Show that

$\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$

Let $I=\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$     ...(1)

$\Rightarrow I=\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}}{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}+\sqrt{\cos \left(\frac{\pi}{2}-x\right)}} d x$     $\left(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right)$

$\Rightarrow I=\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\cos }}{\sqrt{\cos }+\sqrt{\sin x}} d x$    ...(2)

Adding (1) and (2), we obtain

$2 I=\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$

$\Rightarrow 2 I=\int_{0}^{\frac{\pi}{2}} 1 \cdot d x$

$\Rightarrow 2 I=[x]_{0}^{\frac{\pi}{2}}$

$\Rightarrow 2 I=\frac{\pi}{2}$

$\Rightarrow I=\frac{\pi}{4}$