# How do you use integration by parts to evaluate the integral 2xsin(x)dx?

Mar 3, 2018

I tried this:

#### Explanation:

Have a look:

Mar 3, 2018

$\sin \left(x\right) - x \cos \left(x\right) + C$

#### Explanation:

We have $\int 2 x \sin \left(x\right) \mathrm{dx}$

Take the constant out:

$2 \int x \sin \left(x\right) \mathrm{dx}$

According to integration by parts, where $u$ and $v$ are functions,

$\int u v \mathrm{dx} = u \int v \mathrm{dx} - \int u ' \left(\int v \mathrm{dx}\right) \mathrm{dx}$

Here, $u = x$ and $v = \sin \left(x\right)$

So we have:

$x \int \sin \left(x\right) \mathrm{dx} - \int \left(x\right) ' \left(\int \sin \left(x\right) \mathrm{dx}\right) \mathrm{dx}$

$x \left(- \cos \left(x\right)\right) - \int \left(- \cos \left(x\right)\right) \mathrm{dx}$

$- x \cos \left(x\right) + \int \cos \left(x\right) \mathrm{dx}$

$- x \cos \left(x\right) + \sin \left(x\right)$

$\sin \left(x\right) - x \cos \left(x\right) + C$

Mar 3, 2018

$I = 2 \left(\sin x - x \cos x\right) + C$

#### Explanation:

Method of Integration by Parts:
$\int u \cdot v \mathrm{dx} = u \int v \mathrm{dx} - \int \left(\frac{\mathrm{du}}{\mathrm{dx}} \int v \mathrm{dx}\right) \mathrm{dx}$
Let,$u = 2 x \mathmr{and} v = \sin x$
So,$\frac{\mathrm{du}}{\mathrm{dx}} = 2 \mathmr{and} \int v \mathrm{dx} = \int \sin x \mathrm{dx} = - \cos x$
$I = \int 2 x \sin x \mathrm{dx} = 2 x \left(- \cos x\right) - \int \left(2\right) \left(- \cos x\right) \mathrm{dx}$
$I = - 2 x \cos x + 2 \int \cos x \mathrm{dx}$
$I = - 2 x \cos x + 2 \sin x + C$
$I = 2 \left(\sin x - x \cos x\right) + C$