# How do you integrate int 2x sin 4x dx?

Jan 20, 2016

Use integration by parts to find that

$\int 2 x \sin \left(4 x\right) \mathrm{dx} = \frac{1}{8} \sin \left(4 x\right) - \frac{1}{2} x \cos \left(4 x\right) + C$

#### Explanation:

Using integration by parts:

Let $u = x$ and $\mathrm{dv} = \sin \left(4 x\right) \mathrm{dx}$
Then $\mathrm{du} = \mathrm{dx}$ and $v = - \frac{1}{4} \cos \left(4 x\right)$

So, by the integration by parts formula $\int u \mathrm{dv} = u v - \int v \mathrm{du}$

2intxsin(4x)dx = 2(-1/4xcos(4x)-int((-1/4)cos(4x)dx)

$= - \frac{1}{2} \left(x \cos \left(4 x\right) - \int \cos \left(4 x\right) \mathrm{dx}\right)$

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

$= \frac{1}{8} \sin \left(4 x\right) - \frac{1}{2} x \cos \left(4 x\right) + C$