Answer
$-\displaystyle \frac{\pi}{24}+\frac{\sqrt{3}}{8}$
Work Step by Step
$\displaystyle \int_{0}^{\pi/6}t\sin 2tdt=$
by parts, $\displaystyle \left[\begin{array}{ll}
u=t & dv=\sin 2tdt\\
du=dt & v=-\frac{1}{2}\cos t
\end{array}\right]=uv|_{0}^{\pi/6}-\int_{0}^{\pi/6}vdu$
$=\displaystyle \left[-\frac{1}{2}t\cos 2t\right]_{0}^{\pi/6}-\int_{0}^{\pi/6}\left(-\frac{1}{2}\cos 2t\right)dt\quad $
$=\displaystyle \left(-\frac{\pi}{12}\cdot\frac{1}{2}\right)-(0)+\left[\frac{1}{4}\sin 2t\right]_{0}^{\pi/6}$
$=-\displaystyle \frac{\pi}{24}+\frac{1}{4}\cdot\frac{\sqrt{3}}{2}$
$=-\displaystyle \frac{\pi}{24}+\frac{\sqrt{3}}{8}$
