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Integrate cos(x^5)

  1. Oct 19, 2007 #1
    Can you solve this [itex] \int \cos x^5 dx[/itex] ?
  2. jcsd
  3. Oct 19, 2007 #2


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    No, I can't
  4. Oct 19, 2007 #3
    nope ...
  5. Oct 19, 2007 #4


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    It probably can't be done in terms of elementary functions. Mathematica has an answer of 2 different forms:

    Take E(v,z) here to be this function http://functions.wolfram.com/GammaBetaErf/ExpIntegralE/ we have:

    [tex] \int \cos x^5 dx = - \frac{1}{10} \left( E\left( \frac{4}{5} , -i x^5 \right) + E\left( \frac{4}{5} , i x^5 \right) \right)[/tex]

    Or in terms of the gamma function http://functions.wolfram.com/GammaBetaErf/Gamma/ :

    [tex]\int \cos x^5 dx = - \frac{\left( x^{10} \right)^{\frac{4}{5}}}{10x^9}\left( \left(i x^5 \right)^{\frac{1}{5}} \Gamma \left( \frac{1}{5} , -i x^5 \right) + \left(-i x^5 \right)^{\frac{1}{5}} \Gamma \left( \frac{1}{5} , i x^5 \right) \right)[/tex]

    I don't really understand why mathematica didn't simplify it further, so I've tried to keep it to what mathematica outputted.
    Last edited: Oct 19, 2007
  6. Oct 19, 2007 #5
    Thank you!
  7. Oct 19, 2007 #6
    thats why series solutions are so conviennt
  8. Oct 20, 2007 #7

    Gib Z

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    The Taylor series for cos converges for all real x, so you should have just let x^5 = u in the taylor expansion of cos u, integrated term by term and you are left with an even nicer result than what mathematicia gives out.
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