Answer
$ - x\cos x + \frac{1}{3}x{\cos ^3}x + \frac{2}{3}\sin x + \frac{{{{\sin }^3}x}}{9} + C$
Work Step by Step
$$\eqalign{
& \int {x{{\sin }^3}x} dx \cr
& {\text{Split the integrand}}{\text{, recall that }}{a^m}{a^n} = {a^{m + n}} \cr
& = \int {x{{\sin }^2}x\sin x} dx \cr
& {\text{Use the pythagorean identity }}{\sin ^2}\theta + {\cos ^2}\theta = 1 \cr
& = \int {x\left( {1 - {{\cos }^2}x} \right)\sin x} dx \cr
& {\text{Integrate by parts}} \cr
& {\text{Let }}u = x,{\text{ }}dv = dx \cr
& dv = \left( {1 - {{\cos }^2}x} \right)\sin x,{\text{ }}v = - \cos x + \frac{1}{3}{\cos ^3}x \cr
& \int {udv} = uv - \int {vdu} \cr
& = x\left( { - \cos x + \frac{1}{3}{{\cos }^3}x} \right) - \int {\left( { - \cos x + \frac{1}{3}{{\cos }^3}x} \right)dx} \cr
& = - x\cos x + \frac{1}{3}x{\cos ^3}x + \int {\cos x} dx - \frac{1}{3}\int {{{\cos }^3}x} dx \cr
& = - x\cos x + \frac{1}{3}x{\cos ^3}x + \int {\cos x} dx - \frac{1}{3}\int {\left( {1 - {{\sin }^2}x} \right)\cos x} dx \cr
& {\text{Integrating}} \cr
& = - x\cos x + \frac{1}{3}x{\cos ^3}x + \sin x - \frac{1}{3}\left( {\sin x - \frac{{{{\sin }^3}x}}{3}} \right) + C \cr
& {\text{Simplifying}} \cr
& = - x\cos x + \frac{1}{3}x{\cos ^3}x + \sin x - \frac{1}{3}\sin x + \frac{{{{\sin }^3}x}}{9} + C \cr
& = - x\cos x + \frac{1}{3}x{\cos ^3}x + \frac{2}{3}\sin x + \frac{{{{\sin }^3}x}}{9} + C \cr} $$
