use integration by parts to find:
$\displaystyle \int^{\frac{\pi}{2}}_{0} xsin^2xcosx dx$
$\displaystyle u = x$
$\displaystyle du = dx$
$\displaystyle dv = \sin^2{x}\cos{x} \, dx$
$\displaystyle v = \frac{\sin^3{x}}{3}$
$\displaystyle \int x\sin^2{x}\cos{x} \, dx = \frac{x\sin^3{x}}{3} - \int \frac{\sin^3{x}}{3} \, dx$
for the last integral ...
$\displaystyle \sin^3{x} = (1 - \cos^2{x})\sin{x} = \sin{x} + \cos^2{x}(-\sin{x})$