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Annoyingly difficult integral

  1. Feb 21, 2007 #1
    The problem statement, all variables and given/known data

    integrate dx/(1-x^2)^(3/2)

    The attempt at a solution

    I first noticed that the derivative of inverse sin equals (1 x^2)^(-1/2) and that my equation is that derivative cubed. However, I had no idea what to do with the cube, so I dicarded that thought.

    Next I tried to do a u substitution. obviously I can't make u equal 1-x^2 because that leaves me with an extra x in my du that won't go away. I was inspired by remembering some powers of trig integrations to try to make du equal the derivative of inverse sin, but that also failed as I had no inverse sin in the original equation to replace with u and last I checked, I can't integrate du^3.

    I finally checked my tables of integrals and found where the integral of du/(a^2-u^2)^(3/2) equals u/a^2(a^2-u^2). This seems to give me my answer, but I can't figure out how to derive that formula.
  2. jcsd
  3. Feb 21, 2007 #2


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    Homework Helper

    From [tex]I=\int\frac{dx}{(1-x^2)^{\frac{3}{2}}}[/tex] apply the substitution [tex]x=\sin\theta[/tex]. Hint: you should get [tex]I=\int\sec ^2 \theta\, d\theta[/tex].
  4. Feb 21, 2007 #3
    You probably substituted wrongly to get du^3
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