Prove

$\frac{\sin ^{2} x}{1+\cos x}=\frac{\left(2 \sin \frac{x}{2} \cos \frac{x}{2}\right)^{2}}{2 \cos ^{2} \frac{x}{2}}\left[\sin x=2 \sin \frac{x}{2} \cos \frac{x}{2} ; \cos x=2 \cos ^{2} \frac{x}{2}-1\right]$

$=\frac{4 \sin ^{2} \frac{x}{2} \cos ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}$

$=2 \sin ^{2} \frac{x}{2}$

$=1-\cos x$

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

$=x-\sin x+\mathrm{C}$