Answer
$\displaystyle \frac{1}{3}x\cos^{3}x-x\cos x+\frac{2}{3}\sin x+\frac{1}{9}\sin^{3}x+C$
Work Step by Step
$I=\displaystyle \int x\sin^{3}xdx$
We want to integrate by parts, in which $u=x,\ du=dx$ ... loses x.
Since our $dv$ is $\sin^{3}x$ we find
$\displaystyle \int\sin^{3}xdx=\int\left(1-\cos^{2}x\right)\sin xdx$
... From the Strategy for Evaluating $\displaystyle \int\sin^{m}x\cos^{n}xdx$, case (b),
... we use the substitution $\left[\begin{array}{l}
u=\cos x,\\
du=-\sin xdx
\end{array}\right]$
$=\displaystyle \int\left(1-u^{2}\right)(-du)$
$=\displaystyle \int\left(u^{2}-1\right)du$
$=\displaystyle \frac{1}{3}u^{3}-u+C$
$=\displaystyle \frac{1}{3}\cos^{3}x-\cos x+C$
So, for our integration py parts $\displaystyle \int udv=uv-\int vdu,$
we take $\left[\begin{array}{ll}
u=x & dv=\sin^{3}xdx\\
du=x & v=\frac{1}{3}\cos^{3}x-\cos x
\end{array}\right]$
$I=\displaystyle \frac{1}{3}x\cos^{3}x-x\cos x-\int\left(\frac{1}{3}\cos^{3}x-\cos x\right)dx$
$=\displaystyle \frac{1}{3}x\cos^{3}x-x\cos x-\frac{1}{3}\int\cos^{3}xdx+\int\cos xdx$
$=\displaystyle \frac{1}{3}x\cos^{3}x-x\cos x-\frac{1}{3}\int\cos^{3}xdx+\sin x$
... this integral is solved in Example 1.
$=\displaystyle \frac{1}{3}x\cos^{3}x-x\cos x-\frac{1}{3}\left(\sin x-\frac{1}{3}\sin^{3}x\right)+\sin x+C$
$=\displaystyle \frac{1}{3}x\cos^{3}x-x\cos x+\frac{2}{3}\sin x+\frac{1}{9}\sin^{3}x+C$
