Answer
$\displaystyle \frac{\pi}{16}$
Work Step by Step
From the Strategy for Evaluating $\displaystyle \int\sin^{m}x\cos^{n}xdx$
(c) If the powers of both sine and cosine are even, use the half-angle identities
$\displaystyle \sin^{2}x=\frac{1}{2}(1-\cos 2x)\qquad\quad \cos^{2}x=\frac{1}{2}(1+\cos 2x)$
It is sometimes helpful to use the identity $\displaystyle \quad \sin x\cos x=\frac{1}{2}\sin 2x$
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$\displaystyle \int_{0}^{\pi}\sin^{2}t\cos^{4}tdt=\frac{1}{4}\int_{0}^{\pi}\left(4\sin^{2}t\cos^{2}t\right)\cos^{2}tdt$
... we are going to use $ \displaystyle \sin x\cos x=\frac{1}{2}\sin 2x$
$=\displaystyle \frac{1}{4}\int_{0}^{\pi}(2\sin t\cos t)^{2}\frac{1}{2}(1+\cos 2t)dt$
$=\displaystyle \frac{1}{8}\int_{0}^{\pi}(\sin 2t)^{2}(1+\cos 2t)dt$
$=\displaystyle \frac{1}{8}\int_{0}^{\pi}\left(\sin^{2}2t+\sin^{2}2t\cos 2t\right)dt$
$=\displaystyle \frac{1}{8}\int_{0}^{\pi}\sin^{2}2tdt+\frac{1}{8}\int_{0}^{\pi}\sin^{2}2t\cos 2tdt$
... now, $\displaystyle \sin^{2}x=\frac{1}{2}(1-\cos 2x)$ for the first integral, and
... for the second, substituting $\left[\begin{array}{lll}
u=\sin^{2}2t & \text{bounds:} & 0, 0\\
du=4\sin 2t\cos 2t & & \\
& &
\end{array}\right]$
we have
$=\displaystyle \frac{1}{8}\int_{0}^{\pi}\frac{1}{2}(1-\cos 4t)dt+ \displaystyle \frac{1}{4}\int_{0}^{0}udu$
$=\displaystyle \frac{1}{8}\int_{0}^{\pi}\frac{1}{2}(1-\cos 4t)dt+0$
$=\displaystyle \frac{1}{16}\left[t-\frac{1}{4}\sin 4t\right]_{0}^{\pi}$
$=\displaystyle \frac{1}{16}[(\pi-0)-(0-0)]$
$=\displaystyle \frac{\pi}{16}$
