$
$(1+\frac{1}{sin\, x}(cos\, x))dx=dt$
$(1+cot\, x)dx=dt$
Put in (1) we get
$\int \! \frac{dt}{t}=log\left | t \right |+C$
$=log\left | x+log\, sin\, x \right |+C$

Indefinite Integrals exercise 18.8 question 45

$cos(a-b)log\left | sin(x+b) \right |-xsin(a-b)+C$
Hint:
$cos(A+B)=cosAcosB-sinAsinB$
Given:
$\int \! \frac{cos(x+a)}{sin(x+b)}dx$
Explanation:
$=\int \! \frac{cos(x+a-b+b)}{sin(x+b)}dx$ ...adding and subtracting b in numerator
$=\int \! \frac{cos(x+b+a-b)}{sin(x+b)}dx$
$=\int \! \frac{cos(x+b)cos(a-b)-sin(x+b)sin(a-b)}{sin(x+b)}dx$
$=\int \! cot(x+b)cos(a-b)dx-\int \! sin(a-b)dx$
$=cos(a-b)\int \! cot(x+b)dx-sin(a-b)\int \!dx$
$=cos(a-b)log\left | sin(x+b) \right |-xsin(a-b)+C$

Indefinite Integrals exercise 18.8 question 46

$2log\left | \sqrt{x}+1 \right |+C$
Hint:
Let $\sqrt{x}=t$
Given:
$\int \! \frac{1}{\sqrt{x}(\sqrt{x}+1)}dx$.......(1)
Explanation:
Let
$\sqrt{x}=t$
$\frac{1}{2\sqrt{x}}dx=dt$
$\frac{dx}{\sqrt{x}}=2dt$
From (1) we have
$\int \! \frac{2dt}{t+1}=2\int \! \frac{1}{t+1}dt$
$=2log\left | t+1 \right |$
$=2log\left | \sqrt{x}+1 \right |+C$

Indefinite Integrals exercise 18.8 question 47

$-\frac{1}{5}log\left | cos5x \right |+\frac{1}{2}log\left | cos2x \right |+\frac{1}{3}log\left | cos3x \right |+C$
Hint:
$tan(A+B)=\frac{tanA+tanB}{1-tanA\, tanB}$
Given:
$\int \! tan2x\, tan3x\, tan5xdx$.....(1)
Explanation:
$tan5x=tan(2x+3x)=\frac{tan2x+tan3x}{1-tan2x\, tan3x}$
Cross-multiplying
$tan5x-tan5x\, tan2x\, tan3x=tan2x+tan3x$
$tan5x-tan2x-tan3x=tan5x\, tan2x\, tan3x$
Put in (1) we get
$\int \! (tan5x-tan2x-tan3x)dx$
$=-\frac{1}{5}log\left | cos5x \right |+\frac{1}{2}log\left | cos2x \right |+\frac{1}{3}log\left | cos3x \right |+C$

Indefinite Integrals exercise 18.8 question 48

$cot\, \theta log\left |\frac{cos\, x}{cos(x+\theta )} \right |+C$
Hint:
$tan2x=\frac{2tan\, x}{1-tan^{2}x}$
Given:
$\int \! [1+tan\, x\, tan\left | x+\theta \right |]dx$...(1)
Explanation:
$tan\, \theta =tan[x+\theta -x]=\frac{tan(x+\theta )-tan\, x}{1+tan\, x\, tan(x+\theta )}$
$1+tan\, x\, tan(x+\theta )=cot\, \theta [tan(x+\theta )-tan\, x]$
Put in (1)
$\int \! cot\, \theta [tan(x+\theta )-tan\, x]dx$
$=cot\, \theta \int \! tan(x+\theta )dx-cot\, \theta \int \! tan\, xdx$
$=cot\, \theta [-log\left | cos(x+\theta ) \right |+log\left | cos\, x \right |]+C$
$=cot\, \theta log\left |\frac{cos\, x}{cos(x+\theta )} \right |+C$

