# Homework Help: 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?