Answer
$$\frac{{{\pi ^2} - 4}}{8}$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /2} {{\theta ^2}} \sin 2\theta d\theta \cr
& {\text{Using integration by parts method }} \cr
& \,\,\,\,\,{\text{Let }}u = {\theta ^2},\,\,\,\,du = 2\theta d\theta,\,\,\,\,\,dv = \sin 2\theta d\theta,\,\,\,\,v = - \frac{1}{2}\cos 2\theta \cr
& \cr
& {\text{Integration by parts then gives}} \cr
& \int {{\theta ^2}\sin 2\theta } d\theta = - \frac{1}{2}{\theta ^2}\cos 2\theta - \int {\left( { - \frac{1}{2}\cos 2\theta } \right)} \left( {2\theta d\theta } \right) \cr
& \int {{\theta ^2}\sin 2\theta } d\theta = - \frac{1}{2}{\theta ^2}\cos 2\theta + \int {\theta \cos 2\theta } d\theta \cr
& \cr
& {\text{Integrate by parts again}} \cr
& \,\,\,\,\,{\text{Let }}u = \theta,\,\,\,\,du = d\theta,\,\,\,\,\,dv = \cos 2\theta d\theta,\,\,\,\,v = \frac{1}{2}\sin 2\theta \cr
& \int {{\theta ^2}\sin 2\theta } d\theta = - \frac{1}{2}{\theta ^2}\cos 2\theta + \left( {\frac{1}{2}\theta \sin 2\theta - \int {\left( {\frac{1}{2}\sin 2\theta } \right)d\theta } } \right) \cr
& \int {{\theta ^2}\sin 2\theta } d\theta = - \frac{1}{2}{\theta ^2}\cos 2\theta + \frac{1}{2}\theta \sin 2\theta - \frac{1}{2}\int {\sin 2\theta d\theta } \cr
& \int {{\theta ^2}\sin 2\theta } d\theta = - \frac{1}{2}{\theta ^2}\cos 2\theta + \frac{1}{2}\theta \sin 2\theta + \frac{1}{4}\cos 2\theta + C \cr
& {\text{Then}}{\text{,}} \cr
& \int_0^{\pi /2} {{\theta ^2}} \sin 2\theta d\theta = \left( { - \frac{1}{2}{\theta ^2}\cos 2\theta + \frac{1}{2}\theta \sin 2\theta + \frac{1}{4}\cos 2\theta } \right)_0^{\pi /2} \cr
& = \left( { - \frac{1}{2}{{\left( {\frac{\pi }{2}} \right)}^2}\cos 2\left( {\frac{\pi }{2}} \right) + \frac{1}{2}\left( {\frac{\pi }{2}} \right)\sin 2\left( {\frac{\pi }{2}} \right) + \frac{1}{2}\cos 2\left( {\frac{\pi }{2}} \right)} \right) - \left( {0 + 0 + \frac{1}{4}\cos 2\left( 0 \right)} \right) \cr
& = - \frac{{{\pi ^2}}}{8}\left( { - 1} \right) + 0 + \frac{1}{4}\left( { - 1} \right) - \frac{1}{4} \cr
& = \frac{{{\pi ^2}}}{8} - \frac{1}{2} \cr
& = \frac{{{\pi ^2} - 4}}{8} \cr} $$
