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