$\int_0^2 x^{5 / 2} \sqrt{2-x} d x$
Put $x=2 \sin ^2 \theta, d x=4 \sin \theta \cos \theta d \theta$
$$
\begin{aligned}
& =\int_0^{\pi / 2} x^{5 / 2} \sqrt{2-x} d x \\
& =\int_0^{\pi / 2}\left(2 \sin ^2 \theta\right)^{5 / 2} \sqrt{2-2 \sin ^2 \theta} \cdot 4 \sin \theta \cos \theta d \theta \\
& =\int_0^{\pi / 2} 2^3 \times 4 \cos ^2 \theta \sin ^6 \theta d \theta=32 \int_0^{\pi / 2} \sin ^6 \theta \cos ^2 \theta d \theta
\end{aligned}
$$
Using Walli's formula
$$
=32\left[\frac{(5 \cdot 3 \ldots 1)(1)}{8 \cdot 6 \cdot 4 \cdot 2}\right] \cdot \frac{\pi}{2}=\frac{5 \pi}{8}
$$