calculus
posted by Parker .
Integrate sqrt(x^2 + 1) dx over [0,2*pi]
I can substitute u=arctan x to get:
Integrate (sec u)^3 du over [0,arctan(2*pi)]
From there, I'm stuck.
(thanks Count Iblis for your last help)

nevermind, I got it. The trick was using trigonometric reduction formulas.