# Partial Fractions prob

1. Dec 2, 2008

### elitespart

$$\int e^{ax}cosbx$$

This one is driving me insane.

So I used e^ax as u and cosbx dx as dv. And then I did it again using e^ax as u and sinbx as dv which left me with $$\int e^{ax}cosbx = \frac{1}{b}e^{ax}sinbx + \frac{a}{b^{2}}e^{ax}cosbx - \frac{a^{2}}{b^{2}}\int e^{ax}cosbxdx$$

I have no ideas if this is right and no clue what do after if it is. Thanks.

2. Dec 2, 2008

### mutton

Notice that you get back the same integral on the right side, so move it to the left and factor.

3. Dec 2, 2008

### elitespart

ok this is probably a stupid algebra question but when I move the integral from the right side to the left side then how do I get rid of the a^2/b^2 that's stuck to it?

4. Dec 2, 2008

### elitespart

nvm (word count)

5. Dec 3, 2008

### HallsofIvy

Staff Emeritus
And, finally, this problem had nothing at all to do with "partial fractions".