Evaluate the following integrals:
Question:

Evaluate the following integrals:

$\int \frac{\sin ^{3} x-\cos ^{3} x}{\sin ^{2} x \cos ^{2} x} d x$

Solution:

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,

$\int \tan x \sec x d x=\sec x$

$\int \cot x \operatorname{cosec} x d x=-\cot x$

$\Rightarrow \sec x-(-\cot x)+c$

$\Rightarrow \sec x+\cot x+c$

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