How do you integrate (e^x)*cos x dx?
Edit: Excuse me for being boring. xp I'll have to remember to use that.
Integration by parts should work fine.
$\displaystyle \int udv = uv - \int vdu$
u = e^x ... du = e^x
dv = cosxdx ... v = sinx
$\displaystyle \int (e^x)cosxdx = (e^x)sinx - \int (e^x)sinxdx$
Doesn't look like it helps, does it? But, if we use integration by parts again for (e^x)sinxdx
$\displaystyle \int (e^x)cosxdx = (e^x)sinx - (-(e^x)cosx - \int -(e^x)cosxdx)$
$\displaystyle \int (e^x)cosxdx = (e^x)sinx + (e^x)cosx - \int (e^x)cosxdx$
Now some simple algebra, add $\displaystyle \int (e^x)cosxdx$ to both sides
$\displaystyle 2\int (e^x)cosxdx = (e^x)sinx + (e^x)cosx$
I trust you can take it from there.