Complex Analysis Integration

Problem:

Prove ABS(Integral over alpha(sin(x)/x^2)) <or= 2epi

Alpha(t) = e^(2itpi) 0 <or= t <or= 1 (Unit Disk)

Work thus far:

So, I was looking int he book and it seemed like the best way to prove this is using the fact that the integral must be less then or equal to l(alpha)C where l(alpha) is the length of the curve (2pi) in our case and C is defined to be the upper bound on the ABS(sin(x)/x^2). The trouble is that I'm unsure how to show that ABS(sin(x)/x^2) <or= e for all arguments on the unit disk.

Relevant Equations:
Eulers formula sin(x) = (e^ix - e^-ix)/(2i)

Basically, I'm kind of stuck in how to proceed. I've also tried plugging in alpha/alpha dirivative to solve it in a straight forward manner, but It gets ugly pretty quickly and I'm unsure how to resolve sin(e^2itpi) which appears in the end result. Any help you can provide would be appreciated.

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