If $\mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}} \frac{{\sin }^{\frac{3}{2}}\mathrm{x}}{{\sin }^{\frac{3}{2}}\mathrm{x}+{\cos }^{\frac{3}{2}}\mathrm{x}}\mathrm{dx}$, then ${\int }_0^{21} \frac{\mathrm{xsin}\mathrm{xcos}\mathrm{x}}{{\sin }^4\mathrm{x}+{\cos }^4\mathrm{x}}\mathrm{dx}$ equals :

For I 
Apply king (P– 5) and add
$\begin{array}{l}2\mathrm{I}={\int }_0^{\mathrm{\pi }/2} \mathrm{dx}=\frac{\mathrm{\pi }}{2}\Rightarrow \mathrm{I}=\frac{\mathrm{\pi }}{4}\\ {\mathrm{I}}_2={\int }_0^{\mathrm{\pi }/2} \frac{\mathrm{xsin}\mathrm{xcos}\mathrm{x}}{{\sin }^4\mathrm{x}+{\cos }^4\mathrm{x}}\mathrm{dx}\end{array}$
Apply king and add
${\mathrm{I}}_2=\frac{\mathrm{\pi }}{4}{\int }_0^{\mathrm{\pi }/2} \frac{\tan {\mathrm{xsec}}^2\mathrm{xdx}}{{\tan }^4\mathrm{x}+1}$
put ${\tan }^2\mathrm{x}=\mathrm{t}$
$\begin{array}{l}\frac{\mathrm{\pi }}{8}{\int }_0^{\mathrm{\infty }} \frac{\mathrm{dt}}{{\mathrm{t}}^2+1}\\ =\frac{\mathrm{\pi }}{8}\cdot \frac{\mathrm{\pi }}{2}=\frac{{\mathrm{\pi }}^2}{16}\end{array}$