Evaluate the following integrals:

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

$\mathrm{d}(\log (\sec x+\tan x))=\mathrm{dt}$

(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 the given equation we get

$\Rightarrow \int \frac{\mathrm{d} t}{t}$

$=\ln |t|+c$

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

$=\ln |\log (\sec x+\tan x)|+c$