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Math Help - Easy integral?

  1. #1
    MHF Contributor Drexel28's Avatar
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    Easy integral?

    Compute \int_0^{2\pi}e^{\cos(x)}\cos\left(x+\sin(x)\right)  dx
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  2. #2
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    Quote Originally Posted by Drexel28 View Post
    Compute \int_0^{2\pi}e^{\cos(x)}\cos\left(x+\sin(x)\right)  dx

    Is the integrand the derivative of

     e^{\cos(x)} \sin( \sin(x) ) ?
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  3. #3
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    Quote Originally Posted by simplependulum View Post
    Is the integrand the derivative of

     e^{\cos(x)} \sin( \sin(x) ) ?

     (e^{\cos(x)} \sin( \sin(x) ))'=-\sin x \,e^{\cos x}\sin(\sin x)+e^{\cos x}\cos(\sin x) \cos x =e^{\cos x}\left(\cos(\sin x)\cos x-\sin x\,\sin(\sin x)\right)=... e^{\cos x}\cos(x+\sin x) !! Oh, dear hollie mollie: yes, it is!

    A question: how did you come up with it? Some program or some insight?

    Tonio
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    I write e^{\cos(x)} \cos( x + \sin(x) )

     = Re[ e^{\cos(x)}e^{ix} e^{i\sin(x)} ]

     = Re[ e^{e^{ix}} e^{ix} ]

    then substitute  u = e^{ix}

    At first i did it by integration by parts but the steps are quite long so i was not sure at that time ... Now , i am sure it is true with your help .
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tonio View Post
     (e^{\cos(x)} \sin( \sin(x) ))'=-\sin x \,e^{\cos x}\sin(\sin x)+e^{\cos x}\cos(\sin x) \cos x =e^{\cos x}\left(\cos(\sin x)\cos x-\sin x\,\sin(\sin x)\right)=... e^{\cos x}\cos(x+\sin x) !! Oh, dear hollie mollie: yes, it is!

    A question: how did you come up with it? Some program or some insight?

    Tonio
    Just tinkering

    Quote Originally Posted by simplependulum View Post
    I write e^{\cos(x)} \cos( x + \sin(x) )

     = Re[ e^{\cos(x)}e^{ix} e^{i\sin(x)} ]

     = Re[ e^{e^{ix}} e^{ix} ]

    then substitute  u = e^{ix}

    At first i did it by integration by parts but the steps are quite long so i was not sure at that time ... Now , i am sure it is true with your help .
    There is another method that involves complex analysis, and it is the way I was hoping someone would do this.


    We know that \oint_{|z|=1}e^zdz=0 and thus \int_0^{2\pi}e^{e^{it}}\left(e^{it}\right)'dt=\int  _0^{2\pi}e^{\cos(t)}\left(\cos(\sin(t))+i\sin(\sin  (t))\right)\left(-\sin(t)+i\cos(t)\right)=0, in particular the real an imaginary parts of that integral must equal zero. Do a little work with trig identities and you will get our integral is the imaginary part of it.
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  6. #6
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    let  I(a) = \int^{2 \pi}_{0} e^{a \cos x} \cos (x + a \sin x) \ dx = Re \Big(\int^{2 \pi}_{0} e^{ae^{ix}}e^{ix} \ dx \Big)

    then  I'(a) = Re \Big(\int^{2 \pi}_{0} e^{2ix}e^{ae^{ix}} \ dx\Big)

    let  u = ae^{ix}

    EDIT: then  I'(a) = Re \Big( -\frac{i}{a^{2}} \int^{a}_{a} ue^{u} \ du \Big) = Re(0) = 0

    so  I(a) = C

    but  I(0) = 0

    which means  C=0

    so  I(a) = 0

    and  \int^{2 \pi}_{0} e^{\cos x} \cos  (x + \sin x) \ dx = I(1) = 0
    Last edited by Random Variable; May 12th 2010 at 06:16 PM. Reason: correcting mistake
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