$\begin{aligned}
& I=\int_0^{\frac{\pi}{2}} \frac{(\sin x)^{\frac{3}{2}} d x}{(\sin x)^{\frac{3}{2}} x+(\cos x)^{\frac{3}{2}}}=\int_0^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}}\left(\frac{\pi}{2}-x\right) d x}{\sin ^{\frac{3}{2}}\left(\frac{\pi}{2}-x\right)+\cos ^{\frac{3}{2}}\left(\frac{\pi}{2}-x\right)} \
& \Rightarrow \text { Adding } 2 l=\int_0^{\frac{\pi}{2}} \frac{(\sin x)^{\frac{3}{2}}+(\cos x)^{\frac{3}{2}}}{(\sin x)^{\frac{3}{2}}+(\cos x)^{\frac{3}{2}}} d x=\frac{\pi}{2} \
& I_0=\int_0^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x=\int_0^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2}-x\right) \sin x \cos x}{(\sin x)^4+(\cos x)^4} d x
\end{aligned}Adding,2 I_0=\int_0^{\frac{\pi}{2}} \frac{\frac{\pi}{2}(\sin x) \cos x}{\left(\sin ^4 x+\cos ^4 x\right)} d x\Rightarrow I_0=\frac{\pi}{4} \int_0^{\frac{\pi}{2}} \frac{\tan x\left(\sec ^2 x\right) d x}{1+\tan ^4 x}put\tan ^2 x=t \Rightarrow 2 \tan x \sec ^2 x d x=d t\begin{aligned}
& \Rightarrow I_0=\frac{\pi}{4} \int_0^{\infty} \frac{\frac{d t}{2}}{\left(1+t^2\right)}=\left.\frac{\pi}{8}\left(\tan ^{-1} t\right)\right|_0 ^{\infty}=\frac{\pi}{8}\left(\frac{\pi}{2}-0\right) \
& \Rightarrow I_0=\frac{\pi^2}{16}
\end{aligned}$