# How do you integrate int cos2x*e^-x by integration by parts method?

Nov 24, 2016

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

#### Explanation:

We use integration by parts

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

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

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

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

We apply once more the integration by parts

$v = \sin 2 x$, $\implies$,$v ' = 2 \cos 2 x$

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

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

So,

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

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

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

$5 \int {e}^{- x} \cdot \cos 2 x \mathrm{dx} = - {e}^{- x} \cdot \cos 2 x + 2 {e}^{- x} \sin 2 x$

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