We have,
$\begin{aligned} & I=\int_0^\pi x\left\{\sin ^2(\sin x)+\cos ^2(\cos x)\right\} d x \\ & I=\int_0^\pi(\pi-x)\left[\sin ^2(\sin (\pi-x)]+\cos ^2(\cos (\pi-x)) d x\right. \\ & I=\int_0^\pi(\pi-x)\left[\sin ^2(\sin x)+\cos ^2(\cos x)\right] d x\end{aligned}$
$2 I=\pi \int_0^\pi\left[\sin ^2(\sin x)+\cos ^2(\cos x)\right] d x$


$\begin{aligned} & I=\pi \int_0^{\pi / 2}\left[\sin ^2\left(\sin \left(\frac{\pi}{2}-x\right)\right)+\cos ^2\left(\cos \left(\frac{\pi}{2}-x\right)\right)\right] d x \\ & I=\pi \int_0^{\pi / 2} \sin ^2(\cos x)+\cos ^2(\sin x) d x \\ & 2 I=\pi \int_0^{\pi / 2} 2 d x \Rightarrow I=\pi\left(\frac{\pi}{2}\right)=\frac{\pi^2}{2}\end{aligned}$
$[I]=\left[\frac{\pi^2}{2}\right]=4$ $\left[\begin{array}{l}\because \pi^2 \in(9,10) \\ \frac{\pi^2}{2} \in(4.5,5)\end{array}\right]$