Cyclical Integration by Parts, going round and round

1. Feb 2, 2012

GeekPioneer

1. The problem statement, all variables and given/known data

Integrate By Parts (i.e. not using formulas)
∫e3xcos(2x)dx

3. The attempt at a solution
I keep going around in circles, I know at some point I should be able to subtract the original integral across the = and then divide out the coefficient and thats the final answer. But I keep on getting different answers.

If some one could show the proper u=x1 du=x2 and v=x2 dv=x4 and state how many times I will have to repeat this and what u=x1 du=x2 and v=x2 dv=x4 should be for each step in the serial progression that is integration by parts. Please and Thanks

2. Feb 2, 2012

jbunniii

You should be able to integrate by parts twice. Then there is a nice trick. Can you show us what you have tried so far?

3. Feb 2, 2012

GeekPioneer

y=∫e3xcos(2x)dx where u=e3x du=3e3x & v=(1/2)sin(2x) dv=cos(2x)

After integrating by parts twice i get this

y=e3xsin(2x)/2 - e3xsin(2x)/2 +sin2xe3x/2 -∫e3xcos(2x)dx

the 2nd time I integrated by parts it was

u=sin2x du=2cos(2x)

v=(1/3)e3x dv=e3x

4. Feb 2, 2012

jbunniii

OK good. Now for the trick.

You have defined

$$y = \int e^{3x} \cos(2x) dx$$

Now look at the right hand side of the result you got. The same integral appears again. So using the definition of y, you have

y = other stuff - y

Can you see what to do now?

5. Feb 2, 2012

GeekPioneer

Yeah I can see it but my answer didn't match up with what Wolfram gave.

they gave 1/13 e^(3 x) (3 cos(2 x)+2 sin(2 x))

And what am I missing why do your equations look so much...prettier?

6. Feb 2, 2012

Bohrok

It's possible your answer is still equivalent to what WA gives you.

7. Feb 2, 2012

GeekPioneer

Well yes or coarse you silly goose. I guess i need to plug in some numbers and see if they work out. Ill report back shortly.

8. Feb 2, 2012

GeekPioneer

Its not coming out the same, I don't get it?