RD Sharma Class 12 Exercise 18.3 Indefinite Integrals Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 18.3 Indefinite Integrals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 10:39 AM IST

Class 12 is a significant school year for students as it will be their last year in school. Therefore, all their subjects, including maths, will have a tough syllabus. The RD Sharma class 12th exercise 18.3 will be an excellent guide for students in high school to help them in their exam preparations. With board exams near, the answers of RD Sharma class 12 chapter 18 exercise 18.3 will be extremely helpful for students to clear their doubts and practice maths at home.

## Indefinite Integrals Excercise: 18.3

Indefinite Integrals Exercise 18.3 Question 1

Answer:$\frac{(2 x-3)^{\6}}{12}+\frac{2}{9}(3 x+2)^{\frac{3}{2}}+c$
Hint: To solve this equation we use $\int(a x+b)^{n}$ formula
Given: $\int(2 x-3)^{5}+\sqrt{3 x+2} d x$
Solution:$\int(a x+b)^{n} d x=\frac{1}{a(n+1)}(a x+b)^{n+1}+c ; n \neq-1$
\begin{aligned} &=\int\left[(2 x-3)^{5}+(3 x+2)^{\frac{1}{2}}\right] d x \\ &=\int(2 x-3)^{5} d x+\int(3 x+2)^{\frac{1}{2}} d x \end{aligned}
\begin{aligned} &=\frac{1}{2(5+1)}(2 x-3)^{5+1}+\frac{1}{3\left(1+\frac{1}{2}\right)}(3 x+2)^{\frac{1}{2}+1}+c \\ &=\frac{1}{12}(2 x-3)^{6}+\frac{2}{9}(3 x+2)^{\frac{3}{2}}+c \end{aligned}

Indefinite Integrals Exercise 18.3 Question 2

Answer: $\frac{-1}{14}(7 x-5)^{-2}+\frac{2}{5} \sqrt{5 x-4}+c$
Hint:To solve this we take denominator as numerator
Given: $\int \frac{1}{(7 x-5)^{3}}+\frac{1}{\sqrt{5 x-4}} d x$
Solution: $\int(7 x-5)^{-3} d x+\int(5 x+4)^{\frac{1}{2}}$
\begin{aligned} &\int(a x+b)^{n} d x=\frac{1}{a(n+1)}(a x+b)^{n+1}+c ; n \neq-1 \\ &=\frac{1}{7} \frac{(7 x-5)^{-2}}{-2}+\frac{1}{5} \frac{(5 x+4)^{\frac{1}{2}}}{\frac{1}{2}}+c \end{aligned}
\begin{aligned} &=\frac{-1}{14}(7 x-5)^{-2}+\frac{2}{5} \sqrt{5 x-4}+c \\ &=\frac{-1}{14}(7 x-5)^{-2}+\frac{2}{5} \sqrt{5 x-4}+c \end{aligned}

Indefinite Integrals Exercise 18.3 Question 3

Answer: $\frac{-1}{3} \log |2-3 x|+\frac{2}{3} \sqrt{3 x-2}+c$
Hint: To solve this equation we use $\int \frac{1}{(a x+b)^{n}} d x$ formula
Given: $\int \frac{1}{2-3 x}+\frac{1}{\sqrt{3 x-2}} d x$
Solution: $\int \frac{1}{2-3 x} d x+\int(3 x-2)^{\frac{-1}{2}} d x$
$\left\{\begin{array}{l} \int \frac{1}{a x+b} d x=\log (a x+b)+c \\ (a x+b)^{n} d x=\frac{(a x+b)^{n+1}}{a(n+1)} \\ b \neq-1 \end{array}\right\}$
\begin{aligned} &=\frac{\log |2-3 x|}{-3}+\frac{(3 x-2)^{\frac{1}{2}}}{3 \cdot \frac{1}{2}}+c \\ &=-\frac{\log |2-3 x|}{3}+\frac{2}{3} \sqrt{3 x+2}+c \\ &=\frac{-1}{3} \log |2-3 x|+\frac{2}{3} \sqrt{3 x-2}+c \end{aligned}

