Evaluate the following integrals:

Solution:

Given:

$\int \frac{\sin ^{2} x}{1+\cos x} d x$

We know that,

$\sin ^{2} x=1-\cos ^{2} x$

$\Rightarrow \int \frac{1-\cos ^{2} x}{1+\cos x} d x$

We treat $1-\cos ^{2} x a s a^{2}-b^{2}=(a+b)(a-b)$

$\Rightarrow \int \frac{(1)^{2}-(\cos x)^{2}}{1+\cos x} d x$

$\Rightarrow \int \frac{(1+\cos x)(1-\cos x)}{1+\cos x} d x$

$\Rightarrow \int(1-\cos x) d x$

By Splitting, we get,

$\Rightarrow \int \mathrm{dx}-\int \cos \mathrm{x} \mathrm{dx}$

We know that,

$\int \mathrm{kdx}=\mathrm{kx}+\mathrm{c}$

$\int \cos \mathrm{x} \mathrm{d} \mathrm{x}=\sin \mathrm{x}$

$\Rightarrow x-\sin x+c$