Answer
$-2\pi^{2}$
Work Step by Step
Integration by parts:
$\displaystyle \int udv=uv-\int vdu$
The idea is to choose a relatively easy $dv$ to integrate, and
a u whose $u'$ does not complicate matters (best case: makes thing simpler).
----
$\left[\begin{array}{ll}
u=t^{2} & dv=\sin 2tdt\\
& \\
du=2tdt & v=-\frac{1}{2}\cos 2t
\end{array}\right]$
$I=\displaystyle \int_{0}^{2\pi}t^{2}\sin 2tdt=uv|_{0}^{2\pi}-\int_{0}^{2\pi}vdu$
$=[-\displaystyle \frac{1}{2}t^{2}\cos 2t]_{0}^{2\pi}+\int_{0}^{2\pi}t\cos 2tdt$
$=(-\displaystyle \frac{1}{2}4\pi^{2}\cos 4\pi-0)+\int_{0}^{2\pi}t \cos 2tdt$
$=-2\displaystyle \pi^{2}+\int_{0}^{2\pi}t \cos 2tdt$
By parts, again, $\left[\begin{array}{ll}
u=t & dv=\cos 2tdt\\
& \\
du=dt & v=\frac{1}{2}\sin 2t
\end{array}\right]$
$I=-2\displaystyle \pi^{2}+([\frac{1}{2}t\sin 2t]_{0}^{2\pi}-\int_{0}^{2\pi}\frac{1}{2}\sin 2tdt)$
$=-2\displaystyle \pi^{2}+0-[-\frac{1}{4}\cos 2t]_{0}^{2\pi}$
$=-2\displaystyle \pi^{2}-[-\frac{1}{4}\cos 4\pi+\frac{1}{4}\cos 0]$
$=-2\pi^{2}-0$
$=-2\pi^{2}$
