How do you find the integral of e^(2x) cos4x dx?

Oct 10, 2015

Integrate by parts twice using $u = {e}^{2 x}$ both times.

Explanation:

$I = \int {e}^{2 x} \cos 4 x \mathrm{dx}$

$u = {e}^{2 x}$ and $\mathrm{dv} = \cos 4 x \mathrm{dx}$

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

$= \frac{1}{4} {e}^{2 x} \sin 4 x - \frac{1}{2} \int {e}^{2 x} \sin 4 x \mathrm{dx}$

Now take $u = {e}^{2 x}$ and $\mathrm{dv} = \sin 4 x \mathrm{dx}$

$I = \frac{1}{4} {e}^{2 x} \sin 4 x - \frac{1}{2} \left[- \frac{1}{4} {e}^{2 x} \cos 4 x + \frac{1}{2} \int {e}^{2 x} \cos 4 x \mathrm{dx}\right]$

$I = \frac{1}{4} {e}^{2 x} \sin 4 x + \frac{1}{8} {e}^{2 x} \cos 4 x - \frac{1}{4} I$

Solve for $I = \frac{1}{5} {e}^{2 x} \sin 4 x + \frac{1}{10} {e}^{2 x} \cos 4 x + C$