RD Sharma Class 12 Exercise 18.2 Indefinite Integrals Solutions Maths

RD Sharma Class 12 Exercise 18.2 Indefinite Integrals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 12:48 PM IST

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RD Sharma Class 12 Solutions Chapter 18 Indefinite Integrals - Other Exercise
Indefinite Integrals Excercise:18.2
RD Sharma Chapter wise Solutions

RD Sharma Class 12 Solutions Chapter 18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.2

Indefinite Integrals exercise 18.2 question 1

Answer:

$\frac{6}{5} x^{\frac{5}{2}}+\frac{1}{2} x^{\frac{3}{2}}+5 x+c$
Hint: To solve this, we break square root then imply $\int x^{a} d x$ formula
$\begin{aligned} &\text { Given: } \int(3 x \sqrt{x}+4 \sqrt{x}+5) d x \\ &\text { Solution: } 3 x x^{\frac{1}{2}}+4 x^{\frac{1}{2}}+5 \\ &=3 x^{\frac{3}{2}}+4 x^{\frac{1}{2}}+5 \\ &I=\int(3 x \sqrt{x}+4 \sqrt{x}+5) d x=\int\left(3 x . x^{\frac{1}{2}}+4 x^{\frac{1}{2}}+5\right) d x \\ &\left\{\int x^{a} d x=\frac{1}{a+1} x^{a+1}+c, a \neq-1\right\} \end{aligned}$
$\begin{aligned} &=3 \int x^{\frac{3}{2}} d x+4 \int x^{\frac{1}{2}} d x+\int 5 d x \\ &=3 \frac{1}{1+\frac{3}{2}} x^{1+\frac{3}{2}}+4 \frac{1}{1+\frac{1}{2}} x^{1+\frac{1}{2}}+5 x+c \\ &=3 \frac{1}{\frac{5}{2}} x^{\frac{5}{2}}+4 \frac{1}{\frac{3}{2}} x^{\frac{3}{2}}+5 x+c \\ &=\frac{6}{5} x^{\frac{5}{2}}+\frac{8}{3} x^{\frac{3}{2}}+5 x+c \end{aligned}$

Indefinite Integrals exercise 18.2 question 2
Answer:

$\frac{2^{x}}{\log 2}+5 \log x-\frac{3}{2} x^{\frac{2}{3}}+c$
Hint: To solve this equation we differentiate it differently.
$\begin{aligned} &\text { Given: } \int\left(2^{x}+\frac{5}{x}-\frac{1}{x^{\frac{1}{3}}}\right) d x \\ &\text { Solution: } \int 2^{x} d x+5 \int \frac{1}{x} d x-\int \frac{1}{x^{\frac{1}{3}}} d x \\ &\left\{\begin{array}{l} \int a^{x} d x=\frac{a^{x}}{\log a}+c \\ \int \frac{1}{x} d x=\log x+c \end{array}\right\} \\ &=\frac{2^{x}}{\log 2}+5 \log x-\frac{x^{-\frac{3}{2}+1}}{-\frac{3}{2}+1}+c \\ &=\frac{2^{x}}{\log 2}+5 \log x-\frac{3}{2} x^{\frac{2}{3}}+c \end{aligned}$

Indefinite Integrals exercise 18.2 question 3

Answer:

$\frac{2}{7} a x^{\frac{7}{2}}+\frac{2}{5} b x^{\frac{5}{2}}+\frac{2}{3} c x^{\frac{3}{2}}+c$
To solve this we multiply $\sqrt{x}$ by $ax^{2}+bx+c$
$\begin{aligned} &\text { Given: } \int\left\{\sqrt{x}\left(a x^{2}+b x+c\right)\right\} d x \\ &\text { Solution: } \int\left\{\sqrt{x}\left(a x^{2}+b x+c\right)\right\} d x \end{aligned}$
$\begin{aligned} &=\int\left(a x^{2+\frac{1}{2}}+b x^{1+\frac{1}{2}}+c x^{\frac{1}{2}}\right) d x \\ &=a \int x^{\frac{5}{2}} d x+b \int x^{\frac{3}{2}} d x+c \int x^{\frac{1}{2}} d x \end{aligned}$
$\begin{aligned} &=a \frac{x^{\frac{5}{2}+1}}{\frac{5}{2}+1}+b \frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}+c \frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1} \quad\left\{\int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right\} \\ &=\frac{2}{7} a x^{\frac{7}{2}}+\frac{2}{5} b x^{\frac{5}{2}}+\frac{2}{3} c x^{\frac{3}{2}}+c \end{aligned}$

Indefinite Integrals exercise 18.2 question 4
Answer:

$3 x^{4}+\frac{4}{3} x^{3}-\frac{17}{2} x^{2}+6 x+c$

Hint: To solve this we multiply $(2-3 x)(3+2 x)(1-2 x)$ then differentiate

$\begin{aligned} &\text { Given: } \int(2-3 x)(3+2 x)(1-2 x) d x \\ &\text { Solution: }(2-3 x)(3+2 x)(1-2 x) \\ &=\left(6+4 x-9 x-6 x^{2}\right)(1-2 x) \\ &=\left(6-5 x-6 x^{2}\right)(1-2 x) \\ &=6-5 x-6 x^{2}+10 x^{2}-12 x+12 x^{3} \\ &=12 x^{3}+4 x^{2}-17 x+6 \end{aligned}$

$\begin{aligned} &\int(2-3 x)(3+2 x)(1-2 x) d x=\int\left(12 x^{3}+4 x^{2}-17 x+6\right) d x \\ &\text { Using identity } \int x^{n} d x=\frac{x^{n+1}}{n+1}+c \\ &\frac{12 x^{4}}{3+1}+\frac{4 x^{3}}{3}-\frac{17 x^{2}}{1+1}+6 x+c \\ &=\frac{12 x^{4}}{4}+\frac{4 x^{3}}{3}-\frac{17 x^{2}}{2}+6 x+c \\ &=3 x^{4}+\frac{4}{3} x^{3}-\frac{17}{2} x^{2}+6 x+c \end{aligned}$

Indefinite Integrals exercise 18.2 question 5

Answer:

Solution: We have ,
$I=\int (\frac{m}{x}+\frac{x}{m}+m^{x}+x^{m}+mx)\\ \\ \Rightarrow \hspace{1cm}I=m\log \left | x \right |+\frac{x^{2}}{2m}+\frac{m^{x}}{\log m}+\frac{x^{m+1}}{m+1}+\frac{mx^{2}}{2}+c$

Indefinite Integrals exercise 18.2 question 6

Answer:

$\frac{x^{3}}{2}+\log x-2 x+c$
Hint: To solve this equation we use $(a-b)^{2}$ formula then $\int x^{a}dx$
$\begin{aligned} &\text { Given: } \int\left[\sqrt{x}-\frac{1}{\sqrt{x}}\right]^{2} d x \\ &\text { Solution: } \int\left[(\sqrt{x})^{2}-2 \sqrt{x} \frac{1}{\sqrt{x}}-\left(\frac{1}{\sqrt{x}}\right)^{2}\right] d x \quad\left\{(a-b)^{2}=a^{2}-b^{2}+2 a b\right\} \end{aligned}$
$\\ \\ \Rightarrow \hspace{1cm}I=\int (x-2+\frac{1}{x})dx\\ \\ \Rightarrow \hspace{1cm}I=\frac{x^{2}}{2}-2x+\log \left | x \right |+c.$

