- anonymous

integrate by parts e^-x cos 2xdx.

- schrodinger

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- bahrom7893

Did you mean
\[\int\limits_{}^{}e^{-x}*Cos(2x)dx\]

- anonymous

yes

- bahrom7893

oaky working on it

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## More answers

- anonymous

thank you!!

- bahrom7893

So in this case you use integration by parts.
Let:
\[u = e^{-x}\], then
\[du = -e^{-x}*dx\]

- bahrom7893

\[dv = Cos(2x)dx\]
\[v = (Sin(2x))/2\]

- bahrom7893

Integration by parts: u*v - Integral(v*du)
\[e^{-x}*(1/2)*(Sin(2x)) - \int\limits_{}^{}(1/2)(Sin(2x))*(-e^{-x})dx\]

- bahrom7893

Simplify (take 1/2 out):
\[(1/2) (e^{-x}Sin(2x) + \int\limits_{}^{}Sin(2x)e^{-x}dx)\]
Now for now ignore everything except:
\[I =\int\limits_{}^{}Sin(2x)*e^{-x}*dx\]

- bahrom7893

I will just call it I, solve it and then plug it back into the equation above. Do you get it?

- bahrom7893

\[I =\int\limits_{}^{}Sin(2x)*e^{-x}*dx\]
use integration by parts again:
u = e^(-x); du = -e^(-x)*dx

- bahrom7893

hey, sorry I hate typing out integral problems, especially long ones, can I write it out, take a picture and email it to you?

- bahrom7893

my page just crashed and I lost half the stuff I entered

- anonymous

yeah no problem

- bahrom7893

okay workin on it

- anonymous

i understand how to get to:
=e^-x cos 2x + 2e^-x sin 2x -4 integral e^-x cos 2xdx,

- bahrom7893

Oh then its all good. Now move -4 integral e^-x cos 2xdx, to the left

- bahrom7893

So you'll get:
5*integral e^-x cos 2xdx = e^-x cos 2x + 2e^-x sin 2x

- bahrom7893

So for the answer:
integral e^-x cos 2xdx = [ (e^-x cos 2x + 2e^-x sin 2x) / 5 ] + C

- bahrom7893

do you get it? I mean as long as your solution is correct, then that's the answer. Wait, actually I just checked it and you're off by a minus sign somewhere

- bahrom7893

[ (-e^-x cos 2x + 2e^-x sin 2x) / 5 ] + C
See, you were right, but just check your arithmetic. Remember this:
When you have two functions that cannot be differentiated down to 0, you keep integrating, one side until you see something that looks like the original integral.

- anonymous

i think i get it! :)

- bahrom7893

So in this case you keep integrating by parts until one side has the original integral in it. Once you get that original integral, just move it over and solve for it.

- anonymous

awesome thank you very much!!!

- bahrom7893

I remember having trouble with integration by parts back in HS. Especially if you have two functions that can't be differentiated down to 0, and kept wonderin which one to pick for u and dv, the answer is, it doesn't matter, your still goin around in circles. Even if u pick Cos(2x) for u and e^(-x) dx for dv, u will get the same answer. Click on Become a fan if I helped, lol I wanna see who I become after superhero! Thanks.

- anonymous

i did! yeah i never know what to pick for u or dv

- bahrom7893

lol 40 fans... and still Superhero, well maybe at 50 =D!

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