Answer
$$\int_{0}^{2}x^{2}\sqrt{4-x^{2}}dx=\pi$$
Work Step by Step
Table 31:
$\int u^{2}\sqrt{a^{2}-u^{2}}du=\frac{u}{8}(2u^{2}-a^{2})\sqrt{a^{2}-u^{2}}+\frac{a^{4}}{8}sin^{-1}(\frac{u}{a})+C$
$$\int_{0}^{2}x^{2}\sqrt{4-x^{2}}dx=\left [\frac{x}{8}(2x^{2}-4)\sqrt{4-x^{2}}+\frac{16}{8}sin^{-1}(\frac{x}{2}) \right ]\tfrac{0}{2}$$
$$=2sin^{-1}1=\pi$$