Answer
$$\frac{-1}{4}t\cos 2t+\frac{ 1}{8} \sin 2t+c$$
Work Step by Step
Given
$$\int t\sin t \cos t dt =\frac{1}{2}\int t\sin 2t dt$$
Let
\begin{align*}
u&=t\ \ \ \ \ \ \ \ dv=\sin 2t dt\\
du&=dt\ \ \ \ \ \ \ \ v=\frac{-1}{2}\cos 2t
\end{align*}
Then
\begin{align*}
\int t\sin t \cos t dt &=\frac{1}{2}\int t\sin 2t dt\\
&=\frac{1}{2}\left(\frac{-t}{2}\cos 2t+\frac{ 1}{2}\int \cos 2t dt \right)\\
&=\frac{1}{2}\left(\frac{-t}{2}\cos 2t+\frac{ 1}{4} \sin 2t \right)+c\\
&=\frac{-1}{4}t\cos 2t+\frac{ 1}{8} \sin 2t+c
\end{align*}
