Evaluate $\displaystyle \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}} \; dx$
$\begin{aligned}
\text{Let } I & = \displaystyle \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}} \; dx \;\;\; \cdots \; (1) \\\\
& = \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\sqrt{\sin \left(\dfrac{\pi}{2} - x\right)}}{\sqrt{\cos \left(\dfrac{\pi}{2} - x\right)}+ \sqrt{\sin \left(\dfrac{\pi}{2} - x\right)}} \; dx \\\\
& \left[\text{Note: } \int \limits_{0}^{a} f\left(x\right) \; dx = \int \limits_{0}^{a} f\left(x - a\right) \; dx\right] \\\\
\text{i.e. } I & = \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \; dx \;\;\; \cdots \; (2)
\end{aligned}$
Adding equations $(1)$ and $(2)$ gives
$\begin{aligned}
2 I & = \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \; dx \\\\
& = \int \limits_{0}^{\frac{\pi}{2}} dx \\\\
& = \left[x\right]_{0}^{\pi / 2} = \dfrac{\pi}{2}
\end{aligned}$
$\implies$ $I = \dfrac{\pi}{4}$