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

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.

This Story also Contains

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

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

RD Sharma Class 12th Exercise 18.8 material is offered by carrier 360 to help students prepare well for their exams. However, as the chapter ‘Indefinite integrals’ contains hundreds of sums, it gets confusing for students to solve all of them. This is why Career360 has provided solutions for students to refer to and complete their syllabus efficiently. This way, they can ensure that no concept is left behind and every part of the syllabus is covered.

As RD Sharma Class 12th Exercise 18.8 material complies with the CBSE syllabus and is updated to the latest version of the book, students can directly refer to it and stay in line with their classes to mark their progress. They can also complete the portion well ahead of time with the help of this material.

Every solution from RD Sharma Class 12th Exercise 18.8 material is passed through various quality tests to ensure that the material provided has the most accurate and easy-to-understand answers. Moreover, as experts prepare this material, students can rest assured that they have the best solutions available.

RD Sharma Class 12 Chapter 18 Exercise 18.8 material is available on the Career360 website free of cost, which means that students can access this material using any device with a browser and internet connection. RD Sharma solutions This material will also help students for revision purposes as the RD Sharma textbook is vast and it does not contain answers for exercise questions.

As teachers can't cover every question in their lectures, students have the extra load of solving the remaining ones. This material will help you get a head start in your preparation and complete homework as well. Thousands of students have already adopted this method of preparation as it is efficient and productive.

RD Sharma Chapter wise Solutions

Frequently Asked Questions (FAQs)

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

Yes, Class 12 RD Sharma Chapter 18 Exercise 18.8 Solutions contain the best solutions that cover the entire syllabus. You can find all the concepts and the solutions related to them in this material.

2. Do I have to pay any extra charges?

Students can access RD Sharma Class 12 Solutions Indefinite Integrals Ex 18.8 material from the Career360 website for absolutely free without any hidden costs.

3. Can I score good marks through this material?

Preparing from this material will help students better understand the subject with the help of the solutions. This will ensure that they cover all concepts and score good marks in exams.

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.

5. How are RD Sharma books better than NCERT for maths?

NCERT books are meant for a basic understanding of the subject, whereas RD Sharma books contain in-depth knowledge about the concepts. This is why RD Sharma books are exam-oriented and more detailed when compared to NCERT books.