Question

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

Answer 1

$Let\ 1 + cos x = t$
$∴ -sin\ x\ dx = dt$
⇒ ∫$$\frac {sin \ x}{(1+cos\ x)^2} dx = $∫-\frac {dt}{t^2}$
= $- ∫t^{-2}dt$
= $\frac 1t+C$
=$\frac {1}{1+cos \ x} +C$