Indefinite Integrals Exercise 18.3 Question 4

Answer: $\frac{-1}{2(x+1)^{2}}-\frac{2}{3(x+1)^{3}}+c$
Hint: $\text { To solve this equation we will spilt } x+3 \text { and then we use } \int \frac{1}{a x+b} d x \text { formula }$
Given: $\int \frac{x+3}{(x+1)^{4}} d x$
Solution: $\frac{x+3}{(x+1)^{4}}=\frac{x+1+2}{(x+1)^{4}}$
\begin{aligned} &=\frac{x+1}{(x+1)^{4}}+\frac{2}{(x+1)^{4}}=\frac{1}{(x+1)^{3}}+\frac{2}{(x+1)^{4}} \\ &=\int \frac{x+3}{(x+1)^{4}} d x \end{aligned}
\begin{aligned} &=\int \frac{1}{(x+1)^{3}} d x+2 \int \frac{1}{(x+1)^{4}} d x\left[\int(a x+b)^{n}=\frac{1}{a(n+1)}(a x+b)^{n+1}+c, n \neq 1\right] \\ &=\frac{1}{1} \times \frac{1}{1-3}(x+1)^{1-3}+2 \frac{1}{1} \times \frac{1}{1-4}(x+1)^{1-4}+c \\ &=\frac{-1}{2}(x+1)^{-2}-\frac{2}{3}(x+1)^{-3}+c \\ &=\frac{-1}{2(x+1)^{2}}-\frac{2}{3(x+1)^{3}}+c \end{aligned}

Indefinite Integrals Exercise 18.3 Question 5

Answer: $\frac{2}{3}\left\{(x+1)^{\frac{3}{2}}-x^{\frac{3}{2}}\right\}+c$
Hint: $\inline \text { To solve this equation we will multiply } \sqrt{x+1}-\sqrt{x} \text { to numerator and denominator }$
Given: $\inline \int \frac{1}{\sqrt{x+1}+\sqrt{x}} d x$
Solution: $\inline \int \frac{\sqrt{x+1}-\sqrt{x}}{(\sqrt{x+1}+\sqrt{x})(\sqrt{x+1}-\sqrt{x})}$
$\inline \because \text { multiply } \sqrt{x+1}-\sqrt{x} \text { to numerator and denominator }$
\inline \begin{aligned} &=\int \frac{\sqrt{x+1}-\sqrt{x}}{x+1-x} d x \\ &=\int \sqrt{x+1} d x-\int \sqrt{x} d x \end{aligned}
$\inline \left[\int(a x+b)^{n}=\frac{(a x+b)^{n+1}}{a(n+1)}+c, n \neq c\right]$
\inline \begin{aligned} &=\frac{(x+1)^{\frac{1}{2}}+1}{\frac{1}{2}+1}-\frac{(x)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c \\ &=\frac{2}{3}\left\{(x+1)^{\frac{3}{2}}-x^{\frac{3}{2}}\right\}+c \end{aligned}

