Mark M. answered 03/24/23
Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.
Arc length = ∫(from a to b) √ (1 + (dy/dx)2) dx
So, we have ∫(from 0 to π/6) √ (1 + tan2x) dx = ∫(from 0 to π/6) secx dx =
ln(secx + tanx)from 0 to π/6) = ln(2/√3 + 1/√3) = ln√3 = (1/2)ln3