# How do you integrate int xcos4x by integration by parts method?

Oct 22, 2016

$\int x \sin 4 x \mathrm{dx} = \frac{x \sin 4 x}{4} + \frac{\cos 4 x}{16} + C$

#### Explanation:

Use the formula

$\int u ' v = u v - \int u v '$

So here $u ' = \cos 4 x$ so $u = \frac{\sin 4 x}{4}$

and $v = x$ so $v ' = 1$

Putting these in the formula

$\int x \sin 4 x \mathrm{dx} = \frac{x \sin 4 x}{4} - \int \frac{\sin 4 x \mathrm{dx}}{4}$

$= \frac{x \sin 4 x}{4} + \frac{\cos 4 x}{16} + C$