Answer
$$ - \frac{1}{4}\theta \cos 2\theta + \frac{1}{8}\sin 2\theta + C$$
Work Step by Step
$$\eqalign{
& \int {\theta \sin \theta \cos \theta d\theta } \cr
& {\text{Use }} \cr
& \sin mx\cos nx = \frac{1}{2}\left( {\sin \left[ {\left( {m - n} \right)x} \right] + \sin \left[ {\left( {m + n} \right)x} \right]} \right) \cr
& {\text{sin}}\theta {\text{cos}}\theta = \frac{1}{2}\left( {\sin \left[ {\left( {1 - 1} \right)\theta } \right] + \sin \left[ {\left( {1 + 1} \right)\theta } \right]} \right) \cr
& {\text{sin}}\theta {\text{cos}}\theta = \frac{1}{2}\sin 2\theta \cr
& \int {\theta \sin \theta \cos \theta d\theta } = \int {\frac{1}{2}\theta \sin 2\theta d\theta } \cr
& {\text{Integrate by parts,}} \cr
& {\text{Let }}u = \frac{1}{2}\theta ,{\text{ }}du = \frac{1}{2}d\theta \cr
& dv = \sin 2\theta d\theta ,{\text{ }}v = - \frac{1}{2}\cos 2\theta \cr
& \int {udv = uv - \int {vdu} } \cr
& \int {\frac{1}{2}\theta \sin 2\theta d\theta } = \frac{1}{2}\theta \left( { - \frac{1}{2}\cos 2\theta } \right) - \int {\left( { - \frac{1}{2}\cos 2\theta } \right)\left( {\frac{1}{2}} \right)d\theta } \cr
& \int {\frac{1}{2}\theta \sin 2\theta d\theta } = - \frac{1}{4}\theta \cos 2\theta + \frac{1}{4}\int {\cos 2\theta d\theta } \cr
& \int {\frac{1}{2}\theta \sin 2\theta d\theta } = - \frac{1}{4}\theta \cos 2\theta + \frac{1}{8}\sin 2\theta + C \cr} $$
