Expert: Ahmed Salami - 10/23/2009

Question
Why the average value of the square of the sine function taken over one period is    
<[sin(wt)]^2>= w/(2PI) *(integral from o to 2PI/w )[sin(wt)]^2 dt =1/2

Answer
Hi Deike,
The average value of a function f(x) is the definite integral of the function within the interval divided by the length of the interval i.e
Average value (from a to b) = ∫[f(x)dx] / (b-a)  where the integral is taken from a to b
So, for f(t) = sin²(wt)
Average value = ∫sin²(wt) dt / (T-0)  where the integral is taken from 0 to T
The period T of sin²(wt) is π/w, and so
Average value = ∫sin²(wt) dt / (π/w)
             = (w/π)∫sin²(wt) dt       from 0 to π/w
             = (w/π).(1/2w)[(wt - (1/2)sin(wt)]      from 0 to π/w
             = (1/2π)[(wt - (1/2)sin(wt)]      from 0 to π/w
             = (1/2π)[(w.π/w - (1/2)sin(w.π/w)]  - (1/2π)[(w.0 - (1/2)sin(w.0)]
             = (1/2π)[(π - (1/2)sin(π)]  - (1/2π)[(0 - (1/2)sin(0)]
             = (1/2π)[(π - (1/2).0]  - (1/2π)[(0 - (1/2).0]
             = (1/2π)(π)  - (1/2π)(0)
             = π/2π - 0
             = 1/2

Note that the period is not 2π/w as you suggested, but π/w.

Regards