The value of ${\int }_0^{\pi /2} \sqrt{\sin 2\theta }\sin \theta \mathrm{d}\theta$ is

$\begin{array}{l}I={\int }_0^{\pi /2} \sqrt{\sin 2\theta }\sin \theta d\theta \\ ={\int }_0^{\pi /2} \sqrt{\sin (\pi -2\theta )}\sin (\pi /2-\theta )\mathrm{d}\theta \\ ={\int }_0^{\pi /2} \sqrt{\sin 2\theta }\cos \theta \mathrm{d}\theta \\ 2I={\int }_0^{\pi /2} \sqrt{\sin 2\theta }(\sin \theta +\cos \theta )\mathrm{d}\theta \end{array}$

In 2I, put $\sin \theta -\cos \theta =t$, so that $t^2=1-\sin 2\theta \Rightarrow 1-t^2=$

$\sin 2\theta$. Also $\mathrm{d}t=(\cos \theta +\sin \theta )\mathrm{d}\theta$

$\begin{array}{l}\therefore 2I={\int }_{-1}^1 \sqrt{1-t^2}\mathrm{d}t\mid =2{\left[\frac{t}{2}\sqrt{1-t^2}+\frac{1}{2}{\sin }^{-1}t\right]}_0^1\\ =2\frac{\pi }{4}=\frac{\pi }{2}\Rightarrow I=\frac{\pi }{4}.\end{array}$