Evaluate \int \frac{\sin ^{2} x}{1-\cos x} d x.

\int \frac{\sin ^{2} x}{1-\cos x} \cdot d x
\sin ^{2} x=1-\cos ^{2} x
\int\left(\frac{1-\cos ^{2} x}{1-\cos x}\right) \cdot d x
=\int(1+\cos x) d x
=\int 1 d x+\int \cos x d x
=x+\sin x+c
\int \frac{\sin ^{2} x}{1-\cos x} \cdot d x=x+\sin x+c