RD Sharma Class 12 Exercise 18.12 Indefinite Integrals Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 18.12 Indefinite Integrals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 12:09 PM IST

RD Sharma class 12th exercise 18.12 is a top-rated NCERT solution that students will find among study materials. The brand has made a name for itself as numerous students have placed their trust in RD Sharma class 12 chapter 18 exercise 18.12 and greatly benefitted from the answers in the book. High school students who are in class 12 especially should use the RD Sharma class 12th exercise 18.12 for their exam preparations and home practice. Maths is a tough subject and to master it students need to practice answering questions daily. Self-study at home will be pretty easy with the help of RD Sharma solutions to master it. If you want to sharpen your skills in maths, thr class 12 RD Sharma chapter 18 exercise 18.12 solution will assist you in this endeavor.

## Indefinite Integrals Excercise:18.12

Indefinite Integrals Exercise 18.12 Question 1

Answer: - $\frac{1}{5} \sin ^{5} x-\frac{1}{7} \sin ^{7} x+C$
Hint: - Use substitution method to solve this integral
Given:-$\int \sin ^{4} x \cos ^{3} x d x$
Solution: - Let, $I=\int \sin ^{4} x \cos ^{3} x d x$
The exponent of $\cos x$ is odd, so we substitute $\sin x=t\Rightarrow \cos dt$then,
$I=\int t^{4} \cos ^{3} x \frac{d t}{\cos x} \quad\quad\quad\quad\quad\quad\quad[\because \sin x=t]$
$=\int t^{4} \cos ^{2} x d t$
$\begin{array}{ll} =\int t^{4}\left(1-\sin ^{2} x\right) d t &\quad\quad\quad\quad\quad\quad {\left[\because \sin ^{2} x=\cos ^{2} x=1\right]} \\ \\=\int t^{4}\left(1-t^{2}\right) d t & \quad\quad\quad\quad\quad\quad{[\because \sin x=t]} \end{array}$
\begin{aligned}\\ &=\int\left(t^{4}-t^{4} t^{2}\right) d t \\ &=\int\left(t^{4}-t^{6}\right) d t \\ &=\int t^{4} d t-\int t^{6} d t \end{aligned}
$=\frac{t^{4+1}}{4+1}-\frac{t^{6+1}}{6+1}+C$ $\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right]$
$=\frac{t^{5}}{5}-\frac{t^{7}}{7}+C$
$=\frac{\sin ^{5} x}{5}-\frac{\sin ^{7}}{7}+C$ $\left [ \because t=\sin x \right ]$

Indefinite Integrals Exercise 18.12 Question 2

Answer:$-\cos x+\frac{2}{3} \cos ^{3} x-\frac{1}{5} \cos ^{5} x+C$
Hint: Use substitution method to solve this integral
Given:$\int \sin^{5 }xdx$
Solution: Let, $I=\int \sin^{5 }xdx$
Re-writing,
\begin{aligned} &I=\int \sin ^{3} x \sin ^{2} x d x \\ &=\int \sin ^{3} x\left(1-\cos ^{2} x\right) \quad \quad \quad \quad \quad \quad\left[\because \sin ^{2} x+\cos ^{2} x=1\right] \end{aligned}
\begin{aligned} &=\int\left(\sin ^{3} x-\sin ^{3} x \cos ^{2} x\right) d x \\ &=\int\left(\sin ^{2} x \sin x-\sin ^{3} x \cos ^{2} x\right) d x \\ &\left.=\int\left(1-\cos ^{2} x\right) \sin x-\sin ^{3} x \cos ^{2} x\right) d x \quad\quad\quad\quad\quad\quad\left[\because \sin ^{2} x=1-\cos ^{2} x\right] \end{aligned}
\begin{aligned} &=\int\left(\sin x-\sin x \cos ^{2} x-\sin ^{3} x \cos ^{2} x\right) d x \\ &=\int \sin x d x-\int \sin x \cos ^{2} x-\int \sin ^{3} x \cos ^{2} x d x \end{aligned}
Substitute $\cos x=t \Rightarrow-\sin x d x \Rightarrow d t$ in second and third integral, then
\begin{aligned} I &=\int \sin x d x-\int t^{2}(-d t)-\int t^{2} \sin ^{3} x \frac{d t}{-\sin x} \\ &=\int \sin x d x+\int t^{2} d t+\int \sin ^{2} x t^{2} d t \\ &=\int \sin x d x+\int t^{2} d t+\int\left(1-\cos ^{2} x\right) t^{2} d t\quad\quad\quad\quad\quad\quad&\left[\because \sin ^{2} x=1-\cos ^{2} x\right] \\ \end{aligned}
\begin{aligned} &=\int \sin x d x+\int t^{2} d t+\int t^{2}\left(1-t^{2}\right) d t \\ &=\int \sin x d x+\int t^{2} d t+\int t^{2}-t^{2} \cdot t^{2} d t \\ &=\int \sin x d x+\int t^{2} d t+\int t^{2} d t-\int t^{4} d t \end{aligned}
\begin{aligned} &=-\cos x+\frac{t^{2+1}}{2+1}+\frac{t^{2+1}}{2+1}-\frac{t^{4+1}}{4+1}+C \quad\quad\quad\quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c \& \int \sin x d x=-\cos x\right] \\ &=-\cos x+\frac{t^{2+1}}{2+1}+\frac{t^{2+1}}{2+1}-\frac{t^{5}}{5}+C \end{aligned}
\begin{aligned} &=-\cos x+\frac{t^{2+1}}{2+1}+\frac{t^{2+1}}{2+1}-\frac{t^{5}}{5}+C \\ &=-\cos x+\frac{t^{3}}{3}+\frac{t^{3}}{3}-\frac{t^{5}}{5}+C \\ &=-\cos x+\frac{2 t^{3}}{3}-\frac{t^{5}}{5}+C \\ &=-\cos x+\frac{2}{3} \cos ^{3} x-\frac{1}{5} \cos ^{5}+C \quad \quad \quad \quad \quad \quad \quad[\because t=\cos x] \end{aligned}

