# How do you integrate int tsin(2t) by integration by parts method?

Aug 14, 2016

$I = - t \cos \frac{2 t}{2} + \frac{1}{4} \sin \left(2 t\right)$

#### Explanation:

Set $\sin \left(2 t\right) \mathrm{dt} = \mathrm{dv}$
then $v = - \cos \frac{2 t}{2}$
$u = t$
$\mathrm{du} = \mathrm{dt}$

Using $\int u \mathrm{dv} = \left[u v\right] - \int v \mathrm{du}$
now you get
$I = - t \cos \frac{2 t}{2} + \frac{1}{2} \int \cos \left(2 t\right) \mathrm{dt}$
$I = - t \cos \frac{2 t}{2} + \frac{1}{4} \sin \left(2 t\right)$