# How do you integrate int xsin2x by integration by parts method?

Jan 16, 2017

$\int x \sin 2 x \mathrm{dx} = - \frac{1}{2} x \cos 2 x + \frac{1}{4} \sin 2 x$

#### Explanation:

As:

$d \left(\cos 2 x\right) = - 2 \sin 2 x \mathrm{dx}$

we can integrate by parts in this way:

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