Compute $\int_{0}^{\pi / 2}\left(\sin ^{2}(\sin x)+\cos ^{2}(\cos x)\right) d x$.

#
Transform the integrand using the reduction formula $\sin x=\cos (\pi / 2-x)$ and the fundamental trigonometric identity $\sin ^{2} x+\cos ^{2} x=1$.

## Solution

Let's write: $\int_{0}^{\pi / 2}\left(\sin ^{2}(\sin x)+\cos ^{2}(\cos x)\right) d x=\int_{0}^{\pi / 2}\left(\sin ^{2}(\sin x)+1-\sin ^{2}(\sin (\pi / 2-x)) d x\right.$. But $\int_{0}^{\pi / 2} f(x) d x=\int_{0}^{\pi / 2} f(\pi / 2-x) d x$, since the graphs of the functions $f(x)$ and $f(\pi / 2-x)$ are symmetric with respect to the line $x=\pi / 4$, and the integrals of these functions over the interval from 0 to $\pi / 2$ will be equal. Thus, the desired integral is $\pi / 2$.

## Answer

$\pi / 2$.