Indefinite Integrals Exercise 18.3 Question 6

Answer: $\frac{1}{18}\left\{(2 x+3)^{\frac{3}{2}}-(2 x-3)^{\frac{8}{2}}\right\}+c$
Hint:$\text { To solve this equation we multiply } \sqrt{2 x+3}-\sqrt{2 x-3} \text { to numerator and denominator }$
Given: $\int \frac{1}{\sqrt{2 x+3}+\sqrt{2 x-3}} d x$
Solution: $\int \frac{1}{\sqrt{2 x+3}+\sqrt{2 x-3}} d x$
$\int \frac{\sqrt{2 x+3}-\sqrt{2 x-3}}{(\sqrt{2 x+3}+\sqrt{2 x-3})(\sqrt{2 x+3}-\sqrt{2 x-3})} d x$
$\because \text { multiply } \sqrt{2 x+3}-\sqrt{2 x-3} \text { to numerator and denominator }$
\begin{aligned} &=\int \frac{\sqrt{2 x+3}-\sqrt{2 x-3}}{2 x+3-2 x+3} d x \\ &=\frac{1}{6} \int \sqrt{2 x+3} d x-\int \sqrt{2 x-3} d x\left[\int(a x+b)^{n}=\frac{(a x+b)^{n+1}}{a(n+1)}+c, n \neq 1\right] \\ &=\frac{1}{6}\left[\frac{1}{2} \frac{(2 x+3)^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{(2 x-3)^{\frac{1}{2}+1}}{2\left(\frac{1}{2}+1\right)}+c\right] \\ &=\frac{1}{18}\left\{(2 x+3)^{\frac{3}{2}}-(2 x-3)^{\frac{8}{2}}\right\}+c \end{aligned}

Indefinite Integrals Exercise 18.3 Question 7

Answer: $\frac{1}{2} \log |2 x+1|+\frac{1}{2(2 x+1)}+c$
Hint : $\text { To solve this equation we differentiate differently }$
Given: $\int \frac{2 x}{(2 x+1)^{2}} d x$
Solution: $\int \frac{2 x}{(2 x+1)^{2}} d x$
$\int \frac{(2 x+1)-1}{(2 x+1)^{2}} d x$
$=\int \frac{1}{(2 x+1)} d x-\int \frac{1}{(2 x+1)^{2}} d x$
$\left[\begin{array}{l} \int \frac{1}{a x+b} d x=\log \frac{1}{a}|a x+b|+c \\ \int(a x+b)^{n}=\frac{(a x+b)^{n+1}}{a(n+1)}+c, n \neq 1 \end{array}\right]$
\begin{aligned} &=\frac{1}{2} \log |2 x+1|+\frac{1}{2} \frac{1}{(2 x+1)}+c \\ &=\frac{1}{2} \log |2 x+1|+\frac{1}{2(2 x+1)}+c \end{aligned}

Indefinite Integrals Exercise 18.3 Question 8

Answer: $\frac{2}{3(a-b)}\left\{(x+a)^{\frac{3}{2}}-(x+b)^{\frac{3}{2}}\right\}+c$
Hint:$\text { To solve this equation we add } \sqrt{x+a}-\sqrt{x+b} \text { to numerator and denominator }$
Given: $\int \frac{1}{\sqrt{x+a}+\sqrt{x+b}} d x$
Solution: $\int \frac{1}{\sqrt{x+a}+\sqrt{x+b}} d x$
$\text { Multiply } \sqrt{x+a}-\sqrt{x+b} \text { to numerator and denominator }$
\begin{aligned} &\int \frac{\sqrt{x+a}-\sqrt{x+b}}{(\sqrt{x+a}+\sqrt{x+b})(\sqrt{x+a}-\sqrt{x+b})} d x \\ &=\int \frac{\sqrt{x+a}-\sqrt{x+b}}{(x+a)-(x+b)} d x \end{aligned}
\begin{aligned} &=\int \frac{\sqrt{x+a}-\sqrt{x+b}}{a-b} d x \\ &=\frac{1}{a-b} \int \sqrt{x+a}-\sqrt{x+b} d x\left[\int(x)^{n}=\frac{(x)^{n+1}}{n+1}+c\right] \\ &=\frac{1}{a-b}\left[\frac{(x+a)^{\frac{3}{2}}}{\frac{3}{2}}-\frac{(x+b)^{\frac{8}{2}}}{\frac{3}{2}}\right]+c \\ &=\frac{1}{a-b} \times \frac{2}{3}\left[(x+a)^{\frac{3}{2}}-(x+b)^{\frac{8}{2}}\right]+c \\ &=\frac{2}{3(a-b)}\left[(x+a)^{\frac{3}{2}}-(x+b)^{\frac{8}{2}}\right]+c \end{aligned}

