Find the integral of: e^(-x)cos(2x)dx

[MarkSA]

Hello,

1) Find the integral of: e^(-x)cos(2x)dx
with integration by parts.

Let u = e^(-x)
du = -e^(-x)dx
dv = cos2xdx
v = 1/2*sin2x
= 1/2 * e^(-x)sin2x + 1/2 * integral of: sin(2x)e^(-x)dx

Let u = e^(-x)
du = -e^(-x)dx
dv = sin2xdx
v = -1/2 * cos2x
= -1/2 * e^(-x)cos(2x) - 1/2 * integral of: cos(2x)e^(-x)dx

At this point i'm not sure what to do. I'm back exactly where I started. Any suggestions?

 

[o_O]

So you have:

\(\displaystyle \int e^{-x}cos(2x) = \frac{1}{2}e^{-x}sin2x + \frac{1}{2}\left[-\frac{1}{2}e^{-x}cos2x - \frac{1}{2}\int e^{-x}cos(2x)dx\right]\)

Let's continue:

\(\displaystyle \underbrace{\int e^{-x}cos(2x)dx} = \frac{1}{2}e^{-x}sin2x - \frac{1}{4}e^{-x}cos2x - \frac{1}{4} \underbrace{\int e^{-x} cos(2x) dx}_{ \mbox{Look familiar?}}\)

 

[MarkSA]

Yes.. it is the same integral as the one I began with, which seems like trouble. I could integrate on and on and I would never get an answer if I kept at it.

I also know the final answer does not contain an integral.

What I don't understand is how to get rid of the integral. If the integral on the right in your example was + integral: of.. instead of - and a fraction, I could cancel it out by moving it to the other side of the equation I suppose.. but it's not.

I can't integrate that.. and if I substitute the integral for all other instances of it, I still end up with another mess that contains an integral.. right?

 

[o_O]

Or ... you could move it to the other side and solve as if you were doing a simple algebra problem

 

[MarkSA]

Ooooooooooh

Thanks