Indefinite Integrals exercise 18.2 question 7

Answer:

$2 \sqrt{x}+\frac{2}{7} x^{\frac{7}{2}}+2 x^{\frac{3}{2}}+\frac{6}{5} x^{\frac{5}{2}}+c$

Hint: To solve this equation we use $(a+b)^{3}$ formula then find the integral

$\begin{aligned} &\text { Given: } \int \frac{(1+x)^{3}}{\sqrt{x}} d x \\ &\text { Solution: } I=\int \frac{(1+x)^{3}}{\sqrt{x}} d x \end{aligned}$

$\begin{aligned} &\text { Using identity }\\ &\left\{(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)\right\}\\ &=\int \frac{1+x^{3}+3 x(1+x)}{\sqrt{x}} d x\\ &=\int \frac{1+x^{3}+3 x+3 x^{2}}{\sqrt{x}} d x\\ &=\int x^{\frac{-1}{2}}+x^{\frac{5}{2}}+3 x^{\frac{1}{2}}+3 x^{\frac{3}{2}} d x \end{aligned}$

$\begin{aligned} &\text { Using identity }\\ &\left\{\int x^{a} d x=\frac{1}{a+1} x^{a+1}+c, a \neq-1\right\}\\ &=\frac{x^{\frac{-1}{2}}}{\frac{-1}{2}+1}+\frac{x^{\frac{5}{2}}+1}{\frac{5}{2}+1}+\frac{3 x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{3 x^{\frac{3}{2}+1}}{\frac{3}{2}+1}\\ &=\frac{\sqrt{x}}{\frac{1}{2}}+\frac{x^{\frac{7}{2}}}{\frac{7}{2}}+\frac{3 x^{\frac{3}{2}}}{\frac{3}{2}}+\frac{3 x^{\frac{5}{2}}}{\frac{5}{2}}\\ &=2 \sqrt{x}+\frac{2}{7} x^{\frac{7}{2}}+\frac{6}{3} x^{\frac{3}{2}}+\frac{6}{5} x^{\frac{5}{2}}+c\\ &=2 \sqrt{x}+\frac{2}{7} x^{\frac{7}{2}}+2 x^{\frac{3}{2}}+\frac{6}{5} x^{\frac{5}{2}}+c \end{aligned}$

Indefinite Integrals exercise 18.2 question 8

Answer:

Hint: To solve this equation we integral the term separately
$\begin{aligned} &\text { Given: } \int\left\{x^{2}+e^{\log x}+\left(\frac{e}{2}\right)^{x}\right\} d x \\ &\text { Solution: (1) } e^{\log x}=x \end{aligned}$
$\\ \\ \Rightarrow \hspace{1cm}I=\int [x^{2}+x+(\frac{e}{2})^{x}]dx\\ \\ \Rightarrow \hspace{1cm}I=\frac{x^{3}}{3}+\frac{x^{2}}{2}+\frac{1}{\log \frac{e}{2}}\cdot(\frac{e}{2})^{x}+c$

Indefinite Integrals exercise 18.2 question 9

Answer:

Solution:We have ,
$\\ \\ I=\int (x^{e}+e^{x}+e^{e})dx\\ \\ \Rightarrow \hspace{0.5cm}I=\frac{x^{e+1}}{e+1}+e^{x}+x\cdot e^{e}+c$

Indefinite Integrals exercise 18.2 question 10

Answer:

$\frac{2}{9} x^{\frac{9}{2}}-4 x^{\frac{1}{2}}+c$
Hint: To solve this we multiply $\sqrt{x}$ to next term
Solution: We have ,
$I=\int \sqrt{x}(x^{3}-\frac{2}{x})dx$
$\begin{aligned} &\text { Using identity }\\ &\left\{\int x^{a} d x=\frac{1}{a+1} x^{a+1}+c, a \neq-1\right\}\\ &=\frac{1}{\frac{7}{2}+1} x^{\frac{7}{2}+1}-2 \frac{1}{1-\frac{1}{2}} x^{1-\frac{1}{2}}+c\\ &=\frac{1}{\frac{9}{2}} x^{\frac{9}{2}}-\frac{2}{\frac{1}{2}} x^{\frac{1}{2}}+c\\ &=\frac{2}{9} x^{\frac{9}{2}}-4 x^{\frac{1}{2}}+c \end{aligned}$

Indefinite Integrals exercise 18.2 question 11

Answer:

$2 x^{\frac{1}{2}}-\frac{1}{2 x^{\frac{1}{2}}}+c$
Hint: To solve this equation we use $\int x^{n}dx$ formula
$\text { Given: } \int \frac{1}{\sqrt{x}}\left(1-\frac{1}{x}\right) d x$
$\begin{aligned} &\text { Solution: } \int \frac{1}{\sqrt{x}}\left(1-\frac{1}{x}\right) d x \\ &=\int x^{\frac{-1}{2}} d x+\int x^{\frac{-3}{2}} d x \quad\left[\int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \\ &=2 x^{\frac{1}{2}}-\frac{x^{\frac{-1}{2}}}{\frac{-1}{2}}+c \\ &=2 x^{\frac{1}{2}}-\frac{1}{2 x^{\frac{1}{2}}}+c \end{aligned}$

Indefinite Integrals exercise 18.2 question 12

Answer:

$\frac{x^{5}}{5}-\frac{x^{3}}{3}+x+c$
Hint: To solve this equation we use $a^{3}+b^{3}$ formula , then integral it.
$\begin{aligned} &\text { Given: } \int \frac{x^{6}+1}{x^{2}+1}\\ &\text { Solution: } x^{6}+1\\ &=\left(x^{2}\right)^{3}+\left(1^{2}\right)^{3} \quad\left[a^{3}+b^{3}=(a+b)\left(a^{2}-a b-b^{2}\right)\right]\\ &=\left(x^{2}+1\right)\left(x^{4}-x^{2}+1\right)\\ &\Rightarrow \int \frac{\left(x^{2}+1\right)\left(x^{4}-x^{2}+1\right)}{x^{2}+1} d x\\ &=\int\left(x^{4}-x^{2}+1\right) d x\\ &\left[\because \int x^{a}=\frac{1}{a} x^{a+1}+c, a \neq-1\right]\\ &\Rightarrow \frac{x^{4+1}}{4+1}-\frac{x^{2+1}}{2+1}+\frac{x^{0+1}}{0+1}+c\\ &\Rightarrow \frac{x^{5}}{5}-\frac{x^{3}}{3}+x+c \end{aligned}$

Indefinite Integrals exercise 18.2 question 13

Solution: We have ,
$I=\int \frac{x^\frac{-1}{3}+\sqrt{x}+2}{\sqrt[3]{x}}dx\\ \\ \Rightarrow \hspace{1cm}I=\int (x^\frac{-2}{3}+x^\frac{1}{6}+2x^\frac{-1}{3})dx\\ \\ \Rightarrow \hspace{1cm}I=3x^\frac{1}{3}+\frac{6}{7}x^\frac{7}{6}+3x^\frac{2}{3}+c$

