Integral of sin^2(πx / 2)?

Answer 1

I think this gets easier if first we integrate sin^2(x). This is a very classical result, we use the identity sin^2(x) = (1 - cos(2x))/2. So,

Integral sin^2(x) dx = [x - 1/2 sin(2x)]/2 + C.

Now, to integrate sin^2(πx / 2), just put πx / 2 = t, so that x = 2t/π and dx = 2/π. So, we get the integral

Integral sin^2(t) 2/π dt = [t - 1/2 sin(2t)]/π + C. Going back to x, we get

Int sin^2(πx / 2) dx = [πx / 2 - 1/2 sin(πx)]/π + C.

Answer 2

Do the integral using the cosine half angle function.
cos(z/2) = sqrt((cos(z) + 1)/2)

Also use:
sin(z/2)^2 = 1 - cos(z/2)^2 = 1 - (cos(z) + 1)/2 = 1/2 - cos(z)/2

In your case z = πx and dz = π dx so dx = dz/π

Substituting: sin^2(πx / 2) dx = 1/(2π) (1 - cos(z)) dz

Integrating and back substituting:
(z - sin(z))/(2π) = x/2 - sin(πx)/(2π)