Answer
$$\frac{\pi}{16}$$
Work Step by Step
Given $$\int_{0}^{1} t^{2} \sqrt{1-t^{2}} d t$$
Let
$$t=\sin u\ \ \ \ \ \ \ \ \ dt =\cos udu $$
at $ t=0\to \ u=0$, at $ t=1\to \ u=\pi/2$ , so
\begin{align*}
\int_{0}^{1} t^{2} \sqrt{1-t^{2}} d t&=\int_{0}^{\pi/2} \sin^{2}u \sqrt{1-\sin^{2}u} \cos udu\\
&= \int_{0}^{\pi/2} \sin^{2}u \cos^2 udu\\
&=\frac{1}{4}\int_{0}^{\pi/2}(1-\cos 2u)(1+\cos 2u)du\\
&= \frac{1}{4}\int_{0}^{\pi/2}(1-\cos^2 2u)du
\\
&= \frac{1}{4}\int_{0}^{\pi/2}\left(\frac{1}{2} +\frac{1}{2}\cos4u\right)du\\
&= \frac{1}{4}\left(\frac{1}{2}u +\frac{1}{8}\sin4u\right)\bigg|_{0}^{\pi/2}\\
&= \frac{\pi}{16}
\end{align*}
