# How do you integrate int x cos 3 x dx  using integration by parts?

Jan 10, 2016

Step by step working is shown below on how to go about integration by parts.

#### Explanation:

$\int x \cos \left(3 x\right) \mathrm{dx}$

Integration by parts rule is

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$

First we select $u$
Using a mnemonic ILATE which stands for Inverse, Logarithmic, Algebraic, Trigonometric and Exponential we select $u$ in the preferred order given by ILATE.

In our problem, we can select $u = x$ this is because Algebraic function is preferred above the Trigonometric.

$u = x$
Differentiating with respect to $x$
$\mathrm{du} = \mathrm{dx}$

Once $u$ is selected remaining terms for the $\mathrm{dv}$

$\mathrm{dv} = \cos \left(3 x\right) \mathrm{dx}$

To find $v$ we have to integrate this function.

$v = \int \mathrm{dv} = \int \cos \left(3 x\right) \mathrm{dx}$

$v = \sin \frac{3 x}{3}$

Using the integration part rule

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$

$\int x \cos \left(3 x\right) \mathrm{dx} = \frac{1}{3} x \sin \left(3 x\right) - \int \left(\sin \frac{3 x}{3}\right) \mathrm{dx}$

$\int x \cos \left(3 x\right) \mathrm{dx} = \frac{1}{3} x \sin \left(3 x\right) + \frac{1}{9} \cos \left(3 x\right) + C$