Indefinite Integrals Exercise 18.12 Question 3

Hint: Use substitution method to solve this integral
Given:$\int \cos ^{5} x d x$
Solution: Let $I=\int \cos ^{5} x d x$
Re-writing, $I=\int \cos ^{5} x d x$
\begin{aligned} I &=\int \cos ^{3} x \cos ^{2} x d x \\ &=\int \cos ^{3} x\left(1-\sin ^{2} x\right) d x\quad\quad\quad\quad\quad\quad\quad\left[\because \sin ^{2} x+\cos ^{2} x=1\right] \end{aligned}
\begin{aligned} &=\int\left(\cos ^{3} x-\cos ^{3} x \sin ^{2} x\right) d x \\ &=\int \cos ^{3} x d x-\int \cos ^{3} x \sin ^{2} x d x \\ &=\int \cos ^{2} x \cos x d x-\int \cos ^{2} x \cos x \sin ^{2} x d x \end{aligned}
\begin{aligned} &=\int \cos x\left(1-\sin ^{2} x\right) d x-\int\left(1-\sin ^{2} x\right) \sin ^{2} x \cos x d x \quad \quad \quad \quad \quad \quad \quad\left[\because \cos ^{2} x=1-\sin ^{2} x\right] \\ &=\int \cos x d x-\int \cos x \sin ^{2} x d x-\int \sin ^{2} x \cos x d x+\int \sin ^{4} x \cos x d x \\ &=\int \cos x d x-2 \int \sin ^{2} x \cos x d x+\int \sin ^{4} x \cos x d x \end{aligned}
Substitute $\sin x=t\Rightarrow \cos xdx=dt$ in second and third integral, then
$I =\int \cos x d x-2 \int t^{2} d t+\int t^{4} d t \quad\quad\quad\quad\quad\quad\left [ \because \int \cos xdx=\sin x+c \right ]$
\begin{aligned} &=\sin x-2 \frac{t^{2+1}}{2+1}+\frac{t^{4+1}}{4+1}+C \\ &=\sin x-2 \frac{t^{3}}{3}+\frac{t^{5}}{5}+C \\ &=\sin x-2 \frac{\sin ^{3} x}{3}+\frac{\sin ^{5} x}{5}+C \end{aligned}

