Answer
$\frac{1}{4}{t^2} - \frac{1}{4}t\sin 2t - \frac{1}{8}\cos 2t + C$
Work Step by Step
$$\eqalign{
& \int {t{{\sin }^2}t} dt \cr
& {\text{Using the identity }}{\sin ^2}\theta = \frac{{1 - \cos 2\theta }}{2} \cr
& = \int {t\left( {\frac{{1 - \cos 2t}}{2}} \right)} dt \cr
& = \frac{1}{2}\int {\left( {t - t\cos 2t} \right)} dt \cr
& = \frac{1}{2}\int t dt - \frac{1}{2}\int {t\cos 2t} dt \cr
& \cr
& {\text{Integrate by parts }}\int {t\cos 2t} dt \cr
& u = t,{\text{ }}du = dt \cr
& dv = \cos 2t,{\text{ }}v = \frac{1}{2}\sin 2t \cr
& = \frac{1}{2}t\sin 2t - \int {\frac{1}{2}\sin 2t} dt \cr
& = \frac{1}{2}t\sin 2t - \frac{1}{2}\left( { - \frac{1}{2}\cos 2t} \right) \cr
& = \frac{1}{2}t\sin 2t + \frac{1}{4}\cos 2t + C \cr
& \cr
& {\text{Therefore}}{\text{,}} \cr
& = \frac{1}{2}\int t dt - \frac{1}{2}\int {t\cos 2t} dt \cr
& = \frac{1}{2}\left( {\frac{{{t^2}}}{2}} \right) - \frac{1}{2}\left( {\frac{1}{2}t\sin 2t + \frac{1}{4}\cos 2t} \right) + C \cr
& {\text{Simplifying}} \cr
& = \frac{1}{4}{t^2} - \frac{1}{4}t\sin 2t - \frac{1}{8}\cos 2t + C \cr} $$