$$\begin{gathered} \mathrm{I} = {\int }_0^{\frac{\mathrm{\pi }}{2}} \frac{\mathrm{x} \sin \mathrm{x} \cos \mathrm{x}}{{\sin }^4 \mathrm{x} + {\cos }^4 \mathrm{x}} \mathrm{dx} ............( \mathrm{i} ) \\ \mathrm{Using} \mathrm{the} \mathrm{property} {\int }_0^{\mathrm{a}} \mathrm{f} ( \mathrm{x} ) \mathrm{dx} = {\int }_0^{\mathrm{a}} \mathrm{f} ( \mathrm{a}- \mathrm{x} ) \mathrm{dx} \\ \mathrm{I} = {\int }_0^{\frac{\mathrm{\pi }}{2}} \frac{\left({\displaystyle \frac{\mathrm{\pi }}{2}} - \mathrm{x} \right) \sin \left(\frac{\mathrm{\pi }}{2} - \mathrm{x} \right) \cos \left(\frac{\mathrm{\pi }}{2} - \mathrm{x} \right)}{{\sin }^4 \left(\frac{\mathrm{\pi }}{2} - \mathrm{x} \right) + {\cos }^4 \left(\frac{\mathrm{\pi }}{2} - \mathrm{x} \right)} \mathrm{dx} \\ \Rightarrow \mathrm{I} = {\int }_0^{\frac{\mathrm{\pi }}{2}} \frac{\left(\frac{\mathrm{\pi }}{2} - \mathrm{x} \right) \cos \mathrm{x} \sin \mathrm{x} }{{\sin }^4 \mathrm{x} + {\cos }^4 \mathrm{x}} \mathrm{dx} .............( \mathrm{ii} ) \end{gathered}$$

Adding  ( i )  and  ( ii ),

$$\begin{gathered} 2 \mathrm{I} = {\int }_0^{\frac{\mathrm{\pi }}{2}} \frac{\left( \frac{\mathrm{\pi }}{2} . \sin \mathrm{x} \cos \mathrm{x} \right)}{{\sin }^4 \mathrm{x} + {\cos }^4 \mathrm{x} } \mathrm{dx} \\ \Rightarrow \mathrm{I} = \frac{\mathrm{\pi }}{4} {\int }_0^{\frac{\mathrm{\pi }}{2}} \left[ \frac{ \sin \mathrm{x} \cos \mathrm{x} }{{\sin }^4 \mathrm{x} + {\cos }^4 \mathrm{x} } \right] \mathrm{dx} \\ = \frac{\mathrm{\pi }}{4} {\int }_0^{\frac{\mathrm{\pi }}{2}} \left[ \frac{ {\displaystyle \frac{\sin \mathrm{x} \cos \mathrm{x}}{{\cos }^4 \mathrm{x}}}}{ {\displaystyle \frac{{\sin }^4 \mathrm{x} }{{\cos }^4 \mathrm{x}}} + 1} \right] \mathrm{dx} \\ = \frac{\mathrm{\pi }}{4} {\int }_0^{\frac{\mathrm{\pi }}{2}} \frac{ \tan \mathrm{x} {\sec }^2 \mathrm{x}}{{\tan }^4 \mathrm{x} + 1} \mathrm{dx} \end{gathered}$$

Put  tan² x = z

$\therefore$ 2 tan x sec² x  dx =  dz

$$\begin{gathered} \Rightarrow \tan \mathrm{x} {\sec }^2 \mathrm{x} \mathrm{dx} = \frac{\mathrm{dz}}{2} \\ \mathrm{When} \mathrm{x} = 0, \mathrm{z} = 0 \mathrm{and} \mathrm{when} \mathrm{x} = \frac{\mathrm{\pi }}{2}, \mathrm{z} = \infty \\ \therefore \mathrm{I} = \frac{\mathrm{\pi }}{4} {\int }_0^{\infty } \frac{{\displaystyle \frac{\mathrm{dz}}{2}}}{{\mathrm{z}}^2 + 1} \\ \Rightarrow \mathrm{I} = \frac{\mathrm{\pi }}{8} {\int }_0^{\infty } \frac{{\displaystyle \mathrm{dz}}}{1+ {\mathrm{z}}^2 } \\ = \frac{\mathrm{\pi }}{8} {\left| {\tan }^{-1} \left( \mathrm{z} \right) \right|}_0^{\infty } \\ = \frac{\mathrm{\pi }}{8} {\tan }^{-1} \infty - {\tan }^{-1} 0 \\ = \frac{\mathrm{\pi }}{8} \left( \frac{\mathrm{\pi }}{2} - 0 \right) \\ = \frac{{\mathrm{\pi }}^2}{16} \end{gathered}$$