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Math Help - Problem with a complicated Integral

  1. #1
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    Problem with a complicated Integral

    Hello,

    I want to integrate \int_{0}^{2\pi} \cos(t+a)\  \cos(b \cos(t)) \ e^{-c |\cos(t)|}\ dt where a,b,c are constants. I tried in Matlab and Mathematica but both are unable to solve the integral. Is there a trick to simplify this integral to make it integrable? Any help is highly appreciated.

    Thanks and Regards,
    Ameya
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  2. #2
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    Quote Originally Posted by amrasa81 View Post
    Hello,

    I want to integrate \int_{0}^{2\pi} \cos(t+a)\  \cos(b \cos(t)) \ e^{-c |\cos(t)|}\ dt where a,b,c are constants. I tried in Matlab and Mathematica but both are unable to solve the integral. Is there a trick to simplify this integral to make it integrable? Any help is highly appreciated.

    Thanks and Regards,
    Ameya
    cos(t+ a)= cos(t)cos(a)- sin(t)sin(a) so you can write this as
    cos(a)\int_0^{2\pi} cos(bcos(t))e^{-c|cos(t)|}cos(t)dt- sin(a)\int_0^{2\pi} cos(bcos(t))e^{-c|cos(t)|}sin(t)dt

    You can simplify the second integral, with the substitution u= cos(t), to sin(a)\int_0^0 cos(bu)e^{-c|u|}du= 0 simply because the limits of integration are the same. The first integral is a little more complicated but \int_0^0 cos(u)e^{-|u|}\frac{u}{\sqrt{1- u^2}}du is still 0.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    cos(t+ a)= cos(t)cos(a)- sin(t)sin(a) so you can write this as
    cos(a)\int_0^{2\pi} cos(bcos(t))e^{-c|cos(t)|}cos(t)dt- sin(a)\int_0^{2\pi} cos(bcos(t))e^{-c|cos(t)|}sin(t)dt

    You can simplify the second integral, with the substitution u= cos(t), to sin(a)\int_0^0 cos(bu)e^{-c|u|}du= 0 simply because the limits of integration are the same. The first integral is a little more complicated but \int_0^0 cos(u)e^{-|u|}\frac{u}{\sqrt{1- u^2}}du is still 0.
    Hello HallsofIvy,

    Thank you for your reply. So what essentially you are saying is that the value of my original integral is zero?

    Regards,
    Ameya
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    cos(t+ a)= cos(t)cos(a)- sin(t)sin(a) so you can write this as
    cos(a)\int_0^{2\pi} cos(bcos(t))e^{-c|cos(t)|}cos(t)dt- sin(a)\int_0^{2\pi} cos(bcos(t))e^{-c|cos(t)|}sin(t)dt

    You can simplify the second integral, with the substitution u= cos(t), to sin(a)\int_0^0 cos(bu)e^{-c|u|}du= 0 simply because the limits of integration are the same. The first integral is a little more complicated but \int_0^0 cos(u)e^{-|u|}\frac{u}{\sqrt{1- u^2}}du is still 0.
    Ok, I see where the problem is with the substitution. I think your substitution is invalid because within the interval (0,2\pi), x = \cos t does not satisfy the bijection, and hence, I have to break this integral up. Anyway, thanks for your effort again.
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