RD Sharma Class 12 Exercise 18.8 Indefinite Integrals Solutions Maths

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.

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RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise
Indefinite Integrals Excercise:18.8
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RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.8

Indefinite Integrals exercise 18.8 question 1

Answer:
$\frac{1}{\sqrt{2}}log\left | tan\frac{x}{2} \right |+C$
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

Answer:
$\sqrt{2}log\left | tan\left ( \frac{\pi }{4}+\frac{x}{4} \right ) \right |+C$
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 3
Answer:

$log\left | sin\: x \right |+C$
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

Answer:
$-2log\left | cos\frac{x}{2} \right |+C$
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

Answer:
$2sin\: x-log\left | sec\: x+tan\; x \right |+C$
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

Answer:
$log\: log\left | sin\: sin\: x+cos\: cos\: x \right |+C$
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

Answer:
$xcos(b-a)+sin(b-a)log\left | sin(x-b) \right |+C$
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

Answer:
$x\: cos\: cos\: 2a-sin\: sin\: 2a \: log\: log\left | sin\: sin(x+a) \right |+C$
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

Answer:
$-log\left | cos\: x-sin\: x \right |+C$
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

Answer:
$(x-a)cos\: a-sin\: a\: log\left | sec(x-a) \right |+C$
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

Answer:
$log\left | cos(\frac{\pi }{4}-x) \right |+C$
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

Answer:
$\frac{1}{3}log\left | e^{3x}+1 \right |+C$
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

Answer:
$\frac{1}{3}log\left | 3sec\; x+5 \right |C$
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

Answer:
$-log\left | sin\; x+cos\: x \right |+C$
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

Answer:
$log\left | log(tan\: x) \right |+C$
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

Answer:
$log\left | 3+log\: x \right |+C$
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

Answer:
$log\left | e^{x}+x \right |+C$
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

Answer:
$log\left | log\: x \right |+C$
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

Answer:
$\frac{1}{b-a}In\left | a\: cos^{2}\: x+b\: sin^{2}\: x \right |+C$
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

Answer:
$\frac{1}{3}log\left | 2+3\, sin\, x \right |+C$
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

Answer:
$log\left | x+cos\, x \right |+C$
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

Answer:
$-\frac{a}{b}log\left | be^{-x}+c \right |+C$
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

Answer:
$-log\left | e^{-x}+1 \right |+C$
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

Answer:
$log\left | log\: sin\: x \right |+C$
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

Answer:
$\frac{1}{2}log\left | e^{2x}-2 \right |+C$
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

Answer:
$\frac{1}{2}log\left | x^{2}+sin\, 2x+2x \right |+C$
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

Answer:
$\frac{1}{sin(b-a)}[log(\frac{sec(x+b)}{sec(x+a)})]+C$
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

Answer:
$log\left | 2sin\, x+cos\, x \right |+C$
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

Answer:
$\frac{1}{3}log\left | cos\, 3x \right |+C$
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

Answer:
$log\left | log(sec\, x+tan\, x) \right |+C$
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

Answer:
$log\left | log \: tan\frac{x}{2} \right |+C$
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

Answer:
$=log\left | log(log\, x) \right |+C$
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

Answer:
$-log\left | 1+cot\, x \right |+C$
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

Answer:
$log\left | 10^{x}+x^{10} \right |+C$
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

Answer:
$log\left | x+cos^{2}x \right |+C$
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

Answer:
$log\left | x+log\, sec\, x \right |+C$
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

Answer:
$\frac{1}{b^{2}}log(a^{2}+b^{2}sin^{2}x)+C$
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

Answer:
$log\left | x+log\, x \right |+C$
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

Answer:
$\frac{1}{3}log\left | 2+3sin^{-1}x \right |+C$
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

Answer:
$log\left | tan\, x+2 \right |+C$
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

Answer:
$log\left | sin2x+tan\, x-5 \right |+C$
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

Answer:
$\frac{1}{3}log\left | sin3x \right |-\frac{1}{5}log\left | sin5x \right |+C$
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

Answer:
$log\left | x+log\, sin\, x \right |+C$
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

Answer:
$cos(a-b)log\left | sin(x+b) \right |-xsin(a-b)+C$
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

Answer:
$2log\left | \sqrt{x}+1 \right |+C$
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

Answer:
$-\frac{1}{5}log\left | cos5x \right |+\frac{1}{2}log\left | cos2x \right |+\frac{1}{3}log\left | cos3x \right |+C$
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

Answer:
$cot\, \theta log\left |\frac{cos\, x}{cos(x+\theta )} \right |+C$
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

Answer:
$log\left | sin(x+\frac{\pi }{6})sin(x-\frac{\pi }{6}) \right |+C$
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

Answer:
$\frac{1}{e}log\left | e^{x}+x^{e} \right |+C$
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

Answer:
$sec\, x+log\left | tan\frac{x}{2} \right |+C$
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

Answer:
$\frac{1}{4}[cos\, ecx-log\left | sec\, x+tan\, x \right |]+C$
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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