# Evaluate the integral? :  int \ 2x \ sin3x \ dx

May 2, 2017

$\int \setminus 2 x \setminus \sin 3 x \setminus \mathrm{dx} = - \frac{2}{3} x \cos 3 x + \frac{2}{9} \sin 3 x + C$

#### Explanation:

We want to evaluate:

$\int \setminus 2 x \setminus \sin 3 x \setminus \mathrm{dx}$

We can use Integration By Parts (IBP). Essentially we would like to identify one function that simplifies when differentiated, and identify one that simplifies when integrated (or is at least is integrable).

So for the integrand $2 x \setminus \sin 3 x$, hopefully you can see that $2 x$ simplifies when differentiated.

Let $\left\{\begin{matrix}u & = 2 x & \implies & \frac{\mathrm{du}}{\mathrm{dx}} = 2 \\ \frac{\mathrm{dv}}{\mathrm{dx}} & = \sin 3 x & \implies & v = - \frac{1}{3} \cos 3 x\end{matrix}\right.$

Then plugging into the IBP formula:

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

Hence:

$\int \setminus \left(2 x\right) \left(\sin 3 x\right) \setminus \mathrm{dx} = \left(2 x\right) \left(- \frac{1}{3} \cos 3 x\right) - \int \setminus \left(- \frac{1}{3} \cos 3 x\right) \left(2\right) \setminus \mathrm{dx}$

$\therefore \int \setminus 2 x \setminus \sin 3 x \setminus \mathrm{dx} = - \frac{2}{3} x \cos 3 x + \frac{2}{3} \setminus \int \setminus \cos 3 x \setminus \mathrm{dx}$
$\text{ } = - \frac{2}{3} x \cos 3 x + \frac{2}{3} \setminus \frac{1}{3} \sin 3 x + C$
$\text{ } = - \frac{2}{3} x \cos 3 x + \frac{2}{9} \sin 3 x + C$