# How do you find the Integral of e^(3x)* cos( 3 x ) dx?

Jun 18, 2018

$I = \frac{\frac{1}{3} {e}^{3 x} \sin 3 x + \frac{1}{3} {e}^{3 x} \cos 3 x}{2}$

#### Explanation:

$I = \int {e}^{3 x} \cos \left(3 x\right) \mathrm{dx}$

Apply Integration by Parts

$\textcolor{g r e e n}{\int u \mathrm{dv} = u v - \int v \mathrm{du}} \text{ "rarr }$ $\textcolor{b l u e}{\text{the trigonometric part will be " dv " in this case" color(red)( " i.e } \cos 3 x \mathrm{dx} = \mathrm{dv}}$

$I = \frac{1}{3} {e}^{3 x} \sin 3 x - \int \cancel{\frac{1}{3}} \sin 3 x \cdot \left(\cancel{3} {e}^{3 x}\right) \mathrm{dx}$

Apply integration by Parts once more

$I = \frac{1}{3} {e}^{3} x \sin 3 x + \frac{1}{3} {e}^{3 x} \cos 3 x - \cancel{\frac{1}{3}} \int \cancel{3} {e}^{3 x} \cos 3 x \mathrm{dx}$

I=1/3e^(3x)sin3x+1/3e^(3x)cos3x-color(blue)(I

color(blue)("since " I=inte^(3x)cos3xdx

$2 I = \frac{1}{3} {e}^{3 x} \sin 3 x + \frac{1}{3} {e}^{3 x} \cos 3 x$

$I = \frac{\frac{1}{3} {e}^{3 x} \sin 3 x + \frac{1}{3} {e}^{3 x} \cos 3 x}{2}$