Answer
$$\frac{-\pi +3\sqrt{3}}{24}$$
Work Step by Step
Given $$ \int_{0}^{\pi/6}t\sin 2t\ dt $$Let
\begin{align*}
u&= t\ \ \ \ \ \ \ \ \ \ \ \ \ \ dv=\sin 2t\\
u&= dt\ \ \ \ \ \ \ \ \ \ \ \ \ dv= \frac{-1}{2}\cos 2t
\end{align*}
Then using integration by parts
\begin{align*}
\int_{0}^{\pi/6}t\sin 2t\ dt &=uv-\int vdu\\
&= \frac{-1}{2}t\cos 2t\bigg|_{0}^{\pi/6} + \frac{1}{2} \int_{0}^{\pi/6} \cos 2tdt\\
&=\frac{-1}{2}t\cos 2t \bigg|_{0}^{\pi/6}+\frac{1}{4}\sin 2t \bigg|_{0}^{\pi/6}\\
&=\frac{-\pi +3\sqrt{3}}{24}-0\\
&=\frac{-\pi +3\sqrt{3}}{24}
\end{align*}
