Using sin(A) - sin(B) = 2 cos[(1/2)(A + B)] sin[(1/2)(A - B)],
I(n) - I(n - 2)
= (int from 0 to 1) [sin(nx) - sin((n - 2)x)] / sin(x) dx
= (int from 0 to 1) 2 cos((n - 1)x) sin(x) / sin(x) dx
= (int from 0 to 1) 2 cos((n - 1)x) dx
= [ 2 sin((n - 1)x) / (n - 1) ] (from 0 to 1)
= 2 sin(n - 1) / (n - 1)