Question

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

Answer 1

 π/4

$$\text{We have},$$
$$I={\int }_0^{\frac{\pi }{2}}\frac{\sin x}{\sin x+\cos x}dx.....\left(1\right)$$
$$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{\sin (\frac{\pi }{2}-x)}{\sin (\frac{\pi }{2}-x)+\cos (\frac{\pi }{2}-x)}dx$$
$$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{\cos x}{\cos x+\sin x}dx$$
$$\therefore I={\int }_0^{\frac{\pi }{2}}\frac{\cos x}{\sin x+\cos x}dx.....\left(2\right)$$
$$\text{Adding}\left(1\right)and\left(2\right),\text{we get}$$
$$2I={\int }_0^{\frac{\pi }{2}}[\frac{\sin x}{\sin x+\cos x}+\frac{\cos x}{\cos x+\sin x}]dx$$
$$={\int }_0^{\frac{\pi }{2}}\left[\frac{\sin x+\cos x}{\sin x+\cos x}\right]dx$$
$$={\int }_0^{\frac{\pi }{2}}dx$$
$$={\left[x\right]}_0^{\frac{\pi }{2}}$$
$$=\frac{\pi }{2}$$
$$HenceI=\frac{\pi }{4}$$