Evaluate the following integrals:

Solution:

$\int \frac{1}{\sin ^{4} x \cos ^{2} x} d x$

Adding the powers : $-4+-2=-6$

Since all are even nos, we will divide each by $\cos ^{6} x$ to convert into positive power

So, $\int \frac{1}{\sin ^{4} x \cos ^{2} x} d x=\int \frac{\frac{1}{\cos ^{6} x}}{\frac{\sin ^{4} x \cos ^{2} x}{\cos ^{6} x}} d x$

$=\int \frac{\sec ^{6} x}{\frac{\sin ^{4} x}{\operatorname{coc}^{4} x}} d x=\int \frac{\sec ^{6} x}{\tan ^{4} x} d x$

$=\int \frac{\sec ^{4} x \sec ^{2} x}{\tan ^{4} x} d x=\int \frac{\left(\sec ^{2} x\right)^{2} \sec ^{2} x}{\tan ^{4} x} d x$

$=\int \frac{\left(1+\tan ^{2} x\right)^{2} \sec ^{2} x}{\tan ^{4} x} d x\left\{\right.$ since $\left.\sec ^{2} x=1+\tan ^{2} x\right\}$

$=\int \frac{\left(1+\tan ^{4} x+2 \tan ^{2} x\right)^{2} \sec ^{2} x}{\tan ^{4} x} d x\left(\right.$ apply $\left.(a+b)^{2}=a^{2}+b^{2}+2 a b\right)$

Let $\tan x=t$, so $d t=d(\tan x)=\sec ^{2} x d x$

So, $\mathrm{dx}=\frac{\mathrm{dt}}{\sec ^{2} \mathrm{x}}$

Put $\mathrm{t}$ and $\mathrm{dx}$ in the above equation,

$\int \frac{\left(1+\tan ^{4} x+2 \tan ^{2} x\right) \sec ^{2} x}{\tan ^{4} x} d x=\frac{\int\left(1+t^{4}+2 t^{2}\right)}{t^{4}} \sec ^{2} x * \frac{d t}{\sec ^{2} x}$

$=\frac{\int\left(1+t^{4}+2 t^{2}\right)}{t^{4}} d t$

$=\int\left(1+t^{-4}+2 t^{-2}\right) d t$

$=\mathrm{t}-\frac{\mathrm{t}^{-3}}{3}-2 \mathrm{t}^{-1}+\mathrm{c}$

$=\mathrm{t}-\frac{2}{\mathrm{t}}-\frac{1}{3 \mathrm{t}^{3}}+\mathrm{c}$

$=\tan x-\frac{2}{\tan x}-\frac{1}{3 \tan ^{2} x}+c$

$=\tan x-2 \cot x-\frac{1}{3} \cot ^{3} x+c\{1 / \tan x=\cot x)$