# How do you integrate int e^(2x)cosx by integration by parts method?

Sep 4, 2016

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

#### Explanation:

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

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

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

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

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

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

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

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

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

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

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