Question

Evaluate $\int \frac{\sin x}{\sin x+\cos x}dx$.

Answer 1

$I=\int \frac{\sin x}{\sin x+\cos x}dx$

$I=\frac{1}{2}\int \frac{2\sin x}{\sin x+\cos x}dx$

$I=\frac{1}{2}\int \frac{(\sin x+\cos x+\sin x-\cos x)}{(\sin x+\cos x)}dx$

$I=\frac{1}{2}\int 1dx+\frac{1}{2}\int \frac{\sin x-\cos x}{(\sin x+\cos x)}dx$

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

$|\begin{array}{c}Let\sin x+\cos x=t\\ (\cos x-\sin x)dx=dt\\ (\sin x-\cos x)dx=-dt\end{array}$

$I=\frac{1}{2}\left(x\right)-\frac{1}{2}\int \frac{1}{t}dt$

$I=\frac{1}{2}\left(x\right)-\frac{1}{2}\log t$

$I=\frac{1}{2}\left(x\right)-\frac{1}{2}\log (\sin x+\cos x)+c$.