# How do I evaluate inte^(2x)cosx dx by parts?

Feb 1, 2015

I´ll start by integrating ${e}^{2 x}$ and leaving $\cos \left(x\right)$ as it is and then derive it leaving the integrated part as it is.

${e}^{2 x} / 2 \cdot \cos \left(x\right) - \int {e}^{2 x} / 2 \cdot \left(- \sin \left(x\right)\right) \mathrm{dx} =$
$= {e}^{2 x} / 2 \cdot \cos \left(x\right) + \frac{1}{2} \int {e}^{2 x} \cdot \left(\sin \left(x\right)\right) \mathrm{dx} =$

by parts again:
$= {e}^{2 x} / 2 \cdot \cos \left(x\right) + \frac{1}{2} \left[{e}^{2 x} / 2 \cdot \left(\sin \left(x\right)\right) - \int {e}^{2 x} / 2 \cdot \cos \left(x\right) \mathrm{dx}\right] =$
$= {e}^{2 x} / 2 \cdot \cos \left(x\right) + {e}^{2 x} / 4 \cdot \left(\sin \left(x\right)\right) - \frac{1}{4} \int {e}^{2 x} \cdot \cos \left(x\right) \mathrm{dx}$

$\int {e}^{2 x} \cos \left(x\right) \mathrm{dx} = {e}^{2 x} / 2 \cdot \cos \left(x\right) + {e}^{2 x} / 4 \cdot \left(\sin \left(x\right)\right) - \frac{1}{4} \int {e}^{2 x} \cdot \cos \left(x\right) \mathrm{dx}$
Now I can take to the left the integral: $- \frac{1}{4} \int {e}^{2 x} \cdot \cos \left(x\right) \mathrm{dx}$
$\int {e}^{2 x} \cos \left(x\right) \mathrm{dx} + \frac{1}{4} \int {e}^{2 x} \cdot \cos \left(x\right) \mathrm{dx} = {e}^{2 x} / 2 \cdot \cos \left(x\right) + {e}^{2 x} / 4 \cdot \left(\sin \left(x\right)\right)$
$\frac{5}{4} \int {e}^{2 x} \cdot \cos \left(x\right) \mathrm{dx} = {e}^{2 x} / 2 \cdot \cos \left(x\right) + {e}^{2 x} / 4 \cdot \left(\sin \left(x\right)\right)$
$\int {e}^{2 x} \cdot \cos \left(x\right) \mathrm{dx} = \frac{4}{5} \left[{e}^{2 x} / 2 \cdot \cos \left(x\right) + {e}^{2 x} / 4 \cdot \left(\sin \left(x\right)\right)\right] + c$