Question

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

Answer 1

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

Put, $1+\cos x=t$

$⟹-\sin xdx=dt$

Now, $\int \frac{\sin x}{(1+\cos x{)}^2}dx=\int \frac{-1}{t^2}dt$

$=-\int t^{-2}dt=-\left(\frac{t^{-1}}{-2+1}\right)+c$

$=\frac{1}{t}+c$

$=\frac{1}{1+\cos x}+c$