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Math Help - Integral Help

  1. #1
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    Integral Help

    I have the following integral i cant find the solution.
    I have tried differents ways to solve but nothing.



    a is constant and it will take values on the interval [0,1]

    Thanks in advance.
    Attached Thumbnails Attached Thumbnails Integral Help-integral.bmp  
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi

    \sqrt{1+a^2\sin^2x} looks like \sqrt{1+\sinh^2X}=\cosh X. Hence, you may try substituting a\cdot\sin x = \sinh u \Leftrightarrow u=\mathrm{arcsinh}(a\cdot\sin x)\Leftrightarrow x=\arcsin\left(\frac{\sinh u}{a}\right), it should give something easier to evaluate.
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  3. #3
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    Quote Originally Posted by marta_fdez View Post
    I have the following integral i cant find the solution.
    I have tried differents ways to solve but nothing.



    a is constant and it will take values on the interval [0,1]

    Thanks in advance.
    The answer is given below, where E[x | m] is an elliptic integral of the second kind.
    Attached Thumbnails Attached Thumbnails Integral Help-3.gif  
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    The answer is given below, where E[x | m] is an elliptic integral of the second kind.
    I'm searching information about elliptic integrals of second kind. That would make sense as the integral arose when I was trying to calculate the arc length of a curve generated when a point is moving along the surface of a sphere, with determined polar and azimuthal angular speed (both constants).

    flyingsquirrel I also tried that variable change at the beggining (i thought the same thing), but i didn't reach anything.

    Thanks for your answers.
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  5. #5
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by marta_fdez View Post
    I
    flyingsquirrel I also tried that variable change at the beggining (i thought the same thing), but i didn't reach anything.
    \sqrt{1+a^2\sin^2x}

    We subsitute u=\mathrm{arcsinh}(a\cdot\sin x)\Leftrightarrow x=\arcsin\left(\frac{\sinh u}{a}\right)<br />
\Rightarrow \mathrm{d}x=\frac{1}{a}\cdot\frac{\cosh u}{\sqrt{1-\frac{\sinh^2 u}{a^2}}}

    We get \int \sqrt{1+a^2\sin^2x}\,\mathrm{d}x=\frac{1}{a}\int <br />
\sqrt{1+\sinh^2 u}\frac{\cosh u}{\sqrt{1-\frac{\sinh^2 u}{a^2}}}\,\mathrm{d}u=\frac{1}{a}\int <br />
\frac{\cosh^2 u}{\sqrt{1-\frac{\sinh^2 u}{a^2}}}\,\mathrm{d}u

    At this point, I realize that, when I did it earlier, I forgot a \cosh u... that's why it was easy to integrate. Sorry
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