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

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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

Updated on 24 Jan 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
$\begin{aligned} &\int f(x)=\int\left(x^{\frac{1}{2}}+x^{-\frac{1}{2}}\right) d x \\ &=\int x^{\frac{1}{2}} d x+\int x^{-\frac{1}{2}} d x \\ &=\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c\\ &=\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+c \\ &=\frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}+c \end{aligned}$

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All the solved questions of RD Sharma class 12 solutions chapter 4 ex 18.2 will help students test their own knowledge and mark their progress. They will also find their mistakes easily and rectify them accordingly.
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The RD Sharma class 12th exercise 18.2 receives regular updates to include the latest syllabus and changes in the NCERT textbooks. Hence, the latest information is always included.