Indefinite Integrals exercise 18.8 question 49

$log\left | sin(x+\frac{\pi }{6})sin(x-\frac{\pi }{6}) \right |+C$
Hint:
$sin(A+B)=sinAcosB+cosAsinB$
Given:
$\int \! \frac{sin2x}{sin(x-\frac{\pi }{6})sin(x+\frac{\pi }{6})}dx$
Explanation:
$\int \! \frac{sin[(x-\frac{\pi }{6}+\frac{\pi }{6}+x)]}{sin(x-\frac{\pi }{6})sin(x+\frac{\pi }{6})}dx$
$\int \! \frac{sin[(x-\frac{\pi }{6})+(x+\frac{\pi }{6})]}{sin(x-\frac{\pi }{6})sin(x+\frac{\pi }{6})}dx$
$=\int \frac{\sin \left(x-\frac{\pi}{6}\right) \cos \left(x+\frac{\pi}{6}\right)+\cos \left(x-\frac{\pi}{6}\right) \sin \left(x+\frac{\pi}{6}\right)}{\sin \left(x-\frac{\pi}{6}\right) \sin \left(x+\frac{\pi}{6}\right)}$
$=\int\! cot(x+\frac{\pi }{6})dx+\int \! cot(x-\frac{\pi }{6})dx$
$=log\left | sin(x+\frac{\pi }{6}) \right |+log\left | sin(x-\frac{\pi }{6}) \right |+C$
$=log\left | sin(x+\frac{\pi }{6})sin(x-\frac{\pi }{6}) \right |+C$

Indefinite Integrals exercise 18.8 question 50

$\frac{1}{e}log\left | e^{x}+x^{e} \right |+C$
Hint:
$U\! se\: f\! ormula\: o\! f\int \! \frac{1}{t}=log(t)+c$
Given:
$\int \! \frac{e^{x-1}+x^{e-1}}{e^{x}+x^{e}}dx$....(1)
Explanation:
Let
$e^{x}+x^{e}=t$
$(e^{x}+ex^{e-1})dx=dt$
$e(e^{x-1}+x^{e-1})dx=dt$
Put in (1)
$\frac{1}{e}\int \! \frac{dt}{t}=\frac{1}{e}log\left | t \right |+C$
$=\frac{1}{e}log\left | e^{x}+x^{e} \right |+C$

Indefinite Integrals exercise 18.8 question 51

$sec\, x+log\left | tan\frac{x}{2} \right |+C$
Hint:
$sin^{2}x+cos^{2}x=1$
Given:
$\int \! \frac{1}{sin\, x\, cos^{2}x}dx$
Explanation:
$\int \! \frac{sin^{2}x+cos^{2}x}{sin\, x\, cos^{2}x}dx$
$=\int \! \frac{sin^{2}x}{sin\, x\, cos^{2}x}dx+\int \! \frac{1}{sinx}dx$
$=\int \! tan\, xsec\, xdx+\int \! cos\, e\, cxdx$
$=sec\, x+log\left | tan\frac{x}{2} \right |+C$