Indefinite Integrals Exercise 18.3 Question 9

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

Indefinite Integrals Exercise 18.3 Question 10

Answer: $-2 \cot \left(\frac{x}{2}\right)-x+c$
Hint: $\text { To solve this equation we use half angle formula }$
Given: $\int \frac{1+\cos x}{1-\cos x} d x$
Solution: $\int \frac{1+\cos x}{1-\cos x} d x$
$=\int \frac{1+2 \cos ^{2} \frac{x}{2}-1}{1-1+2 \sin ^{2} \frac{x}{2}} d x$\text { [By using half angle formula] }
\begin{aligned} &=\int \frac{2 \cos ^{2} \frac{x}{2}}{2 \sin ^{2} \frac{x}{2}} d x \\ &=\int \cot ^{2} \frac{x}{2} d x \end{aligned}
\begin{aligned} &=\int\left(\cos e c^{2} \frac{x}{2}-1\right) d x \\ &=\int \cos e c^{2} \frac{x}{2} d x-\int d x\left[\int \operatorname{cosec}^{2} x d x=-\cot x+c\right] \\ &=-2 \cot \frac{x}{2}-x+c \end{aligned}
Answer: $2\left[\tan \frac{x}{2}+\sec \frac{x}{2}\right]+c$
Hint: $\text { We will add } 1+\sin \frac{x}{2} \text { to the equation }$
Given: $\int \frac{1}{1-\sin \frac{x}{2}} d x$
Solution: $\int \frac{1}{1-\sin \frac{x}{2}} d x$
$\text { Multiplying } 1+\sin \frac{x}{2} \text { in numerator and denominator }$
$=\int \frac{1+\sin \frac{x}{2}}{\left(1-\sin \frac{x}{2}\right)\left(1-\sin \frac{x}{2}\right)} d x$
\begin{aligned} &=\int \frac{1+\sin \frac{x}{2}}{1-\sin ^{2} \frac{x}{2}} d x \\ &=\int \frac{1+\sin \frac{x}{2}}{\cos ^{2} \frac{x}{2}} d x \end{aligned}
$=\int \frac{1}{\cos ^{2} \frac{x}{2}} d x+\int \frac{\sin \frac{x}{2}}{\cos ^{2} \frac{x}{2}} d x$
$=\int \frac{1}{\cos ^{2} \frac{x}{2}} d x+\int \frac{\sin \frac{x}{2}}{\cos \frac{x}{2} \cdot \cos \frac{x}{2}} d x$
\begin{aligned} &=\int \sec ^{2} \frac{x}{2} d x+\int \tan \frac{x}{2} \sec \frac{x}{2} d x \\ &=2\left[\tan \frac{x}{2}+\sec \frac{x}{2}\right]+c \end{aligned} \quad\left[\int \sec ^{2}(a x+b) d x=\frac{\tan (a x+b)}{a}+c\right]

