# Antiderivative of Secant

1. Sep 19, 2006

### Frillth

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. Sep 19, 2006

### Hurkyl

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