Answer
$$0$$
Work Step by Step
$$\eqalign{
& \int_0^1 {{{\sin }^2}\left( {\pi x} \right)\cos \left( {\pi x} \right)dx} \cr
& {\text{Integrate by substitution }} \cr
& u = \sin \left( {\pi x} \right),\,\,\,du = \cos \left( {\pi x} \right)\pi dx \cr
& \int {{{\sin }^2}\left( {\pi x} \right)\cos \left( {\pi x} \right)dx} = \frac{1}{\pi }\int {{u^2}du} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{{3\pi }}{u^3} + C \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{{3\pi }}{\sin ^3}\left( {\pi x} \right) + C \cr
& {\text{Then}} \cr
& \int_0^1 {{{\sin }^2}\left( {\pi x} \right)\cos \left( {\pi x} \right)dx} = \frac{1}{{3\pi }}\left[ {{{\sin }^3}\left( {\pi x} \right)} \right]_0^1 \cr
& = \frac{1}{{3\pi }}\left[ {{{\sin }^3}\left( \pi \right) - {{\sin }^3}\left( 0 \right)} \right] \cr
& = \frac{1}{{3\pi }}\left[ {0 - 0} \right] \cr
& = 0 \cr} $$
