Evaluate the following integrals:

Solution:

Given:

$I=\int(\tan x+\cot x)^{2} d x$

$\Rightarrow \int\left(\tan ^{2} x+\cot ^{2} x+2 \tan x \cot x\right)^{1} d x$

We know that,

$\tan ^{2} x=\sec ^{2} x-1$

$\cot ^{2} x=\operatorname{cosec}^{2} x-1$

$\tan x=\frac{1}{\cot x}$

$\Rightarrow \int\left(\sec ^{2} x-1+\operatorname{cosec}^{2}-1+\frac{2}{\cot x} \operatorname{cotx}\right) d x$

$\Rightarrow \int\left(\sec ^{2} x+\operatorname{cosec}^{2} x-2+2\right) d x$

$\Rightarrow \int\left(\sec ^{2} x+\operatorname{cosec}^{2} x\right) d x$

$\Rightarrow \int \sec ^{2} x+\int \operatorname{cosec}^{2} x d x$

We know that,

$\int \sec ^{2} x d x=\tan x$

$\int \operatorname{cosec}^{2} x d x=-\cot x$

$I=\tan x-\cot x-c$