# How do you integrate int e^-x*cos2xdx using integration by parts?

Jul 12, 2016

$= - \frac{1}{5} {e}^{-} x \cdot \cos 2 x + 2 {e}^{-} x \sin 2 x + C$

#### Explanation:

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

using IBP

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

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

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

$= - {e}^{-} x \cdot \cos 2 x - 2 J$

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

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

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

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

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

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

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

$5 I = - {e}^{-} x \cos 2 x + 2 {e}^{-} x \sin 2 x + C$

$I = - \frac{1}{5} {e}^{-} x \cos 2 x + 2 {e}^{-} x \sin 2 x + C$