Answer
$\displaystyle \frac{2}{5}$
Work Step by Step
Use the double angle identity $\sin 2x=2\sin x\cos x$
$\displaystyle \int_{0}^{\pi/2}\cos^{3}x\sin 2xdx=\int_{0}^{\pi/2}\cos^{3}x(2\sin x\cos x)dx$
$=\displaystyle \int_{0}^{\pi/2}2\cos^{4}x\sin xdx\qquad\left[\begin{array}{ll}
t=\cos x & \\
dt=-\sin xdx & \\
x=\pi/2\Rightarrow t=0 & x=0\Rightarrow t=1
\end{array}\right]$
$=-2\displaystyle \int_{1}^{0}t^{4}dt$
$=-2\left[\frac{1}{5}t^{5}\right]_{1}^{0}$
$=-2(0-\displaystyle \frac{1}{5})$
$=\displaystyle \frac{2}{5}$
