Question: $\frac{\sin x}{(1+\cos x)^{2}}$
Solution:
Let $1+\cos x=t$
$\therefore-\sin x d x=d t$
$\Rightarrow \int \frac{\sin x}{(1+\cos x)^{2}} d x=\int-\frac{d t}{t^{2}}$
$=-\int t^{-2} d t$
$=\frac{1}{t}+\mathrm{C}$
$=\frac{1}{1+\cos x}+\mathrm{C}$
