$\begin{align}  & \text{Omitting the limit for now} \\ & \int{{{(\sin 2\theta -\sin \theta )}^{2}}d\theta }=\int{({{\sin }^{2}}2\theta -2\sin 2\theta \sin \theta +{{\sin }^{2}}\theta )d\theta } \\ & Note: \\ & \int{{{\sin }^{2}}2\theta d\theta }=\int{\frac{1-\cos 4\theta }{2}d\theta }=\frac{1}{2}\int{(1-\cos 4\theta )d\theta } \\ & \int{{{\sin }^{2}}2\theta d\theta }=\frac{1}{2}\left( \theta -\frac{\sin 4\theta }{4} \right) \\ & \int{{{\sin }^{2}}\theta d\theta }=\int{\frac{1-\cos 2\theta }{2}d\theta =\frac{1}{2}\int{(1-\cos 2\theta )d\theta }} \\ & \int{{{\sin }^{2}}\theta d\theta }=\frac{1}{2}\left( \theta -\frac{\sin 2\theta }{2} \right) \\ & \int{2\sin 2\theta }\sin \theta =\int{2{{\sin }^{2}}\theta \cos \theta d\theta }=2\left( \frac{{{\sin }^{2}}\theta }{3} \right)+C \\ & \int{{{(\sin 2\theta -\sin \theta )}^{2}}d\theta }=\frac{1}{2}\left( \theta -\frac{\sin 4\theta }{4} \right)-2\left( \frac{{{\sin }^{2}}\theta }{3} \right)+\frac{1}{2}\left( \theta -\frac{\sin 2\theta }{2} \right)+C \\ & \int{{{(\sin 2\theta -\sin \theta )}^{2}}d\theta }=\frac{\theta }{2}-\frac{\sin 4\theta }{8}-\frac{2{{\sin }^{2}}\theta }{3}+\frac{\theta }{2}-\frac{\sin 2\theta }{4}+C \\ & \int{{{(\sin 2\theta -\sin \theta )}^{2}}d\theta }=\theta -\frac{\sin 4\theta }{8}-\frac{\sin 2\theta }{4}-\frac{2{{\sin }^{2}}\theta }{3}+C \\ & \text{Put the limits and finish up the problem} \\\end{align}$