Question

$\underset{0}{\overset{\pi /2}{\int }}\frac{\cos x}{1+\cos x+\sin x}dx$

Answer 1

$I={\int }_0^{\pi /2}\frac{\cos x}{1+\cos x+\sin x}$

$={\int }_0^{\pi /2}\frac{{\cos }^{2}x/2-{\sin }^{2}x/2}{2{\cos }^{2}x/2+2\sin x/2\cos x/2}dx$

$={\int }_0^{\pi /2}\frac{({\cos }^{2}x/2-{\sin }^{2}x/2)}{2\cos x/2(\cos x/2+\sin x/2)}dx$

$={\int }_0^{\pi /2}\frac{(\cos x/2-\sin x/2)(\cos x/2+\sin x/2)}{2\cos x/2(\cos x/2+\sin x/2)}dx$

$={\int }_0^{\pi /2}\frac{(\cos x/2-\sin x/2)}{2\cos x/2}dx$

$=\frac{1}{2}{\int }_0^{\pi /2}dx-\frac{1}{2}{\int }_0^{\pi /2}\tan \frac{x}{2}dx$

$={[\frac{1}{2}x-\frac{1}{2}\log \sec \frac{x}{2}.2]}_0^{\pi /2}$

$={[\frac{x}{2}-\log \sec \frac{x}{2}]}_0^{\pi /2}$

$=\frac{1}{2}[\frac{\pi }{2}-\log 2]$