$\begin{align}  & \int{{{\theta }^{2}}\cos 2\theta d\theta } \\ & \text{Using integration by part} \\ & \int{udv}=uv-\int{vdu} \\ & \text{Let }u={{\theta }^{2}};\text{ }du=2\theta d\theta  \\ & dv=\cos 2\theta d\theta ,\text{ }v=\frac{1}{2}\sin 2\theta  \\ & \int{{{\theta }^{2}}\cos 2\theta }d\theta ={{\theta }^{2}}\left( \frac{1}{2}\sin 2\theta  \right)-\int{\frac{1}{2}\sin 2\theta (2\theta d\theta )} \\ & =\frac{1}{2}{{\theta }^{2}}\sin 2\theta -\int{\theta \sin 2\theta d\theta } \\ & =\frac{1}{2}{{\theta }^{2}}\sin 2\theta -\left[ \theta \int{\sin 2\theta d\theta }-\int{\left( (\int \sin 2\theta d\theta )\tfrac{d}{d\theta }(\theta ) \right)d\theta } \right] \\ & =\frac{1}{2}{{\theta }^{2}}\sin 2\theta -\left[ \frac{\theta \cos 2\theta }{2}-\int{\frac{-\cos 2\theta }{2}d\theta } \right] \\ & =\frac{1}{2}{{\theta }^{2}}\sin 2\theta +\frac{\theta \cos 2\theta }{2}+\frac{1}{2}\int{\cos 2\theta d\theta } \\ & =\frac{1}{2}{{\theta }^{2}}\sin 2\theta +\frac{\theta \cos 2\theta }{2}+\frac{1}{4}\sin 2\theta +C \\\end{align}$