# How do you find the antiderivative of e^( -x)*cos2x?

Apr 11, 2018

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

#### Explanation:

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

$= m a t h \boldsymbol{R e} \left(\int \mathrm{dx} \setminus {e}^{- x} {e}^{2 i x}\right)$

$= m a t h \boldsymbol{R e} \left(\frac{1}{- 1 + 2 i} {e}^{\left(- 1 + 2 i\right) x}\right)$

$= {e}^{- x} m a t h \boldsymbol{R e} \left(\frac{- 1 - 2 i}{5} \left(\cos 2 x + i \sin 2 x\right)\right)$

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

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

Apr 11, 2018

$\int \setminus {e}^{- x} \cos \left(2 x\right) \setminus \mathrm{dx} = {e}^{- x} / 5 \left\{2 \sin \left(2 x\right) - \cos \left(2 x\right)\right\} + C$

#### Explanation:

Let:

$I = \int \setminus {e}^{- x} \setminus \cos \left(2 x\right) \setminus \mathrm{dx}$

We can use integration by parts:

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

Then plugging into the IBP formula:

$\int \setminus \left(u\right) \left(\frac{\mathrm{dv}}{\mathrm{dx}}\right) \setminus \mathrm{dx} = \left(u\right) \left(v\right) - \int \setminus \left(v\right) \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) \setminus \mathrm{dx}$

gives us

$\int \setminus \left(\cos 2 x\right) \left({e}^{- x}\right) \setminus \mathrm{dx} = \left(\cos 2 x\right) \left(- {e}^{- x}\right) - \int \setminus \left(- {e}^{- x}\right) \left(- 2 \sin 2 x\right) \setminus \mathrm{dx}$
$\therefore I = - {e}^{- x} \cos 2 x - 2 \setminus \int \setminus {e}^{- x} \setminus \sin 2 x \setminus \mathrm{dx}$ .... [A]

At first it appears as if we have made no progress, as now the second integral is similar to $I$, having exchanged $\sin 2 x$ for $\sin 2 x$, but if we apply IBP a second time then the progress will become clear:

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

Then plugging into the IBP formula, gives us:

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

Inserting this result into [A] we get:

$I = - {e}^{- x} \cos 2 x - 2 \left(- {e}^{x} \sin 2 x + 2 I\right) + A$

Now we can solve fior $I$

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

$\therefore 5 I = {e}^{x} \left\{2 \sin 2 x - \cos 2 x\right\} + A$

$\therefore I = {e}^{x} / 5 \left\{2 \sin 2 x - \cos 2 x\right\} + A$