Single Variable Calculus, Chapter 8, 8.1, Section 8.1, Problem 6

Evaluate $\displaystyle \int t \sin 2t dt$ by using Integration by parts.
If we let $u = t$ and $dv = \sin 2t dt$, then
$\displaystyle du = dt = v = \int \sin 2t dt = -\frac{1}{2} \cos 2t$

So,

$
\begin{equation}
\begin{aligned}
\int t \sin 2t dt = uv - \int vdu &= -\frac{1}{2} t \cos 2t - \int \left( -\frac{1}{2} \cos 2t \right) dt\\
\\
&= \frac{-t \cos 2t}{2} + \frac{1}{4} \sin 2t + c
\end{aligned}
\end{equation}
$