Integral of e^x*cos(2x)
Would I use the u,du,dv,v method? And set dv = e^x, and U=Cos(2x) ?
You need to use integration by parts twice. Read example 2.5 here: Integration by Parts
I use the ILATE mnemonic.
Yes you have to use integration by parts, this is one of the tricky ones though. You have to use parts twice, then realize something you've seen before. Heres the hint, your integral after using parts twice will look something like (forgive the lack of mathematical rigor, but I'm simply trying to give a hint):
$\displaystyle \int e^xcos(2x) = (something)+(something\;else)-(constant)\int e^xcos(2x)$
Moving over the last part we have:
$\displaystyle \int e^xcos(2x) +(constant)\int e^xcos(2x) = (something)+(something\;else)$
Combining like terms on the left side gives:
$\displaystyle (\int e^xcos(2x))(1+(constant)) = (something)+(something\;else)$
And finnally, buy dividing by the constant plus one:
$\displaystyle \int e^xcos(2x) = \frac{(something)+(something\;else)}{1+(constant)}$
And thats my hint, take it for what its worth.
Ok, first off, how do I write in the proper math formulas? (mine always looks so messy
I've almost gotten to the end of the problem right now it's looking like:
Integral of (e^x)cos(2x) = cos(2x)(e^x) + 2(sin(2x)(e^x) - 2[integralsign](e^x)cos(2x))
So I have gotten back to the original function but I'm a little confused about how to distribute the 2, does the 2 go in front of both the sin(2x) and the (e^x)?? Thanks
Your almost there, notice on the left side of the equation that you have two like terms, both being $\displaystyle \ int e^xcos(2x)$, so the following:
$\displaystyle \int e^xcos(2x) +4\int e^xcos(2x) = [\int e^xcos(2x)][1+4] = 5 \int e^xcos(2x) $
Then, in the last step you showed, you simply divide by 5 on both sides of the equation to get:
$\displaystyle \int e^xcos(2x) = \frac{e^xcos(2x) + 2sin(2x)e^x}{5}$