# How do you integrate int e^(5x)cos3x?

Sep 19, 2017

You can integrate by parts twice and algebraically solve for the integral to get $\int \setminus {e}^{5 x} \cos 3 x \setminus \mathrm{dx} = \frac{5}{34} {e}^{5 x} \cos 3 x + \frac{3}{34} {e}^{5 x} \sin 3 x + C$

#### Explanation:

Let $I = \int \setminus {e}^{5 x} \cos 3 x \setminus \mathrm{dx}$. Now let $u = {e}^{5 x}$ and $\mathrm{dv} = \cos 3 x \setminus \mathrm{dx}$ so that $\mathrm{du} = 5 {e}^{5 x} \setminus \mathrm{dx}$ and $v = \frac{1}{3} \sin 3 x$.

Then $I = \frac{1}{3} {e}^{5 x} \sin 3 x - \frac{5}{3} \int \setminus {e}^{5 x} \sin 3 x \setminus \mathrm{dx}$.

For this next integral, let $u = {e}^{5 x}$ and $\mathrm{dv} = \sin 3 x \setminus \mathrm{dx}$ so that $\mathrm{du} = 5 {e}^{5 x}$ and $v = - \frac{1}{3} \cos 3 x$. It follows that

$I = \frac{1}{3} {e}^{5 x} \sin 3 x - \frac{5}{3} \left(- \frac{1}{3} {e}^{5 x} \cos 3 x + \frac{5}{3} \int {e}^{5 x} \cos 3 x \setminus \mathrm{dx}\right)$

$= \frac{1}{3} {e}^{5 x} \sin 3 x + \frac{5}{9} {e}^{5 x} \cos 3 x - \frac{25}{9} I$.

Therefore, $\frac{34}{9} I = \frac{1}{3} {e}^{5 x} \sin 3 x + \frac{5}{9} {e}^{5 x} \cos 3 x$.

Multiplying both sides by $\frac{9}{34}$, rearranging, and tacking on a $+ C$ at the end gives

$I = \int \setminus {e}^{5 x} \cos 3 x \setminus \mathrm{dx} = \frac{5}{34} {e}^{5 x} \cos 3 x + \frac{3}{34} {e}^{5 x} \sin 3 x + C$.