Indefinite Integrals Exercise 18.3 Question 12

Answer: $2\left[\tan \frac{x}{2}+\sec \frac{x}{2}\right]+c$
Hint: $\text { We will add } 1+\sin \frac{x}{2} \text { to the equation }$
Given: $\int \frac{1}{1-\sin \frac{x}{2}} d x$
Solution: $\int \frac{1}{1-\sin \frac{x}{2}} d x$
$\text { Multiplying } 1+\sin \frac{x}{2} \text { in numerator and denominator }$
$=\int \frac{1+\sin \frac{x}{2}}{\left(1-\sin \frac{x}{2}\right)\left(1-\sin \frac{x}{2}\right)} d x$
\begin{aligned} &=\int \frac{1+\sin \frac{x}{2}}{1-\sin ^{2} \frac{x}{2}} d x \\ &=\int \frac{1+\sin \frac{x}{2}}{\cos ^{2} \frac{x}{2}} d x \end{aligned}
$=\int \frac{1}{\cos ^{2} \frac{x}{2}} d x+\int \frac{\sin \frac{x}{2}}{\cos ^{2} \frac{x}{2}} d x$
$=\int \frac{1}{\cos ^{2} \frac{x}{2}} d x+\int \frac{\sin \frac{x}{2}}{\cos \frac{x}{2} \cdot \cos \frac{x}{2}} d x$
\begin{aligned} &=\int \sec ^{2} \frac{x}{2} d x+\int \tan \frac{x}{2} \sec \frac{x}{2} d x \\ &=2\left[\tan \frac{x}{2}+\sec \frac{x}{2}\right]+c \end{aligned} \quad\left[\int \sec ^{2}(a x+b) d x=\frac{\tan (a x+b)}{a}+c\right]

Indefinite Integrals Exercise 18.3 Question 13

Answer: $\frac{1}{3}\left[\frac{1-\cos 3 x}{\sin 3 x}\right]+C$
Hint: $\text { We will add } 1-\cos 3 x \text { to the equation }$
Given: $\int \frac{1}{1+\cos 3 x} d x$
Solution: $\int \frac{1}{1+\cos 3 x} d x$
$\text { Multiplying } 1-\cos 3 x \text { in numerator and denominator }$
\begin{aligned} &\int \frac{1-\cos 3 x}{(1+\cos 3 x)(1-\cos 3 x)} d x \\ &=\int \frac{1-\cos 3 x}{1-\cos ^{2} 3 x} d x \\ &=\int \frac{1-\cos 3 x}{\sin ^{2} 3 x} d x \end{aligned}
\begin{aligned} &=\int \frac{1}{\sin ^{2} 3 x} d x-\int \frac{\cos 3 x}{\sin 3 x \sin 3 x} d x \\ &=\int \operatorname{cosec}^{2} 3 x d x-\int \cot 3 x \operatorname{cosec} 3 x d x \\ &=\frac{-\cot 3 x}{3}+\frac{\operatorname{cosec} 3 x}{3}+c \end{aligned}
\begin{aligned} &=\frac{1}{3}[\operatorname{cosec} 3 x-\cot 3 x]+c \\ &=\frac{1}{3}\left[\frac{1}{\sin 3 x}-\frac{\cos 3 x}{\sin 3 x}\right]+C \\ &=\frac{1}{3}\left[\frac{1-\cos 3 x}{\sin 3 x}\right]+C \end{aligned}

Indefinite Integrals Exercise 18.3 Question 14

Answer: $\frac{e^{3 x}}{3}+e^{2 x}+e^{x}+c$
Hint: $\text { To solve this } \int x^{n} d x \text { formula }$
Given: $\int\left(e^{x}+1\right)^{2} e^{x} d x$
Solution: $\int\left(e^{x}+1\right)^{2} e^{x} d x$
\begin{aligned} &=\int\left(e^{2 x}+1+2 e^{x}\right) e^{x} d x \quad\left[(a+b)^{2}=a^{2}+b^{2}+2 a b\right] \\ &=\int\left(e^{3 x}+2 e^{2 x}+e^{x}\right) d x \end{aligned}
\begin{aligned} &=\int\left(e^{3 x}\right) d x+2 \int\left(e^{2 x}\right) d x+\int\left(e^{x}\right) d x \\ &=\frac{e^{3 x}}{3}+e^{2 x}+e^{x}+c \\ &=\frac{e^{3 x}+e^{2 x}+e^{x}}{3}+c \\ &=\frac{1}{3}\left(e^{x}+1\right)^{2}+c \end{aligned}