Indefinite Integrals exercise 18.8 question 52

$\frac{1}{4}[cos\, ecx-log\left | sec\, x+tan\, x \right |]+C$
Hint:
$Put \; \; 1=sin^{2}x+cos^{2}x$
Given:
$\int \! \frac{1}{cos3x-cos\, x}dx$
Explanation:
$\int \! \frac{sin^{2}x+cos^{2}x}{cos3x-cos\, x}dx$
$=\int \! \frac{sin^{2}x+cos^{2}x}{4cos^{3}x-3cos\, x-cos\, x}dx$ $cos3x=4cos^{3}x-3cos\, x$
$=\int \! \frac{sin^{2}x+cos^{2}x}{4cos^{3}x-4cos\, x}dx$
$=\int \! \frac{sin^{2}x+cos^{2}x}{4cos\, x(cos^{2}x-1)}dx$
$=-\int \! \frac{sin^{2}x+cos^{2}x}{4cos\, x(sin^{2}x)}dx$ $sin^{2}x+cos^{2}x=1$
$=-\int \! \frac{sin^{2}x}{4cos\, x\, sin^{2}x}dx+(-\int \! \frac{cos^{2}x}{4cos\, x\, sin^{2}x}dx)$
$=-\frac{1}{4}\int \! \frac{1}{cos\, x}dx-\frac{1}{4}\int \! \frac{cos\, x}{ sin^{2}x}dx$
$=-\frac{1}{4}\int \! sec\, xdx-\frac{1}{4}\int \! cot\, x\, cos\, e\, cxdx$
$=-\frac{1}{4}log\left | sec\, x+tan\, x \right |-\frac{1}{4}(-cos\, e\, cx)+C$
$=\frac{1}{4}[cos\, ecx-log\left | sec\, x+tan\, x \right |]+C$

Indefinite Integrals exercise 18.9 question 62

Answer: $\frac{\sec ^{3} 2 x}{6}-\frac{\sec 2 x}{2}+c$
Hint: Use substitution method to solve this integral

Given: $\int \tan ^{3} 2 \: x \cdot \sec 2 \: x \; d x$
Solution:
$\text { Let } I=\int \tan ^{3} 2 x \cdot \sec 2 x\; d x$
$\Rightarrow I=\int \tan ^{2} 2 x \cdot \tan 2 x \cdot \sec 2 x\; d x$
\begin{aligned} &=\int\left(\sec ^{2} 2 x-1\right) \tan 2 x \cdot \sec 2 x \; d x \\ &=\int\left(\sec ^{2} 2 x \cdot \tan 2 x \cdot \sec 2 x-\tan 2 x \sec 2 x\right) d x \\ &=\int \sec ^{2} 2 x \cdot \tan 2 x \cdot \sec 2 x\; d x-\int \tan 2 x \cdot \sec 2 x\; d x \end{aligned}
\begin{aligned} &\Rightarrow I=\int \sec ^{2} 2 x \cdot \tan 2 x \cdot \sec 2 x \; d x-\int \tan 2 x \cdot \sec 2 x\; d x \\ &\text { Put } \sec 2 x=t \Rightarrow \sec 2 x . \tan 2 x \cdot 2 \cdot d x=d t \end{aligned}
$\Rightarrow d x=\frac{d t}{2 \sec 2 x \cdot \tan 2 x} \text { then the Ist integral becomes }$
$\Rightarrow I=\int t^{2} \cdot \tan 2 x \cdot \sec 2 x \cdot \frac{d t}{2 \sec 2 x \cdot \tan 2 x}-\int \sec 2 x \cdot \tan 2 x \; d x$
\begin{aligned} &=\int \frac{t^{2}}{2} d t-\int \sec 2 x \cdot \tan 2 x\; d x \\ &=\frac{1}{2} \int t^{2} d t-\int \sec 2 x \cdot \tan 2 x\; d x=\frac{1}{2} \frac{t^{2+1}}{2+1}-\frac{\sec 2 x}{2}+c \end{aligned}
$=\frac{1}{2} \cdot \frac{t^{3}}{3}-\frac{\sec 2 x}{2}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c, \int \sec a x \cdot \tan a x \; d x=\frac{\sec a x}{a}+c\right]$
$=\frac{1}{6} \sec ^{3} 2 x-\frac{\sec 2 x}{2}+c \quad[\because t=\sec 2 x]$

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## RD Sharma Chapter wise Solutions

1. Does this material cover all concepts of Indefinite Integrals?

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4. What is Indefinite Integration?

Indefinite Integration is the process of calculating the integral of a function without using any limits. For more information, check RD Sharma Class 12 Solutions Chapter 18 Ex 18.8.

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