# How do you integrate e^(-x) * cos(2x) dx ?

Apr 3, 2018

The answer is $= {e}^{-} \frac{x}{5} \left(2 \sin \left(2 x\right) - \cos \left(2 x\right)\right) + C$

#### Explanation:

Perform integration by parts $2$ times

$\int u v ' = u v - \int u ' v$

$u = \cos 2 x$, $\implies$, $u ' = - 2 \sin 2 x$

$v ' = {e}^{-} x$, $\implies$, $v = - {e}^{-} x$

Therefore,

$\int {e}^{-} x \cos 2 x \mathrm{dx} = - {e}^{-} x \cos 2 x - \int 2 {e}^{-} x \sin 2 x \mathrm{dx}$

$u = 2 \sin 2 x$, $\implies$, $u ' = 4 \cos 2 x$

$v ' = {e}^{-} x$, $\implies$, $v = - {e}^{-} x$

$\int 2 {e}^{-} x \sin 2 x \mathrm{dx} = - 2 {e}^{-} x \sin 2 x + \int 4 {e}^{-} x \cos 2 x$

So,

$\int {e}^{-} x \cos 2 x \mathrm{dx} = - {e}^{-} x \cos 2 x - \left(- 2 {e}^{-} x \sin 2 x + \int 4 {e}^{-} x \cos 2 x\right)$

$= - {e}^{-} x \cos 2 x + 2 {e}^{-} x \sin 2 x - 4 \int {e}^{-} x \cos 2 x$

$5 \int {e}^{-} x \cos 2 x d = - {e}^{-} x \cos 2 x + 2 {e}^{-} x \sin 2 x$

$\int {e}^{-} x \cos 2 x d = \frac{1}{5} \left(2 {e}^{-} x \sin 2 x - {e}^{-} x \cos 2 x\right) + C$

$= {e}^{-} \frac{x}{5} \left(2 \sin 2 x - \cos 2 x\right) + C$