Integrate the function $\frac{\sin x}{(1+\cos x)^{2}}$.

Let $1+\cos x = t$.
Differentiating both sides with respect to $x$,we get $-\sin x \, dx = dt$,or $\sin x \, dx = -dt$.
Substituting these into the integral:
$\int \frac{\sin x}{(1+\cos x)^{2}} \, dx = \int -\frac{dt}{t^{2}}$.
$= -\int t^{-2} \, dt$.
$= -\left( \frac{t^{-1}}{-1} \right) + C$.
$= \frac{1}{t} + C$.
Substituting back $t = 1+\cos x$,we get:
$= \frac{1}{1+\cos x} + C$,where $C$ is an arbitrary constant.