Just so clarify I figured the problem out.

ABS(Sin(x)/x^2) = ABS(sin(x)) + ABS(1/x^2) and since we are restricted to the unit disk, ABS(1/x^2) = 1, thus ABS(sin(x)) is all we need to calculate.

ABS(sin(x)) = *By Eulers Formula* ABS((e^ix - e^-ix)/(2i))

which is

<or= ABS(e^ix/2i) + ABS(e^-ix/2i) = 1/2 + 1/2 = 1

Thus: Integral over Alpha of (sin(x)/x^2)dx <or= l(alpha)C = 2pi <or= 2epi