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A Integral of cos(2cosx)dx

  1. Apr 20, 2017 #1
    Hi everyone, my friend challenged me to solve this definite integral...integral from -2pi to 2pi ((sin(2sinx)+cos(2cosx))dx, i proved by using definite integral properties that this integral equals to integral from -2pi to 2pi cos(2cosx)dx, can you give me any ideas how to solve this?? I know that i can use the contour integration but how??
     
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  3. Apr 20, 2017 #2

    scottdave

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    Note that cosine oscillates between -1 and 1, but you are taking cosine of 2cos, so you are taking cosine of a function which oscillates between -2 and +2. Since cosine is even, cos(x) is the same as cos(-x) so this new function has a period of pi, rather than 2pi. Analytically how to integrate cosine of cosine, I am not sure. You could do it numerically, for sure.
     
  4. Apr 20, 2017 #3

    scottdave

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    Since it still has the shape of a cosine wave, just shifted up and scaled down, you should be able to rewrite it as something like A*cos(2*x) + B, where A and B are constants to be determined (by the max and min), then integrate that over 1 period, then do for how many periods in your range.
     
  5. Apr 20, 2017 #4
    $$I = -4 \pi + 8 \int_{0}^{\pi}cos^2(cos(x))\, dx$$
    and with Diff Under the Integral Sign we can get the ODE
    $$\frac{\partial }{\partial a}\left ( \int_{0}^{\pi}cos^2(cos(ax))\, dx \right )=\frac{1-\pi\, sin^2(cos(a\pi))}{a}\, -\, \frac{1}{a} \int_{0}^{\pi}cos^2(cos(ax))\, dx$$
    then you can solve for
    $$\int_{0}^{\pi}cos^2(cos(ax))\, dx$$
    and set a=1
     
  6. Apr 24, 2017 #5
    Ammmmm, I understand what you did, but please if you have the final answer give it to me....and thanks
     
  7. Apr 24, 2017 #6
    wolfram alpha is your friend
     
  8. Apr 24, 2017 #7

    DrClaude

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  9. Apr 24, 2017 #8
    No the answer is right, 8 integral from(0>pi) (cos(cosx))^2-4pi=wolframalpha answer.
    Ammmm the answer of the previous integral is given in terms of bessel function....interesting!
    So there's a relationship between them....
     
  10. Apr 24, 2017 #9

    scottdave

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    I am not sure exactly where the cos²(cos(ax)) came from either. Cos(2u) is equal to cos²(u) - sin²(u) or 2cos²(u) - 1
     
  11. Apr 24, 2017 #10

    Dr Transport

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    [itex] \cos(\cos(x))[/itex] is the real part of [itex] e^{ikr}[/itex] which is the definition of the Bessel function.....
     
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