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

Nov 28, 2016

The answer is $= \frac{\sin 2 x}{8} - \frac{x \sin 2 x}{4} + C$

#### Explanation:

We use
$\sin 2 x = 2 \sin x \cos x$

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

The integration by parts is

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

$u = x$, $\implies$, $u ' = 1$

$v ' = \sin 2 x$, $\implies$, $v = - \frac{\cos 2 x}{2}$

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

$= - \frac{x \cos 2 x}{2} + \frac{1}{2} \cdot \frac{\sin 2 x}{2}$

$= \frac{\sin 2 x}{4} - \frac{x \cos 2 x}{2}$

And finally

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

$= \frac{\sin 2 x}{8} - \frac{x \sin 2 x}{4} + C$