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Integral of (cos(2x))^0.5

  1. Mar 20, 2013 #1
    I tried to do a arc length integral s, for y as an elliptic function of x. But as I continued with the integration, I found myself at the above integral(cosine of 2x). I quickly substituted cos(2x) with A and carried on but got stuck after about a step or two. The new problem now became (A^0.5)/(1-A^2)^0.5
    I tried integration by parts, a lot of substitutions, and nothing worked. Then I thought I should give this to wolframalpha. It gave the results but in, what I only recently found out, elliptic functions. So, does this mean there are no closed solutions/expressions to integrals like these? And does that mean there is no exact formula for applied mathematical problems involving these, like here the perimeter of an ellipse?
     
  2. jcsd
  3. Mar 20, 2013 #2
    No. For these type of problems, and in reality in physics especially, closed-form solutions are extremely rare!

    Indeed, the integral as you posted it only has the elliptical function as a solution, which is a "special function" solution.
    There are others for other such problems for example, such as the hypergeometric function, Laguerre polynomials, etc.

    Although, there is no closed-form solution, you can perform numerical integrations. Since Wolfram Alpha is based on Mathematica, research the function NIntegrate to see how this works.

    For example, to evaluate your integral from 0 to 1, you can try:
    NIntegrate[Sqrt[Cos[2x]],{x,0,1}],
    and experiment with that.
     
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