anonymous
  • anonymous
integrate by parts e^-x cos 2xdx.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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bahrom7893
  • bahrom7893
Did you mean \[\int\limits_{}^{}e^{-x}*Cos(2x)dx\]
anonymous
  • anonymous
yes
bahrom7893
  • bahrom7893
oaky working on it

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anonymous
  • anonymous
thank you!!
bahrom7893
  • bahrom7893
So in this case you use integration by parts. Let: \[u = e^{-x}\], then \[du = -e^{-x}*dx\]
bahrom7893
  • bahrom7893
\[dv = Cos(2x)dx\] \[v = (Sin(2x))/2\]
bahrom7893
  • 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
  • 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
  • bahrom7893
I will just call it I, solve it and then plug it back into the equation above. Do you get it?
bahrom7893
  • bahrom7893
\[I =\int\limits_{}^{}Sin(2x)*e^{-x}*dx\] use integration by parts again: u = e^(-x); du = -e^(-x)*dx
bahrom7893
  • 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
  • bahrom7893
my page just crashed and I lost half the stuff I entered
anonymous
  • anonymous
yeah no problem
bahrom7893
  • bahrom7893
okay workin on it
anonymous
  • anonymous
i understand how to get to: =e^-x cos 2x + 2e^-x sin 2x -4 integral e^-x cos 2xdx,
bahrom7893
  • bahrom7893
Oh then its all good. Now move -4 integral e^-x cos 2xdx, to the left
bahrom7893
  • bahrom7893
So you'll get: 5*integral e^-x cos 2xdx = e^-x cos 2x + 2e^-x sin 2x
bahrom7893
  • bahrom7893
So for the answer: integral e^-x cos 2xdx = [ (e^-x cos 2x + 2e^-x sin 2x) / 5 ] + C
bahrom7893
  • 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
  • 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
  • anonymous
i think i get it! :)
bahrom7893
  • 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
  • anonymous
awesome thank you very much!!!
bahrom7893
  • 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
  • anonymous
i did! yeah i never know what to pick for u or dv
bahrom7893
  • bahrom7893
lol 40 fans... and still Superhero, well maybe at 50 =D!

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