Find the area of the surface obtained by rotating the curve about the x-axis:
y=cos 2x, 0<=x<=pi/6
Surface area about the x-axis = Integral of 2pi * f(x) * sqrt(1+[f'(x)]^2) dx
The Attempt at a Solution
I think I set up my integral correctly, so my problem is figuring out how to solve the integral. I have:
2pi * Integral of cos 2x * sqrt(1+4sin^2 2x) dx on [0, pi/6]
I can't figure out how to solve that. And can someone teach me how to use Latex for integrals so it looks better.