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

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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{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}