Indefinite Integrals Exercise 18.3 Question 15

Answer: $\frac{1}{2} e^{2 x}+2 x-\frac{1}{2} e^{-2 x}+c$
Hint: $\text { To solve this equation } \int e^{a x} d x \text { formula }$
Given: $\int\left(e^{x}+\frac{1}{e^{x}}\right)^{2} d x$
Solution: $=\left(e^{x}+\frac{1}{e^{x}}\right)^{2} \quad\left[\begin{array}{l} (a+b)^{2}=a^{2}+b^{2}+2 a b \\ \int e^{a x} d x=\frac{1}{a} e^{a x}+c, a \neq 0 \end{array}\right]$
\begin{aligned} &=\left(e^{x}\right)^{2}+2 \cdot e^{x} \cdot \frac{1}{e^{x}}+\left(\frac{1}{e^{x}}\right)^{2} \\ &=e^{2 x}+2+e^{-2 x} \end{aligned}
\begin{aligned} &\int\left(e^{x}+\frac{1}{e^{x}}\right)^{2} d x=\int\left(e^{2 x}+2+e^{-2 x}\right) d x \\ &=\frac{1}{2} e^{2 x}+2 x-\frac{1}{2} e^{-2 x}+c \\ &=\frac{1}{2}\left(e^{2 x}-e^{-2 x}\right)+2 x+c \end{aligned}

### Indefinite Integrals Exercise 18.3 Question 16

Answer: $\frac{-1}{8} \cos 4 x+c$
Hint: $\text { To solve this equation we use } \int \sin 2 x_{5} \int \cos 2 x d x \text { formula }$
Given: $\int \frac{1+\cos 4 x}{\cot x-\tan x} d x$
Solution: $\int \frac{1+\cos 4 x}{\cot x-\tan x} d x$
\begin{aligned} &=\frac{2 \cos ^{2} 2 x}{\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}} d x \\ &=\frac{2 \cos ^{2} 2 x}{\frac{\cos ^{2} x-\sin ^{2} x}{\sin x \cos x}} d x \\ &=\frac{2(\sin x \cos x)(\cos 2 x)^{2}}{\cos ^{2} x-\sin ^{2} x} d x \end{aligned}
\begin{aligned} &=\frac{\sin 2 x(\cos 2 x)^{2}}{\cos 2 x} \\ &=\sin 2 x \cos 2 x \end{aligned} \quad\left[\begin{array}{c} \because 2 \sin x \cos x=\sin 2 x \\ \cos ^{2} x-\sin ^{2} x=\cos 2 x \end{array}\right]
\begin{aligned} &\text { Multiply } 2 \text { in numerator and denominator }\\ &=\frac{1}{2}(2 \sin 2 x \cos 2 x)\\ &=\frac{1}{2}(\sin 4 x) \quad[\because 2 \sin 2 x \cos 2 x=\sin 4 x] \end{aligned}
\begin{aligned} &\int \frac{1+\cos 4 x}{\cot x-\tan x} d x=\int \frac{1}{2}(\sin 4 x) d x \\ &=\frac{1}{2} \int(\sin 4 x) d x \quad\left[\because \int \sin (a x+b) d x=\frac{-1}{a} \cos (a x+b)+c, a \neq 0\right] \\ &=\frac{1}{2}\left(\frac{-1}{4} \cos 4 x\right)+c \\ &=\frac{-1}{8} \cos 4 x+c \end{aligned}

