Find the average value of f(x)=cos²x over the interval [0,10π]

Thank you very much and please show your steps.
I really appreciate any help!

Average value is:
[1 / (b-a)] ∫[x=a to b] f(x) dx

In your case, a = 0, b = 10π, and f(x) = cos²(x)

Then plug in it:
[1 / (10π - 0)] / ∫[x=0 to 10π] cos²(x) dx
= (1/10π) ∫[x=0 to10π] (1/2)(1 + cos(2x)) dx .... since cos²(x) = (1/2)(1 + cos(2x))
= (1/20π) ∫[x=0 to 10π] (1 + cos(2x)) dx ...... since (1/10π) •(1/2) = (1/20π)
= (1/20π)[x + (1/2)sin(2x)] from x=0 to 10π ...... integrating
= (1/20π)[10π + (1/2)sin(20π) - 0 - 0] ............. plugging in values from 0 to 10π
= (10π/20π)
= 1/2