What is $\int_{0}^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt {\cos x}}dx$?

0 1 $\pi/2$ $\pi/4$

$\pi/4$

Using ,$\int_{0}^{a}f(x)dx$=$\int_{0}^{a}f(a-x)dx$ we get $I=\int_{0}^{\pi/2}\frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt {\cos x}}dx$. Hence $2I=\int_{0}^{\pi/2}\frac{\sqrt{\cos x}+\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt {\cos x}}dx=\pi/2$. Hence $I=\pi/4$