# integrate [E^(2x)](Cos[2x])

• Apr 12th 2006, 08:07 AM
guess
integrate [E^(2x)](Cos[2x])
• Apr 12th 2006, 09:25 AM
CaptainBlack
Quote:

Originally Posted by guess

$\displaystyle e^{2x} \cos(2x)= \mbox{Re}\ (e^{2x} e^{2x \mathrm{i}})=\mbox{Re}\ (e^{2x(1+\mathrm{i})})$

Hence:

$\displaystyle \int e^{2x} \cos(2x) dx= \int \mbox{Re}\ (e^{2x(1+\mathrm{i})})dx=\mbox{Re}\ \left(\int e^{2x(1+\mathrm{i})}dx \right)$

$\displaystyle \int e^{2x} \cos(2x) dx= \mbox{Re}\ \left(\frac{e^{2x(1+\mathrm{i})}}{2(1+\mathrm{i})} \right) + C$.

You should be able to complete the simplification from here.

RonL
• Apr 12th 2006, 09:47 AM
ThePerfectHacker
Quote:

Originally Posted by guess

You have,
$\displaystyle \int e^{2x}\cos 2xdx$
Integrate by parts, $\displaystyle u'=e^{2x}\mbox{ and } v=\cos 2x$
$\displaystyle \frac{1}{2}e^{2x}\cos 2x+\int e^{2x}\sin 2x dx$
Integrate by parts, $\displaystyle u'=e^{2x}\mbox{ and } v=\sin 2x$
$\displaystyle \frac{1}{2}e^{2x}\cos 2x+\left( \frac{1}{2}e^{2x}\sin 2x-\int e^{2x}\cos 2x dx \right)$.
Thus,
$\displaystyle \int e^{2x}\cos 2xdx=\frac{1}{2}e^{2x}\cos 2x+\left( \frac{1}{2}e^{2x}\sin 2x-\int e^{2x}\cos 2x dx \right)$
Thus, (combine integrals),
$\displaystyle 2\int e^{2x}\cos 2xdx=\frac{1}{2}e^{2x}\cos 2x+ \frac{1}{2}e^{2x}\sin 2x$
Thus,
$\displaystyle \int e^{2x}\cos 2xdx=\frac{1}{4}e^{2x}(\sin 2x+\cos 2x)+C$