#### Provide Solutio for RD Sharma Class 12 Chapter 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}