Answer
$$\frac{1}{3}x{\sin ^3}x + \frac{1}{6}{\cos ^3}x - \frac{1}{3}\cos x + C$$
Work Step by Step
$$\eqalign{
& \int {x{{\sin }^2}x\cos x} dx \cr
& {\text{Integrate by parts}} \cr
& {\text{Let }}u = x,{\text{ }}du = dx \cr
& dv = {\sin ^2}x\cos xdx \to v = \frac{1}{3}{\sin ^3}x \cr
& {\text{Use integration by parts formula}} \cr
& \int {udv} = uv - \int {vdu} \cr
& \int {x{{\sin }^2}x\cos x} dx = \frac{1}{3}x{\sin ^3}x - \int {\frac{1}{3}{{\sin }^3}xdx} \cr
& \int {x{{\sin }^2}x\cos x} dx = \frac{1}{3}x{\sin ^3}x - \frac{1}{3}\int {{{\sin }^3}xdx} \cr
& = \frac{1}{3}x{\sin ^3}x - \frac{1}{3}\int {{{\sin }^2}x\sin xdx} \cr
& {\text{Use pythagorean identity}} \cr
& = \frac{1}{3}x{\sin ^3}x - \frac{1}{3}\int {\left( {{{\cos }^2}x - 1} \right)\sin xdx} \cr
& = \frac{1}{3}x{\sin ^3}x - \frac{1}{3}\int {{{\cos }^2}x\sin xdx} + \frac{1}{3}\int {\sin x} dx \cr
& {\text{Integrating}} \cr
& = \frac{1}{3}x{\sin ^3}x + \frac{1}{6}{\cos ^3}x - \frac{1}{3}\cos x + C \cr} $$
