Cyclical Integration by Parts, going round and round

[GeekPioneer]

Homework Statement

Integrate By Parts (i.e. not using formulas)
∫e^3xcos(2x)dx

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 that's 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

 

[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?

 

[GeekPioneer]

y=∫e^3xcos(2x)dx where u=e^3x du=3e^3x & v=(1/2)sin(2x) dv=cos(2x)

After integrating by parts twice i get this

y=e^3xsin(2x)/2 - e^3xsin(2x)/2 +sin2xe^3x/2 -∫e^3xcos(2x)dx


the 2nd time I integrated by parts it was

u=sin2x du=2cos(2x)

v=(1/3)e^3x dv=e^3x

 

[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?

 

[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?

 

[Bohrok]

And what did you get for your answer?
It's possible your answer is still equivalent to what WA gives you.

 

[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.

 

[GeekPioneer]

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