Answer
$$\frac{\pi }{16}$$
Work Step by Step
Given
$$ \int_{0}^{\pi}\sin^2 t\cos^4tdt$$
since
$$\cos^2x=\frac{1}{2}(1+\cos 2x),\ \ \ \sin^2x=\frac{1}{2}(1-\cos 2x)$$
then
\begin{align*}
\int_{0}^{\pi}\sin^2 t\cos^4( t)dt&=\frac{1}{8}\int_{0}^{\pi}[ (1+\cos 2t)^2 (1-\cos 2t)]dt\\
&=\frac{1}{8}\int_{0}^{\pi}[ (1+\cos 2t) (1-\cos^2 2t)]dt\\
&=\frac{1}{8}\int_{0}^{\pi}[ 1-\cos^2 2t + \cos 2t-\cos^3 2t)]dt\\
&=\frac{1}{8}\int_{0}^{\pi}[ \frac{1}{2}-\frac{1}{2}\cos4t + \cos 2t-(1-\sin^2(2t))\cos 2t)]dt\\
&=\frac{1}{8}\int_{0}^{\pi}[ \frac{1}{2}-\frac{1}{2}\cos4t + \sin^2(2t) \cos 2t)]dt\\
&=\frac{1}{8}\left( \frac{1}{2}t-\frac{1}{8}\sin4t + \frac{1}{6} \sin^3(2t) \right)\bigg|_{0}^{\pi}\\
&=\frac{\pi }{16}
\end{align*}