Indefinite Integrals Exercise 18.3 Question 17

Answer: $\frac{2}{3}\left\{(x+3)^{\frac{3}{2}}+(x+2)^{\frac{3}{2}}\right\}+c$
Hint: $\text { To solve this equation we will add }(\mathrm{x}+3)+(\mathrm{x}+2 \text { ) in numerator and denominator }$
Given: $\int \frac{1}{\sqrt{x+3}-\sqrt{x+2}} d x$
Solution: $\int \frac{1}{\sqrt{x+3}-\sqrt{x+2}} d x$
\begin{aligned} &=\int \frac{\sqrt{x+3}+\sqrt{x+2}}{(\sqrt{x+3}-\sqrt{x+2})(\sqrt{x+3}+\sqrt{x+2})} d x \\ &=\int \frac{\sqrt{x+3}+\sqrt{x+2}}{x+3-x-2} d x \\ &=\int \sqrt{x+3} d x+\int \sqrt{x+2} d x \end{aligned}
\begin{aligned} &{\left[\int(a x+b)^{n}=\frac{(a x+b)^{n+1}}{a(n+1)}+c, n \neq 1\right]} \\ &=\frac{2(x+3)^{\frac{3}{2}}}{3}+\frac{2(x+2)^{\frac{3}{2}}}{3}+c \\ &=\frac{2}{3}\left\{(x+3)^{\frac{3}{2}}+(x+2)^{\frac{3}{2}}\right\}+c \end{aligned}

Indefinite Integrals Exercise 18.3 Question 18

Answer: $\frac{1}{2} \tan (2 x-3)-x+c$
HInt: $\text { To solve this we convert tan to sec form }$
Given: $\int \tan ^{2}(2 x-3) d x$
Solution: $\int \tan ^{2}(2 x-3) d x$
\begin{aligned} &=\int \sec ^{2}(2 x-3)-1 d x\left[\because \tan ^{2} x=\sec ^{2} x-1\right] \\ &=\int \sec ^{2}(2 x-3)-\int 1 d x \\ &=\frac{\tan (2 x-3)}{2}-x+c\left[\because \int \sec ^{2}(a x+b) d x=\frac{\tan (a x+b)}{a}+c\right] \\ &=\frac{1}{2} \tan (2 x-3)-x+c \end{aligned}

Indefinite Integrals Exercise 18.3 Question 19

Answer: $\frac{1}{1-\tan (x)}+c$
Hint: $\text { To solve this we use } \tan x \text { as } t \text { and } \frac{1}{\cos x}=\sec x$
Given: $\int \frac{1}{\cos ^{2} x(1-\tan x)^{2}} d x$
Solution: $\int \frac{1}{\cos ^{2} x(1-\tan x)^{2}} d x$ $\left[\tan x=\frac{\sin x}{\cos x}\right]$
$I=\int \frac{1}{\cos ^{2} x\left(1-\frac{\sin x}{\cos x}\right)^{2}} d x$
\begin{aligned} &=\int \frac{1}{(\cos x-\sin x)^{2}} d x \\ &=\int \frac{1}{1-\sin 2 x} d x \\ &=\int \frac{1}{1+\cos \left(\frac{\pi}{2}+2 x\right)} d x \\ &=\int \frac{1}{2 \cos ^{2}\left(\frac{\pi}{4}+x\right)} d x \\ &=\int \sec ^{2}\left(\frac{\pi}{4}+x\right) d x \\ &=\frac{1}{2} \tan \left(\frac{\pi}{4}+x\right)+c \end{aligned}

The 18th chapter of the class 12 maths book contains the topic of Indefinite Integrals, which has a huge syllabus. The concepts that students will have to learn are the Reverse power rule, Graphs of indefinite integrals, Indefinite integrals of common functions, etc. Exercise 18.3 has 19 questions in the third part of the chapter.

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## RD Sharma Chapter wise Solutions

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1. Is class 12 RD Sharma chapter 18 exercise 18.2 solution a recommended book?

The class 12 RD Sharma chapter 18 exercise 18.2 solution is one of the best  NCERT solutions as claimed by students and teachers alike. Experts have created the answers in the book so they contain modern methods and calculations.

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3. What is the 18th Chapter of class 12 maths book?

The 18th Chapter of the Class 12 maths book contains the topic Indefinite Integrals. The concepts which will be taught to students will have the Reverse power rule,, Indefinite integrals of common functions, Graphs of indefinite integrals, and the like.

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