Question

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

Answer 1

Let $I=\int \frac{1}{1+\cot x}dx$
$=\int \frac{\sin x}{\sin x+\cos x}dx$
$=\frac{1}{2}\int \frac{2\sin x}{\sin x+\cos x}dx$
$=\frac{1}{2}\int \frac{(\sin x+\cos x)+(\sin x-\cos x)}{(\sin x+\cos x)}dx$
$=\frac{1}{2}\int 1dx+\frac{1}{2}\int \frac{\sin x-\cos x}{\sin x+\cos x}dx$
$=\frac{1}{2}\left(x\right)+\frac{1}{2}\int \frac{\sin x-\cos x}{\sin x+\cos x}dx$$Put$\sin x+\cos x=t\Rightarrow (\cos x-\sin x)dx=dt\therefore\displaystyle I=\frac {x}{2}+\frac {1}{2}\int \frac {-(dt)}{t}\displaystyle =\frac {x}{2}-\frac {1}{2}\log |t|+C\displaystyle =\frac {x}{2}-\frac {1}{2}\log |\sin x+\cos x|+C$