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Tricky trigonometric integral

  1. Mar 21, 2010 #1
    Hi. I'm having troubles calculating this indefinite integral. I have tried a few things, but none seem to work.

    gif.latex?\int%20\frac{1}{sin%28x%29%282cos^2{x}-1%29}dx.gif

    Actually, universal substitution does make it possible to integrate, but there has to be some shorter, more elegant way. Anyone?
     
  2. jcsd
  3. Mar 21, 2010 #2
    Replacing the 1 in the numerator by sin^2(x) + cos^2(x) does seem to work. You then do get the original integral back, but it is multiplied by 1/2 so you can move it back to the left hand side to solve for it.
     
  4. Mar 21, 2010 #3

    I'm sorry, but I'm not sure that I understand what are you trying to say. Could you go a little bit more in-depth?
     
  5. Mar 21, 2010 #4
    Abbreviation: Sin(x) = S, Cos(x) = C

    [S^2 + C^2]/[S (2 C^2 - 1)] =

    S/[2 C^2 - 1] (easy to integrate as the derivative of C is -S and S is in the numerator)

    +

    C^2/[S(2C^2 - 1)]

    We can rewrite the numerator of the last term as:

    C^2 = 1/2 2 C^2 = 1/2 (2 C^2 - 1 + 1)

    This means that you can write the last term as:

    1/(2S) + 1/2 * Original term you wanted to integrate.

    Then you're done if you can integrate 1/S and that you can do using more or less the same trick:

    1/Sin(x) = 1/(2Sin(1/2 x) Cos(1/2 x)) and then replace the numerator by Cos^2(1/2 x) + Sin^2(1/2 x) and you're done.
     
  6. Mar 21, 2010 #5
    It works. Very nice. Thank you very much.
     
    Last edited: Mar 21, 2010
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