Integrating Square Roots

Question

Evaluate the integral

_{0 ∫} ^2pi √((1-cosx)/2) dx

I have used (1-cosx)/2 = sin²(x/2) and canceled out the square root and the square on the sin function. I am to the point of ₀ ∫ ^2pi abs(sin (x/2))dx and don't really know where to take it from there.

Answer 1

∫₀²ᵖˡ √((1 - cosx) / 2) dx

# From Trig, we have: sin(x/2) = √((1 - cosx) / 2)

= ∫₀²ᵖˡ sin(x/2) dx = 2 ∫₀²ᵖˡ sin(x/2) (1/2)dx # i.e. ∫sin(u)du

= -2cos(x/2) | ₀²ᵖˡ = -2(cos(π) - cos(0)) = (-2)(-1 - 1) = 4

Answer 2

y = l sin(x/2) l has period (2π)/(1/2) = 4π.

Between 0 and 2π, the graph lies above the x-axis and lies below the x-axis from 2π to 4π.

So, ∫_{(0 to 2pi)} l sin(x/2) l dx = ∫_{(0 to 2pi)} sin(x/2)dx Evaluate by substitution with u = x/2.