${\int }_0^{\mathrm{\pi }/2} \frac{{\sin }^{3/2}\mathrm{x}}{{\sin }^{3/2}\mathrm{x}+{\cos }^{3/2}\mathrm{x}}\mathrm{dx}$ is equal to

Let,

$\begin{array}{l}\mathrm{I}={\int }_0^{\mathrm{\pi }/2} \frac{{\sin }^{3/2}\mathrm{x}}{{\sin }^{3/2}\mathrm{x}+{\cos }^{3/2}\mathrm{x}}\mathrm{dx}----\left(\mathrm{i}\right)\\ \mathrm{then} \mathrm{I}={\int }_0^{\mathrm{\pi }/2} \frac{{\sin }^{3/2}\left(\frac{\mathrm{\pi }}{2}-\mathrm{x}\right)}{{\sin }^{3/2}\left(\frac{\mathrm{\pi }}{2}-\mathrm{x}\right)+{\cos }^{3/2}\left(\frac{\mathrm{\pi }}{2}-\mathrm{x}\right)}\mathrm{dx}\\ \left[\because {\int }_0^{\mathrm{a}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_0^{\mathrm{a}} \mathrm{f}(\mathrm{a}-\mathrm{x})\mathrm{dx}\right]\\ \Rightarrow \mathrm{I}={\int }_0^{\mathrm{\pi }/2} \frac{{\cos }^{3/2}\mathrm{x}}{{\cos }^{3/2}\mathrm{x}+{\sin }^{3/2}\mathrm{x}}\mathrm{dx}-----\left(\mathrm{ii}\right)\end{array}$

$\left[\because \sin \left(\frac{\mathrm{\pi }}{2}-\mathrm{x}\right)=\cos \mathrm{x}\text{ and }\cos \left(\frac{\mathrm{\pi }}{2}-\mathrm{x}\right)=\sin \mathrm{x}\right]$

On adding Eqs. (i) and (iil, we get

$\begin{array}{l}2\mathrm{I}={\int }_0^{\mathrm{\pi }/2} \frac{{\sin }^{3/2}\mathrm{x}+{\cos }^{3/2}\mathrm{x}}{{\sin }^{3/2}\mathrm{x}+{\cos }^{3/2}\mathrm{x}}\mathrm{dx}\\ ={\int }_0^{\mathrm{\pi }/2} 1\mathrm{dx}=[\mathrm{x}{]}_0^{\mathrm{\pi }/2}=\frac{\mathrm{\pi }}{2}-0\Rightarrow \mathrm{I}=\frac{\mathrm{\pi }}{4}\end{array}$