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

# 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

NCERT Solutions can be extremely helpful for students who are in class 12. Since board exams are near, students need to put in a lot of practice to score high and top their exams. This is why they should use RD Sharma class 12th exercise 18.2 Solutions for their exam preparations. RD Sharma solutions is a respectable and trusted name among all the NCERT solutions. They contain Solutions to all chapters including the RD Sharma class 12 chapter 18 exercise 18.2. It is not surprising that both students and teachers recommend the book among all others.

## Indefinite Integrals Excercise:18.2

Indefinite Integrals exercise 18.2 question 1

$\frac{6}{5} x^{\frac{5}{2}}+\frac{1}{2} x^{\frac{3}{2}}+5 x+c$
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}

$\frac{2^{x}}{\log 2}+5 \log x-\frac{3}{2} x^{\frac{2}{3}}+c$
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

$\frac{2}{7} a x^{\frac{7}{2}}+\frac{2}{5} b x^{\frac{5}{2}}+\frac{2}{3} c x^{\frac{3}{2}}+c$
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}
$3 x^{4}+\frac{4}{3} x^{3}-\frac{17}{2} x^{2}+6 x+c$

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

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

$\frac{x^{3}}{2}+\log x-2 x+c$
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

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

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

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

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

$\frac{2}{9} x^{\frac{9}{2}}-4 x^{\frac{1}{2}}+c$
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

$2 x^{\frac{1}{2}}-\frac{1}{2 x^{\frac{1}{2}}}+c$
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

$\frac{x^{5}}{5}-\frac{x^{3}}{3}+x+c$
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

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

$2 x^{\frac{3}{2}}-2 x^{\frac{5}{2}}+c$
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

$\frac{2}{5} x^{\frac{5}{2}}-\frac{2}{3} x^{\frac{3}{2}}-4 \sqrt{x}+c$
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

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

$3 x^{3}+12 x^{2}+16 x+c$
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

$\frac{3x^{2}}{2}+\frac{2x^{3}}{3}+c$
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

\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

$x-sin x +c$
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

$\sec x-\cos ec x+c$
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

$\frac{-5}{2} \operatorname{cosec} x+3 \sec x+c$
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 25Answer:

$\tan x-\cot x+c$
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

$\tan x-x+c$
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}
$-(cosec x + cot x +x)+c$
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}
$\\ \\ \frac{\cot x}{\cos ec x -\cot x}\times \frac{\cos ec x+\cot x}{\cos ec x+\cot x}=\cot x\cdot \cos ec x+\cot^{2}x \\ \\ \Rightarrow \hspace{1cm}I=\int (\cot x \cdot \cos ec x)dx+\int (\cos ec^{2}x-1)dx \\ \\ \Rightarrow \hspace{1cm}I=-\cos ec x -\cot x+c$
$\frac{1}{\sqrt{2}}x+c$
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}
$-cot x- cosec x +c$
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}

$\tan x+ sec x+c$
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}
$\sec x-\tan x +x+c$
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}
$-cot x-cosec x+c$
\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

\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}
\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}
\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 36Answer:

\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}
\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}

\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}
\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}
\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}
\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}
\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

\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}

$\frac{x^{2}}{2}+\frac{1}{x}+c\\ Hint: Using\;\; \int x^{n} d x\\ Given: If \;\; f^{\prime}(x)=x-\frac{1}{x^{2}} \;\;\; and \;\;\; f(1)=\frac{1}{2}, \;\;\; find \;\;\;f(x)\\ Solution: f^{\prime}(x)=x-\frac{1}{x^{2}}\\$
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

\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

$2x^{4}-x^{4}-20\\ hint:\;; Using\;;\int x^{n}dx\\ Given:\;\;If\;\;{f}'(x) =8x^{3}-2x\;\;and\;\;f(2)=8,\;\; find \;\;f(x)\\ Solution:\;\;{f}'(x)=8x^{3}-2x$
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

\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

\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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