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".