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anonymous
 5 years ago
integrate by parts e^x cos 2xdx.
anonymous
 5 years ago
integrate by parts e^x cos 2xdx.

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bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0Did you mean \[\int\limits_{}^{}e^{x}*Cos(2x)dx\]

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0So in this case you use integration by parts. Let: \[u = e^{x}\], then \[du = e^{x}*dx\]

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0\[dv = Cos(2x)dx\] \[v = (Sin(2x))/2\]

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0Integration by parts: u*v  Integral(v*du) \[e^{x}*(1/2)*(Sin(2x))  \int\limits_{}^{}(1/2)(Sin(2x))*(e^{x})dx\]

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0Simplify (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
 5 years ago
Best ResponseYou've already chosen the best response.0I will just call it I, solve it and then plug it back into the equation above. Do you get it?

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0\[I =\int\limits_{}^{}Sin(2x)*e^{x}*dx\] use integration by parts again: u = e^(x); du = e^(x)*dx

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0hey, sorry I hate typing out integral problems, especially long ones, can I write it out, take a picture and email it to you?

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0my page just crashed and I lost half the stuff I entered

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i understand how to get to: =e^x cos 2x + 2e^x sin 2x 4 integral e^x cos 2xdx,

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0Oh then its all good. Now move 4 integral e^x cos 2xdx, to the left

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0So you'll get: 5*integral e^x cos 2xdx = e^x cos 2x + 2e^x sin 2x

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0So for the answer: integral e^x cos 2xdx = [ (e^x cos 2x + 2e^x sin 2x) / 5 ] + C

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0do 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
 5 years ago
Best ResponseYou've already chosen the best response.0[ (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.

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0So 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
 5 years ago
Best ResponseYou've already chosen the best response.0awesome thank you very much!!!

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0i did! yeah i never know what to pick for u or dv

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0lol 40 fans... and still Superhero, well maybe at 50 =D!
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