∫ 0 π x sin 2 sin x + cos 2 cos x d x =

$I={\int }_0^{\pi }x\left({\sin }^2\left(\sin x\right)+{\cos }^2\left(\cos x\right)\right)dx$

$={\int }_0^{\pi }\left(\mathrm{\pi }-x\right)\left({\sin }^2\left(\sin x\right)+{\cos }^2\left(\cos x\right)\right)dx$

$\Rightarrow 2I=\mathrm{\pi }{\int }_0^{\pi }\left({\sin }^2\left(\sin x\right)+{\cos }^2\left(\cos x\right)\right)dx$

$=2\mathrm{\pi }{\int }_0^{\frac{\pi }{2}}\left({\sin }^2\left(\sin x\right)+{\cos }^2\left(\cos x\right)\right)dx$

$\Rightarrow I=\mathrm{\pi }{\int }_0^{\frac{\pi }{2}}\left({\sin }^2\left(\sin x\right)+{\cos }^2\left(\cos x\right)\right)dx ...\left(i\right)$

$I=\mathrm{\pi }{\int }_0^{\frac{\pi }{2}}\left({\sin }^2\left(\sin \left(\frac{\mathrm{\pi }}{2}-x\right)\right)+{\cos }^2\left(\cos \left(\frac{\mathrm{\pi }}{2}-x\right)\right)\right)dx$

$I=\mathrm{\pi }{\int }_0^{\frac{\pi }{2}}\left({\sin }^2\left(\cos x\right)+{\cos }^2\left(\sin x\right)\right)dx ...\left(ii\right)$

Adding both the equation, we get,

$2I=\mathrm{\pi }{\int }_0^{\frac{\pi }{2}}2dx\Rightarrow I=\frac{{\mathrm{\pi }}^2}{2}$