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