Indefinite Integrals Exercise 18.12 Question 4

Answer:- $\frac{1}{6} \sin ^{6} x+C$
Hint: - Use substitution method to solve this integral.
Given:- $\int \sin ^{5} x \cdot \cos x d x$
Solution: - Let $\int \sin ^{5} x \cdot \cos x d x$
Substitute,
\begin{aligned} \operatorname{Sin} x &=t \Rightarrow \cos x d x=d t, \text { then } \\ \quad I &=\int t^{5} \cdot d t \\ \quad &=\frac{t^{5}+1}{5+1}+C \quad\quad\quad\quad\quad\quad\quad\quad\left[\therefore \int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right] \end{aligned}
\begin{aligned} &=\frac{t^{6}}{6}+C \\ &=\frac{\sin ^{6} x}{6}+C \quad\quad\quad\quad\quad\quad\quad\quad[\because t=\sin x] \end{aligned}

Indefinite Integrals Exercise 18.12 Question 5

Answer:-$\frac{-1}{7} \cos ^{7} x+\frac{1}{9} \cos ^{9} x+c$
Hint: - We substitute method to solve this integral.
Given:- $\int \sin ^{3} x \cdot \cos ^{6} x d x$
Solution: - Let $I=\int \sin ^{3} x \cdot \cos ^{6} x d x$
The exponent of $\sin x$ is odd, so we substitute
$\operatorname{Cos} x=t \Rightarrow-\operatorname{Sin} x d x=d t$, then
\begin{aligned} &I=\int \sin ^{3} x t^{6} \cdot \frac{d t}{-\operatorname{Sin} x}\quad\quad\quad\quad\quad\left [ t=Cos x \right ]\\ &I=-\int \operatorname{Sin}^{2} x t^{6} d t \end{aligned}
\begin{aligned} &=-\int\left(1-\operatorname{Cos}^{2} x\right) \cdot t^{6} d t \quad\quad\quad\quad\quad\quad\left[\because \operatorname{Sin}^{2} x+\operatorname{Cos}^{2} x=1\right] \\ &=-\int t^{6}\left(1-t^{2}\right) d t \end{aligned}
\begin{aligned} &=-\int t^{6}\left(1-t^{2}\right) d t \\ &=-\int\left(t^{6}-t^{6} \cdot t^{2}\right) d t \\ &=-\int\left(t^{6}-t^{8}\right) d t \\ &=-\int t^{6} d t+\int t^{8} d t \end{aligned}
\begin{aligned} &=-\frac{t^{6+1}}{6+1}+\frac{t^{8+1}}{8+1}+C \quad\quad\quad\quad\quad\quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right] \\ &=-\frac{t^{7}}{7}+\frac{t^{9}}{9}+C \\ &=-\frac{\operatorname{Cos}^{7} x}{7}+\frac{\operatorname{Cos}^{9} x}{9}+C \quad\quad\quad\quad\quad[\because t=\operatorname{Cos} x] \end{aligned}

Indefinite Integrals Exercise 18.12 Question 6

Answer:- $\operatorname{Sin} x-\operatorname{Sin}^{3} x+\frac{3}{5} \operatorname{Sin}^{5} x-\frac{1}{7} \operatorname{Sin}^{7} x+c$
Hint: - Use substitute method to solve this integral.
Given:-$\int \operatorname{Cos}^{7} x d x$
Solution: - Let $I=\int \operatorname{Cos}^{7} x d x$
Re-writing,
\begin{aligned} &I=\int \operatorname{Cos}^{6} x \cdot \operatorname{Cos} x d x \\ &\Rightarrow I=\int\left(\operatorname{Cos}^{2} x\right)^{3} \cdot \operatorname{Cos} d x \end{aligned}
$\begin{array}{ll} \Rightarrow I=\int\left(1-\operatorname{Sin}^{2} x\right)^{3} \cdot \operatorname{Cos} d x &\quad\quad\quad\quad\quad {\left[\because \operatorname{Sin}^{2} x+\operatorname{Cos}^{2} x=I\right]} \\\\ \Rightarrow I=\int\left[1-\operatorname{Sin}^{6} x-3 \operatorname{Sin}^{2} x+3 \operatorname{Sin}^{4} x\right] \cdot \operatorname{Cos} d x &\quad\quad\quad\quad {\left[\because(a-b)^{3}=a^{3}-b^{3}-3 a^{2} b+3 a b^{b}\right]} \end{array}$
\begin{aligned} &\Rightarrow I=\int\left(\operatorname{Cos} x-\operatorname{Sin}^{6} x \cdot \operatorname{Cos} x-3 \operatorname{Sin}^{2} x \cdot \operatorname{Cos} x+3 \operatorname{Sin}^{4} x \cdot \operatorname{Cos} x\right) d x \\ &\Rightarrow I=\int \operatorname{Cos} d x-\int \operatorname{Sin}^{6} x \cdot \operatorname{Cos} d x-\int 3 \operatorname{Sin}^{2} x \cdot \operatorname{Cos} d x+3 \int \sin ^{4} x \cdot \cos x d x \end{aligned}
Substitute $\operatorname{Sin} x=t \Rightarrow \operatorname{Cos} x d x=d t$ in 2nd, 3rd and 4th integral, then
\begin{aligned} &I=\int \operatorname{Cos} d x-\int t^{6} d t-3 \int t^{2} d t+3 \int t^{4} \cdot d t \\ &=\operatorname{Sin} x-\frac{t^{6}+1}{6+1}-3 \frac{t^{2+1}}{2+1}+3 \frac{t^{4+1}}{4+1}+C \quad \quad \quad \quad \quad \quad \quad\left[\because \int \operatorname{Cos} x d x=\operatorname{Sin} x+c \& \int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right] \end{aligned}
\begin{aligned} &=\operatorname{Sin} x-\frac{t^{7}}{7}-3 \frac{t^{3}}{3}+3 \cdot \frac{t^{5}}{5}+C \\ &=\operatorname{Sin} x-\frac{\operatorname{Sin}^{7} x}{7}-\operatorname{Sin}^{3} x+\frac{3}{5} \operatorname{Sin}^{5} x+C \\ &=\operatorname{Sin} x-\operatorname{Sin}^{3} x+\frac{3}{5} \operatorname{Sin}^{5} x-\frac{1}{7} \operatorname{Sin}^{7} x+C \quad\quad\quad\quad\quad\quad\quad[\because \operatorname{Sin} x=t] \end{aligned}

