Answer
$\frac{\sqrt 3}{8}$
Work Step by Step
lets calculate the integral of:
$\int_{0}^{\frac{\pi}{3}}\sin^2x\cos xdx$
by using substitution ;
take, $\sin x=t$
differentiate $wrt .x$ on both sides
we get, $\cos xdx=dt$
And limits are from $0$ to $\frac{\sqrt 3}{2}$
$=>\int_{0}^{\frac{\sqrt 3}{2}}t^2dt=[\frac {t^3}{3}]_{0}^{\frac {\sqrt 3}{2}}$
$\frac{1}{3}$ $[(\frac{\sqrt 3}{8})^3-0]$
$\frac{1}{3}[\frac{3\sqrt 3}{8}]=\frac{\sqrt 3}{8}$
