# How do you integrate int xsin(2x) by integration by parts method?

Aug 8, 2016

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

#### Explanation:

For $u \left(x\right) , v \left(x\right)$

$\int u v ' \mathrm{dx} = u v ' - \int u ' v \mathrm{dx}$

$u \left(x\right) = x \implies u ' \left(x\right) = 1$

$v ' \left(x\right) = \sin \left(2 x\right) \implies v \left(x\right) = - \frac{1}{2} \cos \left(2 x\right)$

$\int x \sin \left(2 x\right) \mathrm{dx} = - \frac{x}{2} \cos \left(2 x\right) + \frac{1}{2} \int \cos \left(2 x\right) \mathrm{dx}$

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