how do you evaluate the integral of sqrt(2 - 2sin(theta)) from 0 to 2pi?

Answer 1

Let's use the trigonometric identity 1 - sin(t) = (sin(t/2) - cos(t/2))^2. We get

Int(0 to 2pi) sqrt(2) (sin(t/2) - cos(t/2) dt = sqrt(2) [-2 cos(t/2) + 2 sin(t/2)] (0 to 2pi)] = 2 sqrt(2) [1 + 0 -(-1 + 0)] =4 sqrt(2)