Trigonometric integral $\int{\frac{\sin x}{\cos x+\sin x}dx}$ & $\int{\frac{\cos x}{\cos x+\sin x}dx}$ Credit to Imad zak

Exercise:

Compute , $\int{\frac{\sin x}{\cos x+\sin x}dx}$

Solution: we have $\frac{\sin x}{\cos x+\sin x}=\frac{2}{2}\times \frac{\sin x}{\cos x+\sin x}=\frac{1}{2}\times \frac{2\sin x}{\cos x+\sin x}$

$=\frac{1}{2}\times \frac{\sin x+\cos x-\cos x+\sin x}{\cos x+\sin x}=\frac{1}{2}\left( 1+\frac{-\cos x+\sin x}{\cos x+\sin x} \right)$

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

Take $u=\cos x+\sin x\Leftrightarrow du=\left( \cos x-\sin x \right)dx$

So $\int{\frac{\sin x}{\cos x+\sin x}dx}=\frac{1}{2}\int{dx}-\frac{1}{2}\int{\frac{\cos x-\sin x}{\cos x+\sin x}dx}=\frac{1}{2}x-\frac{1}{2}\ln \left| \cos x+\sin x \right|+c$ 

Exercise:

Compute, $\int{\frac{\cos x}{\cos x+\sin x}dx}$

Solution: we have $\frac{\cos x}{\cos x+\sin x}=\frac{1}{2}\times \frac{2\cos x}{\cos x+\sin x}=\frac{1}{2}\times \frac{\cos x+\sin x-\sin x+\cos x}{\cos x+\sin x}$

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

So $\int{\frac{\sin x}{\cos x+\sin x}dx}=\frac{1}{2}\int{\left( 1+\frac{\cos x-\sin x}{\cos x+\sin x} \right)dx}=\frac{1}{2}x+\frac{1}{2}\int{\frac{d\left( \cos x+\sin x \right)}{\cos x+\sin x}}$

                          $=\frac{1}{2}x+\frac{1}{2}\ln \left| \cos x+\sin x \right|+c$