Single Variable Calculus, Chapter 8, 8.1, Section 8.1, Problem 16

Evaluate $\displaystyle \int t \sin h mt dt$
If we let $u = t$ and $dv =\sin h (mt) dt$, then...
$du = dt$ and $\displaystyle v = \int \sin h (mt) dt = \frac{1}{m} \cos h (mt)$

So,

$
\begin{equation}
\begin{aligned}
\int t \sin h mt dt &= uv - \int v du = \frac{t \cos h (mt)}{m} - \int \frac{\cos h(mt) dt}{m}\\
\\
&= \frac{t \cos h (mt)}{m} - \frac{1}{m} \int \cos h (mt) dt\\
\\
&= \frac{t \cos h (mt)}{m} - \frac{1}{m} \left( \sin h (mt) \left( \frac{1}{m} \right) + c\right)\\
\\
&= \frac{t \cos h (mt)}{m} - \frac{\sin h (mt)}{m^2} + c
\end{aligned}
\end{equation}
$