Indefinite Integrals Exercise 18.12 Question 7

Answer: -$-\frac{1}{8} \operatorname{Cos}^{4} x^{2}+C$
Hint: - Use substitution method to solve this integral.
Given: -$\int x \operatorname{Cos}^{3} x^{2} \operatorname{Sin} x^{2} d x$
Solution: - Let $I=\int x \operatorname{Cos}^{3} x^{2} \operatorname{Sin} x^{2} d x$
Substitute, $\operatorname{Cos} x^{2}=t \Rightarrow-\operatorname{Sin} x^{2} 2 x d x=d t \text { , then }$
\begin{aligned} I &=\int t^{3} x \operatorname{Sin} x^{2} \cdot \frac{d t}{-\left(\sin x^{2}\right)(x)^{2}}\quad\quad\quad\quad &\left[\because \operatorname{Cos} x^{2}=t\right] \\ &=-\frac{1}{2} \int t^{3} d t=\frac{-1}{2} \cdot \frac{t^{3+1}}{3+1}+C &\quad\quad\quad\quad\quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right] \end{aligned}
\begin{aligned} &=\frac{-1}{2} \cdot \frac{t^{4}}{4}+C=\frac{-1}{8} t^{4}+C \\ &=\frac{-1}{8} \cos ^{4} x^{2}+C \quad \quad \quad \quad \quad \quad\left[\because t=\cos x^{2}\right] \end{aligned}

