# How do you integrate intx cos 2x dx?

Jun 5, 2015

$\int x \cos \left(2 x\right) \mathrm{dx} =$ by parts:
$= x \sin \frac{2 x}{2} - \left[\int \sin \frac{2 x}{2} \mathrm{dx}\right] =$
$= x \sin \frac{2 x}{2} + \frac{1}{4} \cos \left(2 x\right) + c$

Jun 5, 2015

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

Let $u = x$, $\setminus \quad \setminus \implies \mathrm{du} = \mathrm{dx}$

and let $\setminus \quad \setminus \quad \mathrm{dv} = \cos \left(2 x\right) \mathrm{dx}$, $\setminus \implies v = \frac{1}{2} \sin \left(2 x\right)$

Now integrate by parts

$\int x \cos \left(2 x\right) \mathrm{dx} = \int u \setminus \quad \mathrm{dv} = u v - \int v \mathrm{du}$

$\setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \quad = x \cdot \frac{1}{2} \sin \left(2 x\right) - \int \frac{1}{2} \sin \left(2 x\right) \mathrm{dx}$

$\setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \quad = \frac{x}{2} \sin \left(2 x\right) + \frac{1}{4} \cos \left(2 x\right) + C$

where $C$ is the constant of integration.

Quick note on how to get integration by parts formula:

The differential of $u v$ is

$d \left[u v\right] = u \mathrm{dv} + v \mathrm{du}$

$u \mathrm{dv} = d \left[u v\right] - v \mathrm{du}$

Integrate both sides

$\setminus \int u \mathrm{dv} = \int d \left[u v\right] - \int v \mathrm{du}$

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