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Integral of cos(e^x)

  1. Oct 4, 2007 #1
    [tex]\int\cos{e^x}dx[/tex]

    let

    [tex]u=\cos{e^x}[/tex]
    [tex]du=-e^{x}\sin{e^x}dx[/tex]

    [tex]dv=dx[/tex]
    [tex]v=x[/tex]

    [tex]\int\cos{e^x}dx=x\cos{e^x}+\int xe^{x}\sin{e^x}dx[/tex]

    advice on any other approach?
     
    Last edited: Oct 4, 2007
  2. jcsd
  3. Oct 4, 2007 #2

    Hurkyl

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    Well, that's generally what happens when you do the same integration by parts first forwards and then backwards: you wind up right where you started.
     
  4. Oct 5, 2007 #3

    Avodyne

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    This does not have a closed-form solution in terms of simple functions. Are you sure you copied it correctly?
     
  5. Oct 5, 2007 #4
    yep, i'm in Calc 2, so i prob wouldn't be able to solve it even with a little guidance?
     
  6. Oct 5, 2007 #5

    Avodyne

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    Well, there's not much to do. Make the substitution x=ln(u), dx=du/u, and you get cos(u)/u. The integral of this is the Cosine Integral function:
    http://mathworld.wolfram.com/CosineIntegral.html
     
  7. Oct 5, 2007 #6
    What about expanding the cosine out and then integrating the infinite series. At least that gives you a result, however ugly.
     
  8. Oct 5, 2007 #7
    i was thinking of using this approach but i wasn't sure if i could do that

    i also tried that website after class and i was like huh :D but thanks.
     
  9. Oct 5, 2007 #8
    i'm not on that section yet so i'll have to wait.
     
  10. Oct 5, 2007 #9
    ar u sure it's not [tex]cos(x)*e^x[/tex] ?? for Calc 2 cos(e^x) seems kindof hard given that u havent done series yet...
     
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