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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

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.

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

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RD Sharma Class 12 Exercise 18.12 Indefinite Integrals Solutions Maths

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
$\begin{aligned} I &=\int t d t+\int_{t}^{1} d t \\ &=\frac{t^{1+1}}{1+1}+\log |t|+C \quad \quad \quad \quad \quad \quad \quad \quad\left[\because \int x^{2} d x=\frac{x^{n+1}}{n+1}+c, \int \frac{1}{x} d x=1\right] \end{aligned}$
$\begin{aligned} &=\frac{t^{2}}{2}+\log |t|+C \\ &=\frac{\tan ^{2} x}{2}+\log |\tan x|+C \quad\quad\quad\quad\quad\quad\quad[\because t=\tan x] \end{aligned}$

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