Numerical integration vs. integrals that are zero

[Albert Chan]

Hi, ijabbott

Transformation is actually a bit more involved, not just x, but dx too

x = (a+b)/2 + (b-a)/2 * (3u/2 - u³/2)

dx = 3/4 * (b-a) * (1 - u²) du

With a=0, b=2*Pi, after simplication, we got:

integrate(sin(x), x = 0 to 2*Pi) = integrate(3/2 pi (1-u^2) * sin(pi/2 u(u^2-3)), u = -1 to 1) = 0

integrate(sin(x), x = 0° to 360°) = 180/Pi * 0 = 0