Indefinite Integrals Exercise 18.12 Question 8

Answer: -$-\cos x+\cos ^{3} x-\frac{3}{5} \cos ^{5} x+\frac{1}{7} \cos ^{7} x+C$
Hint: - Use substitution method to solve this integral.
Given: -$\int \sin ^{7} x d x$
Solution: - Let $I=\int \sin ^{7} x d x$
Re-writing,
\begin{aligned} &I=\int \sin ^{6} x \cdot \sin x d x \\ &\Rightarrow I=\int\left(\sin ^{2} x\right)^{3} \sin x d x \\ &\Rightarrow I=\int\left(1-\cos ^{2} x\right) \sin x d x \quad\quad\quad\quad\quad\quad\quad\left[\because \sin ^{2} x+\cos ^{2} x=I\right] \end{aligned}
\begin{aligned} &\Rightarrow I=\int\left[1-\cos ^{6} x-3 \cos ^{2} x+3 \cos ^{4} x\right] \sin x d x \\ &\Rightarrow I=\int\left[\sin x-\cos ^{6} x \sin x-3 \cos ^{2} x \sin x+3 \cos ^{4} x \sin x\right] \\ &\Rightarrow I=\int \sin d x-\int \cos ^{6} \cdot \sin x d x-3 \int \cos ^{2} x \sin x d x+3 \int \cos ^{4} x \sin x d x \end{aligned}
Substitute,$\cos x=t \Rightarrow-\sin x d x=d t$ in 2nd, 3rd and 4th integral, then
\begin{aligned} I &=\int \sin x d x-\int t^{6}(-d t)-3 \int t^{2}(-d t)+3 \int t^{4}(-d t) \\ &=-\cos x+\int t^{6} d t+3 \int t^{2} d t-3 \int t^{4} d t &\quad\quad\quad\quad\quad\left [ \because \int \sin xdx=-\cos x +c \right ]\\ \end{aligned}
\begin{aligned} &=-\cos x+\frac{t^{6+1}}{6+1}+3 \frac{t^{2+1}}{2+1}-3 \frac{t^{4+1}}{4+1}+C \quad\quad\quad\quad\quad\quad\quad\quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right] \\ &=-\cos x+\frac{t^{7}}{7}+3 \cdot \frac{t^{3}}{3}-3 \frac{t^{5}}{5}+C \end{aligned}
\begin{aligned} &=-\cos x+\frac{\cos ^{7} x}{7}-\frac{3}{5} \cos ^{5} x+\cos ^{3} x+C \quad\quad\quad\quad\quad\quad\quad[\because t=\cos x] \\ &=-\cos x+\cos ^{3} x-\frac{3}{5} \cos ^{5} x+\frac{\cos ^{7} x}{7}+C \end{aligned}

Indefinite Integrals Exercise 18.12 Question 10

Answer: -$\frac{-1}{3} \cot ^{3} x-2 \cot x+\tan x+C$
Hint: - Use substitution method to solve this integration.
Given: -$\int \frac{1}{\sin ^{4} x \cdot \cos ^{2} x} d x$
Solution: - Let $I=\int \frac{1}{\sin ^{4} x \cdot \cos ^{2} x} d x$
The power of denominator is $4+2=6$so we divide both numerator and denominator by $\cos ^{6}x$so that the integral $I$can come in integral able form.
\begin{aligned} &\therefore I=\int \frac{\frac{1}{\cos ^{6} x}}{\frac{\sin ^{4} x \cdot \cos ^{2} x}{\cos ^{6} x}} d x=\int \frac{\sec ^{6} x}{\frac{\sin ^{4} x}{\cos ^{4} x}} d x \\ &\Rightarrow=\int \frac{\sec ^{6} x}{\tan ^{4} x} d x \end{aligned}
Re-writing,
\begin{aligned} I &=\int \frac{\sec ^{4} x \cdot \sec ^{2} x}{\tan ^{4} x} d x \\ &=\int \frac{\left(\sec ^{2} x\right)^{2} \sec ^{2} x}{\tan ^{4} x} d x \end{aligned}
\begin{aligned} &=\int \frac{\left(1+\tan ^{2} x\right)^{2} \sec ^{2} x}{\tan ^{4} x} d x \quad\quad\quad\quad\quad\quad\quad\quad\left[\because \sec ^{2} x-\tan ^{2} x=1\right] \\ &=\int \frac{\left(1+\tan ^{4} x+2 \tan ^{2} x\right) \sec ^{2} x}{\tan ^{4} x} \mathrm{dx} \quad\quad\quad\quad\quad\quad\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right] \end{aligned}
Substitute $\tan x=t \Rightarrow \sec ^{2} x d x=d t$ then
\begin{aligned} I &=\int \frac{\left(1+t^{4}+2 t^{2}\right)}{t^{4}} d t \\ &=\int\left(\frac{1}{t^{4}}+\frac{t^{4}}{t^{4}}+\frac{2 t^{2}}{t^{4}}\right) d t \\ &=\int\left(\frac{1}{t^{4}}+1+2 \cdot \frac{1}{t^{2}}\right) d t=\int\left(t^{-4}+t^{0}+2 t^{-2}\right) d t \end{aligned}
\begin{aligned} &=\int\left(\frac{1}{t^{4}}+1+2 \cdot \frac{1}{t^{2}}\right) d t=\int\left(t^{-4}+t^{0}+2 t^{-2}\right) d t \\ &=\frac{t^{-4+1}}{-4+1}+\frac{t^{0+1}}{0+1}+2 \frac{t^{-2+1}}{-2+1}+C \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right] \\ &=\frac{t^{-3}}{-3}+t+2 \frac{t^{-1}}{(-1)}+C \end{aligned}
\begin{aligned} &=-\frac{t^{3}}{3}+t-2 t^{-1}+C=-\frac{1}{3 t^{3}}+t-\frac{2}{t}+C \\ &=-\frac{1}{3 \tan ^{3} x}+\tan x-\frac{2}{\tan x}+C \quad\quad\quad\quad\quad\quad\left [ \because t=\tan x \right ]\\ &=-\frac{1}{3} \cot ^{3} x+\tan x-2 \cot x+C\quad\quad\quad\quad\quad\quad\left [ \because \tan x=\frac{1}{\cot x} \right ] \\ &=-\frac{1}{3} \cot ^{3} x-2 \cot x+\tan x+C \end{aligned}