Indefinite Integrals exercise 18.2 question 14

Answer:

$2 \sqrt{x}+2 x+\frac{2}{3} x^{\frac{3}{2}}+c$
Hint:To solve this equation $\left ( 1+\sqrt{x} \right )$ will be differentiate first
Given: $\int \frac{\left ( 1+\sqrt{x} \right )^{2}}{\sqrt{x}}$
Solution: We have
$\begin{aligned} &\int \frac{(1+\sqrt{x})^{2}}{\sqrt{x}}^{\frac{1}{x^{\frac{1}{2}}}} d x=\int \frac{1+x+2 \sqrt{x}}{x^{\frac{1}{2}}}+\left[\because(a+b)^{2}=a^{2}+2 a b+b^{2}\right] \\ &=\int \left ( \frac{1}{x^{\frac{1}{2}}}+\frac{x}{x^{\frac{1}{2}}}+\frac{2\sqrt{x}}{\sqrt{x}} \right )dx\\ &=\int x^{\frac{-1}{2}} d x+\int x^{\frac{1}{2}} d x+2 \int d x \\ &=\frac{x^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}+\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+2 x+c \\ &=2 \sqrt{x}+\frac{2}{3} x^{\frac{3}{2}}+2 x+c \end{aligned}$

Indefinite Integrals exercise 18.2 question 15

Answer:

$2 x^{\frac{3}{2}}-2 x^{\frac{5}{2}}+c$
Hint: To solve this equation use multiply $\sqrt{x}=x^{\frac{1}{2}} \text { by } 3-5 x$
$\begin{aligned} &\text { Given: } \int \sqrt{x}(3-5 x) d x \\ &\text { Solution: } \int \sqrt{x}(3-5 x) d x \\ &\int(x)^{\frac{1}{2}}(3-5 x) d x \\ &\int 3 x^{\frac{1}{2}}-5 x^{\frac{3}{2}} d x \\ &=3 \int x^{\frac{1}{2}} d x-5 \int x^{\frac{3}{2}} d x \\ &{\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c, n \neq-1\right]} \end{aligned}$
$\begin{aligned} &=\frac{3 x^{\frac{3}{2}}}{\frac{3}{2}}-5 \frac{x^{\frac{5}{2}}}{\frac{5}{2}}+c \\ &=2 x^{\frac{3}{2}}-2 x^{\frac{5}{2}}+c \end{aligned}$

Indefinite Integrals exercise 18.2 question 16

Answer:

$\frac{2}{5} x^{\frac{5}{2}}-\frac{2}{3} x^{\frac{3}{2}}-4 \sqrt{x}+c$
Hint: To solve this equation we multiply the upper term
$\begin{aligned} &\text { Given: } \int \frac{(x+1)(x-2)}{\sqrt{x}} d x \\ &\text { Solution: } \int \frac{x(x-2)+1(x+2)}{\sqrt{x}} d x \\ &=\int \frac{x^{2}-2 x+x+2}{\sqrt{x}} d x \\ &=\int \frac{x^{2}-x+2}{\sqrt{x}} d x \\ &=\int \frac{x^{2}}{\sqrt{x}}-\frac{x}{\sqrt{x}}-\frac{2}{\sqrt{x}} d x \end{aligned}$
$\begin{aligned} &=\int x^{\frac{3}{2}} d x-\int \sqrt{x} d x-2 \int d x \\ &{\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c, n \neq-1\right]} \\ &=2 \frac{x^{\frac{5}{2}}}{5}-2 \frac{x^{\frac{3}{2}}}{3}-4 \sqrt{x}+c \\ &=\frac{2}{5} x^{\frac{5}{2}}-\frac{2}{3} x^{\frac{3}{2}}-4 \sqrt{x}+c \end{aligned}$

Indefinite Integrals exercise 18.2 question 17

Answer:

Solution: We have ,
$I=\int \frac{x^{5}+x^{-2}+2}{x^{2}}dx \\ \\ \Rightarrow \hspace{1cm}I=\int (x^{3}+x^{-4}+2x^{-2})dx \\ \\ \Rightarrow \hspace{1cm}I=\frac{x^{4}}{4}-\frac{1}{3x^{3}}-\frac{2}{x}+c$

Indefinite Integrals exercise 18.2 question 18

Answer:

$3 x^{3}+12 x^{2}+16 x+c$
Hint: To solve this equation we do middle term equation
$\begin{aligned} &\text { Given: } \int(3 x+4)^{2} d x \\ &\text { Solution: }(3 x+4)^{2} \\ &=(3 x)^{2}+(4)^{2}+2(3 x)(4) \\ &=9 x^{2}+16+24 x \\ &=9 x^{2}+24 x+16 \\ &=\int (3 x+4)^{2} d x=\int 9 x^{2}+24 x+16 d x \\ &=9 \frac{1}{1+2} x^{1+2}+24 \frac{1}{1+1} x^{1+1}+16 x+c \quad\left[\int x^{a} d x=\frac{1}{a} x^{a+1}+c, a \neq-1\right] \\ &=\frac{9}{3} x^{3}+\frac{24}{2} x^{2}+16 x+c \end{aligned}$
$3 x^{3}+12 x^{2}+16 x+c$

Indefinite Integrals exercise 18.2 question 19

Answer:

$\frac{3x^{2}}{2}+\frac{2x^{3}}{3}+c$
Hint: To solve this equation we do middle term equation
$\begin{aligned} &\text { Given: } \int \frac{2 x^{4}+7 x^{3}+6 x^{2}}{x^{2}+2 x} d x \\ &\text { Solution: } \int \frac{2 x^{4}+4 x^{3}+3 x^{3}+6 x^{2}}{x(x+2)} d x \\ &=\int \frac{2 x^{3}(x+2)+3 x^{2}(x+2)}{x(x+2)} d x \\ &=\int \frac{2 x^{3}+3 x^{2}}{x} d x \\ &=\int \frac{2 x^{3}+3 x^{2}}{x} d x \\ &=\int \frac{x\left(2 x^{2}+3 x\right)}{x} d x \\ &=\int 3 x+2 x^{2} d x\left[\int x^{a} d x=\frac{1}{a} x^{a+1}+c, a \neq-1\right] \\ &=3 \frac{x^{1+1}}{1+1}+2 \frac{x^{2+1}}{2+1}+c \\ &=\frac{3 x^{2}}{2}+\frac{2 x^{3}}{3}+c \end{aligned}$

Indefinite Integrals exercise 18.2 question 20

Answer:

$\begin{aligned} & \frac{5 x^{3}}{3}+\frac{7 x^{2}}{2}+c\\ &\text { Hint: To solve this equation we do middle term spilt }\\ &\text { Given: } \int \frac{5 x^{4}+12 x^{3}+7 x^{2}}{x^{2}+x} d x \end{aligned}$ $\begin{aligned} &\text { Solution: } \int \frac{5 x^{4}+12 x^{3}+7 x^{2}}{x^{2}+x} d x \\ &=\int \frac{x^{2}\left(5 x^{2}+12 x+7\right)}{x(x+1)} d x \\ &=\int \frac{x\left(5 x^{2}+12 x+7\right)}{(x+1)} d x \\ &=\int \frac{x\left(5 x^{2}+5 x+7 x+7\right)}{(x+1)} d x \\ &=\int \frac{x(5 x(x+1)+7(x+1))}{(x+1)} d x \\ &=\int \frac{x(5 x+7)(x+1)}{(x+1)} d x \\ &=\int x(5 x+7) d x \\ &=\int 5 x^{2}+7 x d x\left[\int x^{a} d x=\frac{1}{a} x^{a+1}+c, a \neq-1\right] \\ &=5 \frac{x^{2+1}}{2+1}+7 \frac{x^{1+1}}{1+1}+c \\ &=\frac{5 x^{3}}{3}+\frac{7 x^{2}}{2}+c \end{aligned}$

