Inside Itself!?

We know that $\displaystyle \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$ , and $\displaystyle \int_{0}^{2a} f(x) dx = 2 \int_{0}^{a} f(x) dx$ for $f(x) = f(2a - x)$ . Let us name these basic identities as $P _ {1}$ and $P_{2}$ respectively.

$I = \displaystyle \int_{0}^{\pi} x(\sin^2(\sin x) + \cos^2 (\cos x)) dx$ ... (i)

$\Rightarrow I = \displaystyle \int_{0}^{\pi} (\pi - x)(\sin^2(\sin x) + \cos^2 (\cos x)) dx$ (using $P_{1}$ ) ... (ii)

Adding (i) and (ii) , $2I = \pi \displaystyle \int_{0}^{\pi} (\sin^2(\sin x) + \cos^2 (\cos x)) dx$

$\Rightarrow 2I = 2 \pi \displaystyle \int_{0}^{\frac{\pi}{2}} (\sin^2(\sin x) + \cos^2 (\cos x)) dx$ (using $P_{2}$ ).... (iii)

$\Rightarrow 2I = \displaystyle 2\pi \int_{0}^{\frac{\pi}{2}} (\sin^2(\cos x) + \cos^2 (\sin x)) dx$ (using $P_{1}$ ) ... (iv)

Adding (iii) and (iv), $\displaystyle 4I = 2\pi \int_{0}^{\frac{\pi}{2}} 2 dx$ (using $\sin^2 + \cos^2 = 1$ )

$\Rightarrow \boxed{I = \frac{\pi^2}{2}}$

Clearly, answer is $607$

_ Feel free to ask the proof of $P_{1}$ and $P_{2}$ in case you are not aware of them _