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Antiderivative of Secant

  1. Sep 19, 2006 #1
    For one of my homework assignments, I had to find the integral of a function. I got my function simplified to sec(t)^(8/3). I tried to use the reduction formula for sec(t)^n, but I believe that it only works if the power of sec is an integer. Could somebody help me out, please?

    Edit: I figured that it might be a good idea if I showed how I got to sec(t)^(8/3)

    My initial problem was the following: integral of cube root(1+x^2) dx.

    First of all, I made the substitution x=tan(t) and dx = sec(t)^2 dt. This gave me:

    integral of cube root(1+tan(t)^2) * sec(t)^2 dt.

    I changed 1+tan(t)^2 to sec^2 to get the following:

    integral of sec(t)^(2/3) * sec(t)^2 dt, or sec(t)^8/3.

    Did I take this problem in the wrong direction, or am I on the right track?
    Last edited: Sep 19, 2006
  2. jcsd
  3. Sep 19, 2006 #2


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    It probably doesn't have an elementary antiderivative. Mathematica expresses it in terms of a hypergeometric function.
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