Evaluate the following integrals:

Assume $\log (\sec x+\tan x)=t$

$d(\log (\sec x+\tan x))=d t$

(use chain rule to differentiate first differentiate $\log (\sec x+\tan x)$ then $(\sec x+\tan x)$

$\Rightarrow \frac{\sec x \tan x+\sec ^{2} x}{\sec x+\tan x} d x=d t$

$\Rightarrow \frac{\sec x(\tan x+\sec x)}{\sec x+\tan x} d x=d t$

$\Rightarrow \sec x d x=d t$

Put $\mathrm{t}$ and $\mathrm{dt}$ in given equation we get

Substituting the values oft and dt we get

$\Rightarrow \int \mathrm{tdt}$

$\Rightarrow \frac{t^{2}}{2}+c$

But $t=\log (\sec x+\tan x)$

$\Rightarrow \frac{\log ^{2}(\sec x+\tan x)}{2}+c$