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Math Help - help rewriting an integral

  1. #1
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    help rewriting an integral

    hey...so my problem is to rewrite the [integral from 0 to pi/2] of SQRT(1 + (asinx)^2)dx in terms of the elliptic integral E(k, phi) = [integral from 0 to phi] SQRT(1-(ksinu)^2)du.

    please note i am aware elliptic integrals are unsolvable in closed form...i am not looking for a solution, I am just attempting to rewrite the integral to make it look like the form of the elliptic integral E

    using the substitution tanu = asinx, I have worked it out to be: [integral from 0 to arctan(asinx)] du/((cosu)^2*SQRT(a^2 - (a^2 + 1)(sinu)^2))

    but I can't figure out where to go from here. Please help!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by minivan15 View Post
    hey...so my problem is to rewrite the [integral from 0 to pi/2] of SQRT(1 + (asinx)^2)dx in terms of the elliptic integral E(k, phi) = [integral from 0 to phi] SQRT(1-(ksinu)^2)du.
    Assuming no typos:

    E(a{\bf{i}},\pi/2)=\int_0^{\pi/2}\sqrt{1 + (a\sin(x))^2}dx

    an alterative approach if you don't like the imaginary argument is to replace \sin(x) by \cos(x-\pi/2) and use the identity \cos^2(x-\pi/2)=1-\sin^2(x-\pi/2)

    RonL
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  3. #3
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    thanks, but unfortunately I need to solve it a specific way, because I'll use the results in later problems. I've worked it down to:

    subst. tanu = asinx

    so x = arcsin((tanu)/a)

    and dx = 1/ ROOT(1-((tanu)/a)^2) * 1/a(cosu)^2 * du

    the integral becomes: [integral from 0 to arctan(a)] du / ((cosu)^2 * ROOT((acosu)^2 - (sinu)^2))

    (after a few simple steps of manipulation)

    I need to finish it with this method to solve the problem, but am stuck here. Can someone please help? I have been stuck for 3 days and it's driving me crazy!!
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