Indefinite Integrals exercise 18.2 question 21

Answer:

$x-sin x +c$
Hint:To solve this equation we replace $sin^{2}x$ by $1-cos^{2}x$
$\begin{aligned} &\text { Given: } \int \frac{\sin ^{2} x}{1+\cos ^{2} x} d x \\ &\text { Solution: } \frac{\sin ^{2} x}{1+\cos x} \end{aligned}$
$\begin{aligned} &=\frac{1-\cos ^{2} x}{1+\cos x}\left[\begin{array}{l} \operatorname{Sin}^{2} x+\cos ^{2} x=1 \\ \operatorname{Sin}^{2} x=1-\cos ^{2} x \end{array}\right] \\ &=\frac{\left(1-\cos x\right)\left(1+\cos x\right)}{1+\cos x} \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right] \\ &=1-\cos x \\ &\int \frac{\sin ^{2} x}{1+\cos x} d x=\int 1-\cos x d x \\ &=x-\sin x+c \end{aligned}$

Indefinite Integrals exercise 18.2 question 23

Answer:

$\sec x-\cos ec x+c$
Hint:To solve this equation we separate the term then differentiate
$\begin{aligned} &\text { Given: } \int \frac{\sin ^{3} x-\cos ^{3} x}{\sin ^{2} x \cdot \cos ^{2} x} d x \\ &\text { Solution: } \int \frac{\sin ^{3} x-\cos ^{3} x}{\sin ^{2} x \cdot \cos ^{2} x} d x \\ &\int \frac{\sin ^{3} x}{\sin ^{2} x \cdot \cos ^{2} x} d x-\int \frac{\cos ^{3} x}{\sin ^{2} x \cdot \cos ^{2} x} d x \\ &=\int \tan x \sec x d x-\int \cot x \cdot \cos \operatorname{ccx} d x \end{aligned}$
$\sec x-\cos ec x+c$

Indefinite Integrals exercise 18.2 question 24

Answer:

$\frac{-5}{2} \operatorname{cosec} x+3 \sec x+c$
Hint: To solve this equation we separate the terms
$\begin{aligned} &\text { Given: } \int \frac{5 \cos ^{3} x+6 \sin ^{3} x}{2 \sin ^{2} x \cdot \cos ^{2} x} d x \\ &\text { Solution: } \int \frac{5 \cos ^{3} x+6 \sin ^{3} x}{2 \sin ^{2} x \cdot \cos ^{2} x} d x \\ &\int \frac{5 \cos ^{3} x}{2 \sin ^{2} x \cdot \cos ^{2} x} d x-\int \frac{6 \sin ^{3} x}{2 \sin ^{2} x \cdot \cos ^{2} x} d x \\ &=\frac{5}{2} \int \frac{\cos x}{\sin ^{2} x} d x-3 \int \frac{\sin x}{\cos ^{2} x} d x \\ &=\frac{5}{2}\int(\cot x \cos e c x)dx+3\int \tan x \sec xdx \\ &=\frac{-5}{2} \cos e c x+3 \sec x+c \end{aligned}$

Indefinite Integrals exercise 18.2 question 25
Answer:

$\tan x-\cot x+c$
Hint:To solve this equation we use $\int \tan x$ ; $tan^{2}x$ formula
$\begin{aligned} &\text { Given: } \int(\tan x+\cot x)^{2} d x \\ &\text { Solution: } \int(\tan x+\cot x)^{2} d x \\ &(a+b)^{2}=a^{2}+b^{2}+2 a b \\ &=\int \tan ^{2} x+\cot ^{2} x+2 \tan x \cot x d x \\ &=\int \tan ^{2} x+\cot ^{2} x+2 \frac{1}{\cot x} \cot x d x \quad\left[\tan x=\frac{1}{\cot x}\right] \end{aligned}$ $\begin{aligned} &=\int( \sec ^{2} x-1+\cos e c^{2} x-1+2)dx+c \quad\left[\begin{array}{l} \tan ^{2} x=\sec ^{2} x-1 \\ \operatorname{cosec}^{2} x=\cot ^{2} x-1 \end{array}\right] \\ &=\int \sec ^{2} x d x+\int \cos e c^{2} x d x \quad\left[\begin{array}{l} \int \sec ^{2} x d x=\tan x+c \\ \int \operatorname{cosec}^{2} x d x=\cot x+c \end{array}\right] \\ &=\tan x-\cot x+c \end{aligned}$

Indefinite Integrals exercise 18.2 question 26

Answer:

$\tan x-x+c$
Hint: To solve this equation we use $\cos 2x, \int sec^{2} xdx$ formula
$\begin{aligned} &\text { Given: } \int \frac{1-\cos 2 x}{1+\cos 2 x} d x\\ &\text { Solution: }\\ &\int \frac{1-\cos 2 x}{1+\cos 2 x} d x\\ &\int \frac{2 \sin ^{2} x}{2 \cos ^{2} x} d x \quad\left[\begin{array}{l} \cos 2 x=1-2 \sin ^{2} x \\ \cos 2 x=2 \cos ^{2} x-1 \end{array}\right]\\ &\int \tan ^{2} x d x\\ &\int \sec ^{2} x d x-\int 1 d x \quad\left[\begin{array}{l} \sec ^{2} x-\tan ^{2} x=1 \\ \sec ^{2} x-1=\tan ^{2} x \\ \int \sec ^{2} x d x=\tan x+c \end{array}\right]\\ &\tan x-x+c \end{aligned}$

Indefinite Integrals exercise 18.2 question 27
Answer:

$-(cosec x + cot x +x)+c$
Hint: To solve this equation we add $cosec x + cot x$
$\begin{aligned} &\text { Given: } \int \frac{\cot x}{\cos e c x-\cot x} d x \\ &\text { Solution: } \int \frac{\cot x}{\cos e c x-\cot x} d x \end{aligned}$
$\\ \\ \frac{\cot x}{\cos ec x -\cot x}\times \frac{\cos ec x+\cot x}{\cos ec x+\cot x}=\cot x\cdot \cos ec x+\cot^{2}x \\ \\ \Rightarrow \hspace{1cm}I=\int (\cot x \cdot \cos ec x)dx+\int (\cos ec^{2}x-1)dx \\ \\ \Rightarrow \hspace{1cm}I=-\cos ec x -\cot x+c$

Indefinite Integrals exercise 18.2 question 28
Answer:

$\frac{1}{\sqrt{2}}x+c$
Hint: To solve this equation we use $2 \cos 2x$ and $\cos 2x$ formula
$\begin{aligned} &\text { Given: } \int \frac{\cos ^{2} x-\sin ^{2} x}{\sqrt{1+\cos 4 x}} d x \\ &\text { Solution: } \frac{\cos 2 x}{\sqrt{2 \cos ^{2} 2 x}} d x \quad\left[\begin{array}{l} \cos ^{2} x-\sin ^{2} x=\cos 2 x \\ 1+\cos 4 x=2 \cos ^{2}2 x \end{array}\right] \\ &=\frac{1}{\sqrt{2}}\left(\frac{\cos 2 x}{\cos 2 x}\right) \\ &=\frac{1}{\sqrt{2}} \\ &\int \frac{\cos ^{2} x-\sin ^{2} x}{\sqrt{1+\cos 4 x}} d x=\int \frac{1}{\sqrt{2}} d x \\ &=\frac{1}{\sqrt{2}} x+c \end{aligned}$

Indefinite Integrals exercise 18.2 question 29
Answer:

$-cot x- cosec x +c$
Hint: To solve this equation we will add $1+cos x$
$\begin{aligned} &\text { Given: } \int \frac{1}{1-\cos x} d x\\ &\text { Solution: } \int \frac{1}{1-\cos x} d x\\ &\text { Multiplying } 1+\cos x \text { to numerator and denominator }\\ &=\int \frac{1+\cos x}{(1-\cos x)(1+\cos x)} d x\\ &=\int \frac{1+\cos x}{\left(1-\cos ^{2} x\right)} d x \quad\left[\begin{array}{l} (a-b)(a+b)=a^{2}-b^{2} \\ 1-\cos ^{2} x=\sin ^{2} x \end{array}\right]\\ &=\int \frac{1+\cos x}{\sin ^{2} x} d x\\ &=\int \frac{1}{\sin ^{2} x} d x+\int \frac{\cos x}{\sin ^{2} x} d x\\ &=\int \cos e c^{2} x d x+\int \cot x \cos e c x d x\\ &=-\cot x+(-\cos e c x)+c\\ &I=-\cot x-\cos e c x+c \end{aligned}$

Indefinite Integrals exercise 18.2 question 30
Answer:

$\tan x+ sec x+c$
Hint: To solve this equation we will add $1+sin x$
$\begin{aligned} &\text { Given: } \int \frac{1}{1-\sin x} d x\\ &\text { Solution: } \int \frac{1}{1-\sin x} d x\\ &\text { Multiplying } 1+\sin x \text { to numerator and denominator }\\ &=\int \frac{1+\sin x}{(1-\sin x)(1+\sin x)} d x \end{aligned}$
$\begin{aligned} &=\int \frac{1+\sin x}{\left(1-\sin ^{2} x\right)} d x \quad\left[\begin{array}{l} (a-b)(a+b)=a^{2}-b^{2} \\ 1-\sin ^{2} x=\cos ^{2} x \end{array}\right] \\ &=\int \frac{1+\sin x}{\cos ^{2} x} d x \quad\left[\begin{array}{l} \frac{1}{\sin x}=\sec x \\ \frac{\sin x}{\cos x}=\tan x \end{array}\right] \\ &=\int \frac{1}{\cos ^{2} x} d x+\int \frac{\sin x}{\cos ^{2} x} d x \\ &=\int \sec ^{2} x d x+\int \sec x \tan x d x \\ &I=\tan x+\sec x+c \end{aligned}$

Indefinite Integrals exercise 18.2 question 31
Answer:

$\sec x-\tan x +x+c$
Hint:Using $\int \sec x \tan x d x \text { and } \int \sec ^{2} x d x$
$\begin{aligned} &\text { Given: } \int \frac{\tan x}{\sec x+\tan x} d x\\ &\text { Solution: } I=\int \frac{\tan x}{\sec x+\tan x} d x\\ &\text { Multiply and divide by } \sec x-\tan x\\ &I=\int \frac{\tan x(\sec x-\tan x)}{(\sec x+\tan x)(\sec x-\tan x)} d x\\ &=\int \frac{\tan x \sec x-\tan ^{2} x}{\sec ^{2} x-\tan ^{2} x} d x \quad\left[\begin{array}{l} (a-b)(a+b)=a^{2}-b^{2} \\ \sec ^{2} x-\tan ^{2} x=1 \end{array}\right]\\ &=\int \frac{\tan x \sec x-\tan ^{2} x}{1} d x\\ &=\int \tan x \sec x d x-\int \tan ^{2} x d x\\ &=\int \tan x \sec x d x-\int\left(\sec ^{2} x-1\right) d x\\ &=\int \tan x \sec x d x-\int \sec ^{2} x d x+\int 1 d x\\ &=\sec x-\tan x+x+c \end{aligned}$

Indefinite Integrals exercise 18.2 question 32
Answer:

$-cot x-cosec x+c$
$\begin{aligned} &\text { Hint: } \mathrm{U} \operatorname{sing} \int \cos e c^{2} x d x \text { and } \int \operatorname{cosec} x \cot x d x\\ &\text { Given: } \int \frac{\cos e c x}{\cos e c x-\cot x} d x\\ &\text { Solution: } I=\int \frac{\cos e c x}{\cos e c x-\cot x} d x\\ &\text { Multiply and divide by } \operatorname{cosec} x+\cot x\\ &I=\int \frac{\operatorname{cosec} x(\operatorname{cosec} x+\cot x)}{(\cos e c x-\cot x)(\operatorname{cosec} x+\cot x)} d x\\ &=\int \frac{\operatorname{cosec}^{2} x+\operatorname{cosec} x \cot x}{\left(\operatorname{cosec}^{2} x-\cot ^{2} x\right)} d x \quad\left[\begin{array}{l} (a-b)(a+b)=a^{2}-b^{2} \\ \operatorname{cosec}^{2} x-\cot ^{2} x=1 \end{array}\right]\\ &=\int \frac{\operatorname{cosec}^{2} x+\operatorname{cosec} x \cot x}{1} d x\\ &=\int \operatorname{cosec}^{2} x d x-\int \operatorname{cosec} x \cot x d x\\ &=-\cot x-\cos e c x+c \end{aligned}$

Indefinite Integrals exercise 18.2 question 33

Answer:

$\begin{aligned} &\frac{1}{2} \tan x+c \\ &\text { Hint: Using } \int \sec ^{2} x d x \\ &\text { Given: } \int \frac{1}{1+\cos 2 x} d x \\ &\text { Solution: } I=\int \frac{1}{1+\cos 2 x} d x \\ &=\int \frac{1}{2 \cos ^{2} x} d x \quad\left[\because 1+\cos 2 \theta=2 \cos ^{2} \theta\right] \\ &=\frac{1}{2} \int \sec ^{2} x d x \quad\left[\because \frac{1}{\cos x}=\sec x\right] \end{aligned}$
$\begin{aligned} &\frac{1}{2} \tan x+c \\ \end{aligned}$

Indefinite Integrals exercise 18.2 question 34
Answer:

$\begin{aligned} &\frac{-\cot x}{2}+c \\ &\text { Hint: Using } \int \operatorname{cosec}^{2} x d x \\ &\text { Given: } \int \frac{1}{1-\cos 2 x} d x \\ &\text { Solution: } I=\int \frac{1}{1-\cos 2 x} d x \\ &=\int \frac{1}{2 \sin ^{2} x} d x \quad\left[\because 1-\cos 2 \theta=2 \sin ^{2} \theta\right] \\ &=\frac{1}{2} \int \operatorname{cosec}^{2} x d x \quad\left[\because \frac{1}{\sin x}=\operatorname{cosec} x\right] \\ &=\frac{1}{2}(-\cot x)+c \\ &=\frac{-\cot x}{2}+c \end{aligned}$

Indefinite Integrals exercise 18.2 question 35
Answer:

$\begin{aligned} & \frac{x^{2}}{2}+c\\ &\text { Hint: Using } \int x d x\\ &\text { Given: } \int \tan ^{-1}\left(\frac{\sin 2 x}{1+\cos 2 x}\right) d x\\ &\text { Solution: } I=\int \tan ^{-1}\left(\frac{\sin 2 x}{1+\cos 2 x}\right) d x \end{aligned}$
$\begin{aligned} &=\int \tan ^{-1}\left(\frac{2 \sin x \cos x}{2 \cos ^{2} x}\right) d x \quad\left[\begin{array}{l} \because \sin 2 \theta=2 \sin \theta \cos \theta \\ 1+\cos 2 \theta=2 \cos ^{2} \theta \end{array}\right. \\ &=\int \tan ^{-1}(\tan x) d x \quad\left[\because \frac{\sin x}{\cos x}=\tan x\right] \\ &=\int x d x=\frac{x^{2}}{2}+c \end{aligned}$

Indefinite Integrals exercise 18.2 question 36
Answer:

$\begin{aligned} & \frac{\pi}{2} x-\frac{x^{2}}{2}+c\\ &\text { Hint: Using } \int x d x\\ &\text { Given: } \int \cos ^{-1}(\sin x) d x\\ &\text { Solution: } I=\int \cos ^{-1}(\sin x) d x\\ &=\int \cos ^{-1}\left(\cos \left(\frac{\pi}{2}-x\right)\right) d x \quad\left[\because \sin x=\cos \left(\frac{\pi}{2}-x\right)\right]\\ &=\int\left(\frac{\pi}{2}-x\right) d x\\ &=\frac{\pi}{2} \int 1 d x-\int x d x\\ &=\frac{\pi}{2} x-\frac{x^{2}}{2}+c \end{aligned}$

Indefinite Integrals exercise 18.2 question 37
Answer:

$\begin{aligned} & \frac{x^{2}}{2}+c\\ &\text { Hint: Using } \int x d x\\ &\text { Given: } \int \cot ^{-1}\left(\frac{\sin 2 x}{1-\cos 2 x}\right) d x \end{aligned}$
$\begin{aligned} &\text { Solution: } I=\int \cot ^{-1}\left(\frac{\sin 2 x}{1-\cos 2 x}\right) d x \\ &=\int \cot ^{-1}\left(\frac{2 \sin x \cos x}{2 \sin ^{2} x}\right) d x \quad\left[\begin{array}{l} \because \sin 2 \theta=2 \sin \theta \cos \theta \\ 1-\cos 2 \theta=2 \sin ^{2} \theta \end{array}\right] \\ &=\int \cot ^{-1}\left(\frac{\cos x}{\sin x}\right) d x=\int \cot ^{-1}(\cot x) d x \quad\left[\because \cot x=\frac{\cos x}{\sin x}\right] \\ &=\int x d x \\ &=\frac{x^{2}}{2}+c \end{aligned}$

Indefinite Integrals exercise 18.2 question 38
Answer:

$\begin{aligned} &x^{2}+c \end{aligned}$
$\begin{aligned} &\text { Hint: Using } \int x d x\\ &\text { Given: } \int \sin ^{-1}\left(\frac{2 \tan x}{1+\tan ^{2} x}\right) d x\\ &\text { Solution: } I=\int \sin ^{-1}\left(\frac{2 \tan x}{1+\tan ^{2} x}\right) d x\\ &=\int \sin ^{-1}(\sin 2 x) d x \quad\left[\because \sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right]\\ &=\int 2 x d x=\frac{2 x^{2}}{2}+c\\ &=x^{2}+c \end{aligned}$

Indefinite Integrals exercise 18.2 question 39
Answer:

$\begin{aligned} &\frac{x^{3}}{3}+\frac{x^{2}}{2}-2 x+c\\ &\text { Hint: Using } \int x^{n} d x\\ &\text { Given: } \int \frac{\left(x^{3}+8\right)(x-1)}{x^{2}-2 x+4} d x \end{aligned}$ $\begin{aligned} &\text { Solution: } I=\int \frac{\left(x^{3}+8\right)(x-1)}{x^{2}-2 x+4} d x \\ &=\int \frac{\left(x^{3}+2^{3}\right)(x-1)}{x^{2}-2 x+4} d x \\ &=\int \frac{(x+2)\left(x^{2}-2 x+4\right)(x-1)}{x^{2}-2 x+4} d x \quad\left[\because a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)\right] \\ &=\int(x+2)(x-1) d x \\ &=\int\left(x^{2}+x-2\right) d x \quad\left[\begin{array}{l} \because(x+2)(x-1)=x(x-1)+2(x-1) \\ =x^{2}-x+2 x-2=x^{2}+x-2 \end{array}\right] \\ &=\int x^{2} d x+\int x d x+2 \int 1 d x \\ &=\frac{x^{3}}{3}+\frac{x^{2}}{2}-2 x+c \end{aligned}$

Indefinite Integrals exercise 18.2 question 40
Answer:

$\begin{aligned} & a^{2} \tan x-b^{2} \cot x-(a-b) x\\ &\text { Hint: Using } \int \sec ^{2} x d x \text { and } \int \cos e c^{2} x d x\\ &\text { Given: } \int(a \tan x+b \cot x)^{2} d x\\ &\text { Solution: }\\ &I=\int(a \tan x+b \cot x)^{2} d x\\ &=\int\left(a^{2} \tan ^{2} x+b^{2} \cot ^{2} x+2 a b \tan x \cot x\right) d x \quad\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right]\\ &=\int\left(a^{2} \tan ^{2} x+b^{2} \cot ^{2} x+2 a b(1)\right) d x \quad\left[\begin{array}{l} \because \tan x=\frac{1}{\cot x} \\ \Rightarrow \tan x \cot x=1 \end{array}\right] \end{aligned}$ $\begin{aligned} &=a^{2} \int \tan ^{2} x d x+b^{2} \int \cot ^{2} x d x+2 a b \int 1 d x\left[\begin{array}{l} \because \tan ^{2} \theta=\sec ^{2} \theta-1 \\ \cot ^{2} \theta=\cos e c^{2} x-1 \end{array}\right] \\ &=a^{2} \int\left(\sec ^{2} x-1\right) d x+b^{2} \int\left(\cos e c^{2} x-1\right) d x+2 a b \int 1 d x \\ &=a^{2} \tan x-a^{2} x+b^{2}(-\cot x)-b^{2} x+2 a b x \\ &I=a^{2} \tan x-b^{2} \cot x-x\left(a^{2}+b^{2}-2 a b\right) \\ &=a^{2} \tan x-b^{2} \cot x-(a-b)^{2} x \quad\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right] \end{aligned}$

Indefinite Integrals exercise 18.2 question 41
Answer:

$\begin{aligned} &\frac{1}{2}\left[\frac{x^{2}}{2}-3 x+5 \log |x|+\frac{7}{x}+\frac{a^{x}}{\log a}\right]+c\\ &\text { Hint: Using } \int x^{n} d x \text { and } \int a^{x} d x\\ &\text { Given: } \int \frac{x^{3}-3 x^{2}+5 x-7+x^{2} a^{x}}{2 x^{2}} d x\\ &\text { Solution: } I=\int \frac{x^{3}-3 x^{2}+5 x-7+x^{2} a^{x}}{2 x^{2}} d x\\ &=\int \frac{x^{3}}{2 x^{2}} d x-\int \frac{3 x^{2}}{2 x^{2}} d x+\int \frac{5 x}{2 x^{2}} d x-7 \int \frac{1}{2 x^{2}} d x+\int \frac{x^{2} a^{x}}{2 x^{2}} d x\\ &=\frac{1}{2} \int x d x-\frac{3}{2} \int 1 d x+\frac{5}{2} \int \frac{1}{x} d x-\frac{7}{2} \int \frac{1}{x^{2}} d x+\frac{1}{2} \int a^{x} d x\\ &=\frac{1}{2}\left[\frac{x^{2}}{2}\right]-\frac{3}{2} x+\frac{5}{2} \log |x|+\frac{7}{2 x}+\frac{1}{2} \frac{a^{x}}{\log a}+c\\ &=\frac{1}{2}\left[\frac{x^{2}}{2}-3 x+5 \log |x|+\frac{7}{x}+\frac{a^{x}}{\log a}\right]+c \end{aligned}$

Indefinite Integrals exercise 18.2 question 42
Answer:

$\begin{aligned} & x-\tan \frac{x}{2}+c \\ &\text { Hint: } \mathrm{U} \operatorname{sing} \int \sec ^{2} x d x \\ &\text { Given: } \int \frac{\cos x}{1+\cos x} d x \\ &\text { Solution: } I=\int \frac{\cos x}{1+\cos x} d x \end{aligned}$
$\begin{aligned} &=\int \frac{(\cos x+1-1)}{1+\cos x} d x \\ &=\int \frac{\cos x+1}{1+\cos x} d x-\int \frac{1}{1+\cos x} d x \\ &=\int 1 d x-\int \frac{1}{2 \cos ^{2} \frac{x}{2}} d x \\ &=\int 1 d x-\frac{1}{2} \int \sec ^{2} \frac{x}{2} d x \\ &=x-\frac{1}{2}\left[\frac{\tan \frac{x}{2}}{\frac{1}{2}}\right]+c \\ &=x-\tan \frac{x}{2}+c \end{aligned}$

Indefinite Integrals exercise 18.2 question 43

$\begin{aligned} &\text { Answer: } 2 \tan \frac{x}{2}-x+c \\ &\text { Hint: Using } \int \sec ^{2} x d x \\ &\text { Given: } \int \frac{1-\cos x}{1+\cos x} d x \\ &\text { Solution: } I=\int \frac{1-\cos x}{1+\cos x} d x \\ &=\int \frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}} d x \quad\left[\begin{array}{l} \because 1-\cos 2 \theta=2 \sin ^{2} \theta \\ 1+\cos 2 \theta=2 \cos ^{2} \theta \end{array}\right] \\ &=\int \tan ^{2} \frac{x}{2} d x \\ &=\int\left(\sec ^{2} \frac{x}{2}-1\right) d x \quad\left[\because 1+\tan ^{2} \theta=\sec ^{2} \theta\right] \end{aligned}$
$\begin{aligned} &=\int \sec ^{2} \frac{x}{2} d x-\int 1 d x \\ &=\frac{\tan \frac{x}{2}}{\frac{1}{2}}-x+c \\ &=2 \tan \frac{x}{2}-x+c \end{aligned}$

Indefinite Integrals exercise 18.2 question 44

Answer:

$\begin{aligned} &=-3 \cos x-4 \sin x+6 \tan x+7 \cot x+c \end{aligned}$
Hint: You must know about integration of all trigonometry function
$\begin{aligned} &\text { Given: } \int\left\{3 \sin x-4 \cos x+\frac{5}{\cos ^{2} x}-\frac{6}{\sin ^{2} x}+\tan ^{2} x-\cot ^{2} x\right\} d x\\ &\text { Solution: } \int 3 \sin x d x-\int 4 \cos x d x+5 \int \frac{1}{\cos ^{2} x} d x-6 \int \frac{1}{\sin ^{2} x} d x+\int \tan ^{2} x d x-\int \cot ^{2} x d x\\ &=\int 3 \sin x d x-\int 4 \cos x d x+5 \int \sec ^{2} x d x-6 \int \operatorname{cosec}^{2} x d x+\int\left(\sec ^{2} x-1\right) d x-\int\left(\operatorname{cosec}^{2} x-1\right) d x\\ &\left[\because \sec \theta=\frac{1}{\cos \theta} ; \operatorname{cosec} \theta=\frac{1}{\sin \theta} ; \tan ^{2} \theta=\sec ^{2} \theta-1 ; \cot ^{2} \theta=\operatorname{cosec}^{2} \theta-1\right]\\ &=3(-\cos x)-4 \sin x+5 \tan x-6(-\cot x)+\tan x-x-(-\cot x)+x+c\\ &=-3 \cos x-4 \sin x+6 \tan x+7 \cot x+c \end{aligned}$

Indefinite Integrals exercise 18.2 question 45
Answer:

$\frac{x^{2}}{2}+\frac{1}{x}+c\\ Hint: Using\;\; \int x^{n} d x\\ Given: If \;\; f^{\prime}(x)=x-\frac{1}{x^{2}} \;\;\; and \;\;\; f(1)=\frac{1}{2}, \;\;\; find \;\;\;f(x)\\ Solution: f^{\prime}(x)=x-\frac{1}{x^{2}}\\$
Integrating both side
$f(x)=\int\left(x-\frac{1}{x^{2}}\right) d x$
$\begin{aligned} &=\int x d x-\int \frac{1}{x^{2}} d x\\ &=\int x d x-\int x^{-2} d x\\ &=\frac{x^{2}}{2}-\frac{x^{-2+1}}{-2+1}+c\\ &\therefore f(x)=\frac{x^{2}}{2}+x^{-1}+c=\frac{x^{2}}{2}+\frac{1}{x}+c \end{aligned}$
$\begin{aligned} &\text { We have } f(1)=\frac{1}{2}\\ &\Rightarrow \frac{(1)^{2}}{2}+\frac{1}{1}+c=\frac{1}{2}\\ &\Rightarrow \frac{1}{2}+\frac{1}{1}+c=\frac{1}{2}\\ &\Rightarrow c=-1\\ &\text { Put in (1) }\\ &\therefore f(x)=\frac{x^{2}}{2}+\frac{1}{x}-1 \end{aligned}$

Indefinite Integrals exercise 18.2 question 46

Answer:

$\begin{aligned} &\frac{x^{2}}{2}+\frac{13}{2} x-2\\ &\text { Hint: } \mathrm{U} \operatorname{sing} \int x^{n} d x\\ &\text { Given: If } f^{\prime}(x)=x+b \text { and } f(1)=5, \text { find } f(x) \end{aligned}$
$\begin{aligned} &\text { Solution: } f^{\prime}(x)=x+b\\ &\text { Integrating both sides. }\\ &f(x)=\int(x+b) d x\\ &f(x)=\frac{x^{2}}{2}+b x+c \quad \text { ..... } \end{aligned}$
$\begin{aligned} &\text { We have } f(1)=5, f(2)=13\\ &\therefore f(x)=\frac{(1)^{2}}{2}+b(1)+c=5\\ &\Rightarrow \frac{1}{2}+b+c=5 \Rightarrow b+c=5-\frac{1}{2} \Rightarrow b+c=\frac{9}{2} \quad \ldots \ldots .(2)\\ &\text { Also } f(2)=13\\ &\Rightarrow \frac{2^{2}}{2}+b(2)+c=13\\ &=2+2 b+c=13 \Rightarrow 2 b+c=11 \end{aligned}$ $\begin{aligned} &\text { Solving (2) and (3) }\\ &b+c=\frac{9}{2}\\ &2 b+c=11\\ &-b \quad=\frac{9}{2}-11\\ &\Rightarrow-b=\frac{-13}{2}\Rightarrow b=\frac{13}{2} \\ &\text { Put in (3) } \\ &2\left(\frac{13}{2}\right)+c=11 \Rightarrow 13+c=11 \Rightarrow c=-2 \end{aligned}$
$\begin{aligned} &\text { Put the values in (1); We get }\\ &f(x)=\frac{x^{2}}{2}+\frac{13}{2} x-2 \end{aligned}$

Indefinite Integrals exercise 18.2 question 47

Answer:

$2x^{4}-x^{4}-20\\ hint:\;; Using\;;\int x^{n}dx\\ Given:\;\;If\;\;{f}'(x) =8x^{3}-2x\;\;and\;\;f(2)=8,\;\; find \;\;f(x)\\ Solution:\;\;{f}'(x)=8x^{3}-2x$
Integration both side
$\begin{aligned} &f(x)=\int\left(8 x^{3}-2 x\right) d x \\ &f(x)=\frac{8 x^{4}}{4}-\frac{2 x^{2}}{2}+c \\ &f(x)=2 x^{4}-x^{2}+c \quad \ldots \ldots(1) \\ &\text { We have } f(2)=8 \\ &\therefore f(2)=2(2)^{4}+(2)^{2}+c=8 \\ &\Rightarrow 32-4+c=8 \Rightarrow c=8-28 \Rightarrow c=-20 \end{aligned}$
Put in(1) we get
$f(x)=2x^{4}-x^{4}-20\\$

Indefinite Integrals exercise 18.2 question 48

Answer:

$\begin{aligned} &2 \cos x+4 \sin x+1\\ &\text { Hint: Using } \int \sin x d x \text { and } \int \cos x d x\\ &\text { Given: } f^{\prime}(x)=a \sin x+b \cos x ; f^{\prime}(0)=4 ; f(0)=3, f\left(\frac{\pi}{2}\right)=5\\ &\text { Solution: } f^{\prime}(x)=a \sin x+b \cos x\\ &\text { Integrating both sides. }\\ &f(x)=\int a \sin x d x+\int b \cos x d x\\ &f(x)=a[-\cos x]+b[\sin x]+c\\ &f(x)=-a \cos x+b \sin x+c \quad \ldots \ldots .(*) \end{aligned}$
$\begin{aligned} &\mathrm{f}(0)=-a \cos 0+b \sin 0+c=3 \\ &\Rightarrow-a(1)+b(0)+c=3 \\ &\Rightarrow-a+c=3 \quad \ldots \ldots(1) \\ &f\left(\frac{\pi}{2}\right)=-a \cos \frac{\pi}{2}+b \sin \frac{\pi}{2}+c=5 \\ &=-a(0)+b(1)+c=5 \Rightarrow b+c=5 \end{aligned}$
$\begin{aligned} &\text { Also } f^{\prime}(0)=4 \\ &\text { i.e } a \sin 0+b \cos 0=4 \Rightarrow a(0)+b(1)=4 \Rightarrow b=4\\ &Put \;\;in (2)\\ &\text { i.e } b+c=5 \\ &=4+c=5 \Rightarrow c=1\\ &Put\;\; in(1)\\ &\text { i.e }-a+c=3 \\ &\Rightarrow-a+1=3 \Rightarrow-a=2 \Rightarrow a=-2\\ & \therefore \;\; By\;\; putting \;\;all\;\; the \;\;values\;\; in(*) ,\;\; we \;\;get\\ &f(x)=-(-2) \cos x+4 \sin x+c \\ &\Rightarrow f(x)=2 \cos x+4 \sin x+1 \end{aligned}$

Indefinite Integrals exercise 18.2 question 49

Answer:

$\begin{aligned} & \frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}+c \\ &\text { Hint: } \mathrm{U} \operatorname{sing} \int x^{n} d x \end{aligned}$
Given: Write the primitive or antiderivative of $f(x)=\sqrt{x}+\frac{1}{\sqrt{x}}$
$\begin{aligned} &\text { Solution: } f(x)=\sqrt{x}+\frac{1}{\sqrt{x}}=x^{\frac{1}{2}}+\frac{1}{x^{\frac{1}{2}}} \\ &f(x)=x^{\frac{1}{2}}+x^{-\frac{1}{2}} \end{aligned}$
Integrating both sides
$\begin{aligned} &\int f(x)=\int\left(x^{\frac{1}{2}}+x^{-\frac{1}{2}}\right) d x \\ &=\int x^{\frac{1}{2}} d x+\int x^{-\frac{1}{2}} d x \\ &=\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c\\ &=\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+c \\ &=\frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}+c \end{aligned}$

The 18th Chapter of the Class 12 maths book is Indefinite Integrals which is a lengthy and complex topic. In order to ace this chapter, students need to seek the help of the class 12 RD Sharma chapter 18 exercise 18.2 solution. The chapter covers concepts like Reverse power rule, Graphs of indefinite integrals, Indefinite integrals of common functions, formulas of integrals and so on. The exercise 18.2 has 49 questions including subparts, based on the evaluation of integrals.

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2. What are the concepts covered in chapter 18 of the NCERT class 12 maths book?

The NCERT maths book contains various concepts and the ones covered in the 18th chapter are Reverse power rule, Graphs of indefinite integrals, Indefinite integrals of common functions, etc.

3. How can I use RD Sharma class 12th exercise 18.2 to solve homework?

Teachers often make use of the RD Sharma class 12th exercise 18.2 to give and check homework of their students. Students can take help from this book when they are stuck with answering complex questions.

4. Is class 12 RD Sharma chapter 18 exercise 18.2 solution a good book?

The class 12 RD Sharma chapter 18 exercise 18.2 solution book is a top rated NCERT solution. Reviews from hundreds of students and teachers in India have proved this fact. The most lucrative feature of the book is that it contains expert-created answers with modern and fast calculations.

5. Can I use RD Sharma class 12 solutions for JEE preparations?

Students who will appear for the JEE mains exams after their boards can definitely use RD Sharma class 12 solutions for home practice. The RD Sharma solutions books follow a syllabus that is compatible with the syllabus of JEE mains.