# How do you integrate e^2xcosx dx?

Feb 7, 2015

Starting from:
$\int {e}^{2} x \cos \left(x\right) \mathrm{dx} =$ I would take the constant ${e}^{2}$ out of the integral:
${e}^{2} \int x \cos \left(x\right) \mathrm{dx} =$ then I would use integration by parts where you have:
$\int f \left(x\right) \cdot g \left(x\right) \mathrm{dx} = F \left(x\right) g \left(x\right) - \int F \left(x\right) g ' \left(x\right) \mathrm{dx}$
Where:
$F \left(x\right) = \int f \left(x\right) \mathrm{dx}$
$g ' \left(x\right)$ is the derivative of $g \left(x\right)$

In your case choosing $g \left(x\right) = x$ you get:

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

hope it helps