The integral of (sin x)^2024 is done using a reduction formula a million times until you find a pattern.
The sin(x)^2024 term is only going to be significant when the cos(x)^2012
term is insignificant, and vice-versa. Therefore I would say you could just
integrate them separately and add; and obviously the definite integrals of the two terms will be the same over periods of 2*Pi although Pi/2 out of phase.
The definite integral over 2*Pi of sin(x)^y over x=0..2*Pi seems to settle
down to a limit L = approx 0.11175 as y increases. (According to Maple;
unexpected, I would have guessed it would go to 0).
Therefore a good numerical approximation to your integral from x= 0 to any
real number s would seem to be 2*(s/2*Pi)*0.11175
or (0.11175)*s/Pi.
If the rigor police come around, you haven't seen me.