2. Compute

$$
\int_{0}^{\pi / 2} \cos ^{2}(\cos x)+\sin ^{2}(\sin x) d x
$$
2. We have

$$
\begin{aligned}
\int_{0}^{\pi / 2} \cos ^{2}(\cos x)+\sin ^{2}(\sin x) d x & = \\
= & \int_{0}^{\pi / 2} \cos ^{2}(\cos x) d x+\int_{0}^{\pi / 2} \sin ^{2}(\sin x) d x .
\end{aligned}
$$

Perform the substitution (see fact 28) $y=\frac{\pi}{2}-x$ in the second integral, then $d y=-d x$, and

$\int_{0}^{\pi / 2} \sin ^{2}(\sin x) d x=\int_{\pi / 2}^{0} \sin ^{2}\left(\sin \left(\frac{\pi}{2}-y\right)\right)(-d y)=\int_{0}^{\pi / 2} \sin ^{2}(\cos y) d y$.

Since $\cos ^{2}(\cos x)+\sin ^{2}(\cos x)=1$, we have

$$
\begin{aligned}
& \int_{0}^{\pi / 2} \cos ^{2}(\cos x)+\sin ^{2}(\sin x) d x= \\
& =\int_{0}^{\pi / 2} \cos ^{2}(\cos x) d x+\int_{0}^{\pi / 2} \sin ^{2}(\cos x) d x=\int_{0}^{\pi / 2} d x=\frac{\pi}{2}
\end{aligned}
$$