Indefinite Integrals Exercise 18.12 Question 12

Answer: -$-\frac{1}{2 \tan ^{2} x}+\log |\tan x|+C$
Hint: - Use substitution method to solve this integral.
Given: -$\int \frac{1}{\sin ^{3} x \cdot \cos x} d x$
Solution: - Let $I=\int \frac{1}{\sin ^{3} x \cdot \cos x} d x$
The exponent of denominator is $3+1=4$ to solve this type of integral; uses have divided both numerator and denominator by $\cos ^{4}x$ then
\begin{aligned} I &=\int \frac{\frac{1}{\cos ^{4} x}}{\frac{\sin ^{3} x \cdot \cos x}{\cos ^{4} x}} d x \\ &=\int \frac{\left(\sec ^{4} x\right)}{\frac{\sin ^{3} x}{\cos ^{3} x}} d x \end{aligned}
\begin{aligned} &=\int \frac{\sec ^{4} x}{\tan ^{3} x} d x=\int \frac{\sec ^{2} x \cdot \sec ^{2} x}{\tan ^{3} x} d x \\ &=\int \frac{\left(1+\tan ^{2} x\right) \sec ^{2} x}{\tan ^{3} x} d x\quad\quad\quad\quad\quad\quad\quad\quad\left[\because 1+\tan ^{2} x=\sec ^{2} x\right] \\ \end{aligned}
Substitute $\tan x=t \Rightarrow \sec ^{2} x d x=d t$ then
\begin{aligned} I &=\int \frac{1+t^{2}}{t^{3}} d t \\ &=\int\left(\frac{1}{t^{3}}+\frac{t^{2}}{t^{3}}\right) d t \\ &=\int\left(t^{-3}+\frac{1}{t}\right) d t \end{aligned}
$=\frac{t^{-3+1}}{-3+1}+\log |t|+C \quad\quad\quad\quad\quad\quad\quad\left[\because \int x^{n} d x=x \frac{n+1}{n+1}+c \text { and } \int \frac{1}{x} d x=\log |x|+C\right]$
\begin{aligned} &=\frac{t^{-2}}{-2}+\log |t|+C \\ &=-\frac{1}{2 t^{2}}+\log |t|+C \\ &=-\frac{1}{2 \tan ^{2} x}+\log |\tan x|+C \quad\quad\quad\quad\quad\quad\quad[\because t=\tan x] \end{aligned}

