Evaluate the following integrals:

Assume $\log \left(\tan \frac{x}{2}\right)=\mathrm{t}$

$d\left(\log \left(\tan \frac{x}{2}\right)\right)=d t$

(use chain rule to differentiate)

$\Rightarrow \frac{\sec ^{2} \frac{x}{2}}{\tan \frac{\frac{x}{2}}{2}} \mathrm{dx}=\mathrm{dt}$

$\Rightarrow \frac{1}{2 \sin \frac{x}{2} \cos \frac{x}{2}} d x=d t$

$\Rightarrow \operatorname{cosec} x d x=d t$

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

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

$=\ln |\mathrm{t}|+\mathrm{c}$

But $t=\log \left(\tan \frac{x}{2}\right)$

$=\ln \left|\log \left(\tan \frac{x}{2}\right)\right|+c$