Integration by Parts: Where Did I Go Wrong?

[stripes]

Homework Statement

Gosh I've been asking a lot of questions lately...anyways...

I tried this question two separate times and couldn't manage to figure out where i went wrong...

\int e^{-x}cos2x dx

Homework Equations

uv - \int v du = \int u dvdx<br />

The Attempt at a Solution

let u = e^{-x} and dv = cos2x dx
so du = -e^{-x} and v = \frac{1}{2}sin2x

\int e^{-x}cos2x dx = \frac{1}{2}e^{-x}sin2x + \frac{1}{2}\int e^{-x}sin2x dx

use integration by parts again, this time let u = e^{-x} and dv = sin2x dx so du = -e^{-x} and v = \frac{-1}{2}cos2x

so now we have:

\int e^{-x}cos2x dx = \frac{1}{2}e^{-x}sin2x - \frac{1}{2}e^{-x}cos2x - \frac{1}{2}\int e^{-x}cos2x dx

bring the \frac{1}{2}\int e^{-x}cos2x dx to the LHS and solve for the integral.

\frac{3}{2}\int e^{-x}cos2x dx = \frac{1}{2}e^{-x}sin2x - \frac{1}{2}e^{-x}cos2x

so finally:

\int e^{-x}cos2x dx = \frac{1}{3}e^{-x}(sin2x - cos2x)

which is incorrect... at which step did I go wrong?

thank you all in advance!

 

[Cyosis]

You forgot a factor 1/2 the second time you used integration by parts.

 

[stripes]

ahh, in which case i would get

<br /> \int e^{-x}cos2x dx = \frac{1}{5}e^{-x}(2sin2x - cos2x)<br />

 

[stripes]

thanks for pointing out my silly mistake...I have one more question that i will post up in a few minutes, this assignment is really killing me. Thanks again cyosis.

 

[Mark44]

ahh, in which case i would get

<br /> \int e^{-x}cos2x dx = \frac{1}{5}e^{-x}(2sin2x - cos2x)<br />

Plus the constant of integration.