Indefinite Integrals Exercise 18.12 Question 13

Answer:$\frac{1}{2} \tan ^{2} x+\log |\tan x|+C$
Hint: Use substitution method to solve this integral.
Given:$\int \frac{1}{\sin x \cdot \cos ^{3} x} d x$
Solution: Let $I=\int \frac{1}{\sin x \cdot \cos ^{3} x} d x$
Re-writing,
\begin{aligned} I &=\int \frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cdot \cos ^{3} x} d x \quad\left[\therefore 1=\sin ^{2} x+\cos ^{2} x\right] \\ &=\int\left(\frac{\sin ^{2} x}{\sin x \cos ^{3} x}+\frac{\cos ^{2} x}{\sin x \cdot \cos ^{3} x}\right) d x \end{aligned}
\begin{aligned} &=\int\left(\frac{\sin x}{\cos x} \cdot \frac{1}{\cos ^{2} x}+\frac{1}{\sin x \cdot \cos x}\right) d x \\ &=\int\left(\tan x \cdot \sec ^{2} x+\frac{\cos ^{2} x}{\cos ^{2} x} \cdot \frac{1}{\sin x \cdot \cos x}\right) d x \end{aligned}
$=\int\left(\tan x \cdot \sec ^{2} x+\frac{1}{\cos ^{2} x} \cdot \frac{1}{\frac{\sin x \cdot \cos x}{\cos ^{2} x}}\right) d x$
$=\int\left(\tan x \cdot \sec ^{2} x+\sec ^{2} x \cdot \frac{1}{\frac{\sin x}{\cos x}}\right) d x$
\begin{aligned} I &=\int\left(\tan x \cdot \sec ^{2} x+\frac{\sec ^{2} x}{\tan x}\right) d x \\ & \Rightarrow I=\int \tan x \cdot \sec ^{2} x d x+\int \frac{\sec ^{2} x}{\tan x} d x \end{aligned}
Substitute $\tan x=t \Rightarrow \sec ^{2} x d x=d t$ then
\begin{aligned} I &=\int t d t+\int_{t}^{1} d t \\ &=\frac{t^{1+1}}{1+1}+\log |t|+C \quad \quad \quad \quad \quad \quad \quad \quad\left[\because \int x^{2} d x=\frac{x^{n+1}}{n+1}+c, \int \frac{1}{x} d x=1\right] \end{aligned}
\begin{aligned} &=\frac{t^{2}}{2}+\log |t|+C \\ &=\frac{\tan ^{2} x}{2}+\log |\tan x|+C \quad\quad\quad\quad\quad\quad\quad[\because t=\tan x] \end{aligned}

The RD Sharma class 12 solutions Indefinite Integrals 18.12 is a must-have solution that must be followed by all aspiring students who want to improve their chances of becoming a topper. Chapter 18 of the NCERT maths book is going to talk about various complicated concepts like the Reverse power rule, Graphs of indefinite integrals, Indefinite integrals of common functions,integrals formulae etc. Exercise 18.12 of the textbook will have a total of 13 questions which are divided into subsections and cover concepts from the whole chapter. The RD Sharma class 12th exercise 18.12 will provide advanced solutions to all these NCERT questions.

Now, students may wonder why the RD Sharma solutions are so popular among other NCERT solutions. If you want to know why RD Sharma class 12th exercise 18.12 is favorite of students then you need to take a look at these features:-

• The RD Sharma class 12th exercise 18.12 contains answers that correspond to all the questions in the latest version of NCERT maths textbooks. The pdf is updated with every latest edition of the book.

• Experts in mathematics were responsible for drafting the answers of the class 12 RD Sharma chapter 18 exercise 18.12 solution. Therefore, the answers are accurate and teach students some unique methods of calculation for skill enhancement.

• The home assignments given to students by their school teachers contain questions from the RD Sharma class 12 solutions chapter 18 ex 18.12. Therefore, students can use these solutions to solve and answer their homework questions correctly.

• The RD Sharma class 12 chapter 18 exercise 18.12 can be used by students for home practice. Students can compare answers provided in the book to check their performance and mark themselves accordingly. If they practice the book thoroughly, they will find common questions in their board exams.

• Students can avail the free copy of RD Sharma class 12 solutions Indefinite Integrals ex 18.12 pdf from the Career360 website.

## RD Sharma Chapter wise Solutions

JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%

1. How to access the free copy of the class 12 RD Sharma chapter 13 exercise MCQ solution pdf online?

Students who want to access the free copy of the class 12 RD Sharma chapter 13 exercise MCQ solution pdf can do so from the Career360 website. There is no need to pay any money or use third party apps

2. What are the concepts covered in Indefinite Integrals?

The concepts that are covered in Indefinite Integrals are:-

• Reverse power rule

• Graphs of indefinite integrals,

Indefinite integrals of common functions

3. What questions are there in RD Sharma class 12th exercise 18.12?

The RD Sharma class 12th exercise 18.12 contains a total of 13 questions on various concepts of Indefinite Integrals.

4. Why should avail RD Sharma solutions?

Experts with immense knowledge in maths have created the answers in the RD Sharma solutions book. The answers are detailed and contain some new methods of calculations which would be useful for all students.

5. How is RD Sharma class 12 solutions chapter 18 ex 18.12 useful for home practice?

Students can use the RD Sharma class 12 solutions chapter 18 ex 18.12 for maths practice by solving questions and comparing answers to check their performance and progress.

## Upcoming School Exams

#### National Means Cum-Merit Scholarship

Application Date:22 July,2024 - 31 August,2024

Exam Date:19 September,2024 - 19 September,2024