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anonymous

  • 5 years ago

integrate by parts e^-x cos 2xdx.

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  1. bahrom7893
    • 5 years ago
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    Did you mean \[\int\limits_{}^{}e^{-x}*Cos(2x)dx\]

  2. anonymous
    • 5 years ago
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    yes

  3. bahrom7893
    • 5 years ago
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    oaky working on it

  4. anonymous
    • 5 years ago
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    thank you!!

  5. bahrom7893
    • 5 years ago
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    So in this case you use integration by parts. Let: \[u = e^{-x}\], then \[du = -e^{-x}*dx\]

  6. bahrom7893
    • 5 years ago
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    \[dv = Cos(2x)dx\] \[v = (Sin(2x))/2\]

  7. bahrom7893
    • 5 years ago
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    Integration by parts: u*v - Integral(v*du) \[e^{-x}*(1/2)*(Sin(2x)) - \int\limits_{}^{}(1/2)(Sin(2x))*(-e^{-x})dx\]

  8. bahrom7893
    • 5 years ago
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    Simplify (take 1/2 out): \[(1/2) (e^{-x}Sin(2x) + \int\limits_{}^{}Sin(2x)e^{-x}dx)\] Now for now ignore everything except: \[I =\int\limits_{}^{}Sin(2x)*e^{-x}*dx\]

  9. bahrom7893
    • 5 years ago
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    I will just call it I, solve it and then plug it back into the equation above. Do you get it?

  10. bahrom7893
    • 5 years ago
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    \[I =\int\limits_{}^{}Sin(2x)*e^{-x}*dx\] use integration by parts again: u = e^(-x); du = -e^(-x)*dx

  11. bahrom7893
    • 5 years ago
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    hey, sorry I hate typing out integral problems, especially long ones, can I write it out, take a picture and email it to you?

  12. bahrom7893
    • 5 years ago
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    my page just crashed and I lost half the stuff I entered

  13. anonymous
    • 5 years ago
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    yeah no problem

  14. bahrom7893
    • 5 years ago
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    okay workin on it

  15. anonymous
    • 5 years ago
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    i understand how to get to: =e^-x cos 2x + 2e^-x sin 2x -4 integral e^-x cos 2xdx,

  16. bahrom7893
    • 5 years ago
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    Oh then its all good. Now move -4 integral e^-x cos 2xdx, to the left

  17. bahrom7893
    • 5 years ago
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    So you'll get: 5*integral e^-x cos 2xdx = e^-x cos 2x + 2e^-x sin 2x

  18. bahrom7893
    • 5 years ago
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    So for the answer: integral e^-x cos 2xdx = [ (e^-x cos 2x + 2e^-x sin 2x) / 5 ] + C

  19. bahrom7893
    • 5 years ago
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    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

  20. bahrom7893
    • 5 years ago
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    [ (-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.

  21. anonymous
    • 5 years ago
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    i think i get it! :)

  22. bahrom7893
    • 5 years ago
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    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.

  23. anonymous
    • 5 years ago
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    awesome thank you very much!!!

  24. bahrom7893
    • 5 years ago
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    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.

  25. anonymous
    • 5 years ago
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    i did! yeah i never know what to pick for u or dv

  26. bahrom7893
    • 5 years ago
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    lol 40 fans... and still Superhero, well maybe at 50 =D!

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