Evaluate the following integrals:
$\int \frac{\sin ^{3} x-\cos ^{3} x}{\sin ^{2} x \cos ^{2} x} d x$
Given:
$\int \frac{\sin ^{3} x-\cos ^{3} x}{\sin ^{2} x \cos ^{2} x} d x$
By Splitting, we get,
$\Rightarrow \int\left(\frac{\sin ^{3} x}{\sin ^{2} x \cos ^{2} x}-\frac{\cos ^{3} x}{\sin ^{2} x \cos ^{2} x}\right) d x$
By cancelling the $\sin ^{2} x$ on first and $\cos ^{2} x$ on second,
$\Rightarrow \int\left(\frac{\sin x}{\cos ^{2} x}-\frac{\cos x}{\sin ^{2} x}\right) d x$
We know that,
$\frac{\sin x}{\cos x}=\tan x$
$\frac{\cos x}{\sin x}=\cot x$
$\frac{1}{\cos x}=\sec x$
$\frac{1}{\sin x}=\operatorname{cosec} x$
$\Rightarrow \int(\tan x \sec x-\cot x \operatorname{cosec} x) d x$
We know that,
$\int \tan x \sec x d x=\sec x$
$\int \cot x \operatorname{cosec} x d x=-\cot x$
$\Rightarrow \sec x-(-\cot x)+c$
$\Rightarrow \sec x+\cot x+c$