Question

$\int \frac{\sin x}{{\left(1+\cos x\right)}^2}dx$

Answer 1

$$\begin{gathered} \int \frac{\sin x}{{\left(1+\cos x\right)}^2}dx \\ \mathrm{Let}1+\cos x=t \\ \Rightarrow -\sin x=\frac{dt}{dx} \\ \Rightarrow \sin xdx=-dt \\ \mathrm{Now},\int \frac{\sin x}{{\left(1+\cos x\right)}^2}dx \\ =\int -\frac{dt}{t^2} \\ =-\int t^{-2}dt \\ =-\left[\frac{t^{-2+1}}{-2+1}\right]+C \\ =\frac{1}{t}+C \\ =\frac{1}{1+\cos x}+C \end{gathered}$$