Find the integral of the function $\frac{\sin ^{2} x}{1+\cos x}$.

We are given the integral $I = \int \frac{\sin ^{2} x}{1+\cos x} dx$.
Using the trigonometric identities $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$ and $1 + \cos x = 2 \cos ^{2} \frac{x}{2}$,we can simplify the integrand:
$\frac{\sin ^{2} x}{1+\cos x} = \frac{(2 \sin \frac{x}{2} \cos \frac{x}{2})^{2}}{2 \cos ^{2} \frac{x}{2}}$
$= \frac{4 \sin ^{2} \frac{x}{2} \cos ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}$
$= 2 \sin ^{2} \frac{x}{2}$
Using the identity $2 \sin ^{2} \theta = 1 - \cos 2\theta$,we have $2 \sin ^{2} \frac{x}{2} = 1 - \cos x$.
Therefore,$I = \int (1 - \cos x) dx$
$= x - \sin x + C$,where $C$ is an arbitrary constant.