Average of $\sin^2 \omega t$, $\cos^2 \omega t$ for long times I don't understand how the averages mentioned in excerpt below work out. The question is regarding calculating the power spectrum of an oscillating classical dipole.
IS averaging over long periods, for example, For $\sin^2 \omega_o t$ given by 
$$<\sin^2 \omega_o t> = \frac{\int t \sin^2 \omega_o t}{\int \sin^2 \omega_o t} =?? \frac{1}{2}$$
Also, how is the time important in doing this average. What I mean is, how is it justified to replace the function with its average value for times longer than its time period, how does this make sense physically?
Thanks

 A: One way to think about this is in the frequency domain in terms of filtering.
The measuring apparatus has in effect a low-pass filter at the input with a corner frequency far below $\omega_0$.  No physical measuring apparatus has infinite bandwidth and thus, there is a high-frequency "roll-off" in the frequency response of the instrument..
The square of a sinusoid contains a DC component and a component at twice $\omega_0$:
$\cos^2(\omega_0 t) = \frac{1}{2}[1 + \cos (2\omega_0 t)]$
but the only component that is measured is the DC component since the high-frequency component is "filtered" out.
A: Just to add a comment on the physical reason for performing this average. The justification is that our measuring instruments & procedures only have a finite temporal resolution. This is because a measurement takes some time: it is not instantaneous. Likewise, the spatial resolution is finite because we cannot make infinitely small probes: they occupy some space. This means that our probes only measure average quantities over distances and times that are large compared to the resolution of the probe. In this example, we are dealing with optical frequency oscillations, which change over a timescale of $10^{-15}$ s. The electronics in a typical experimenter's lab react on timescales of order $10^{-9}$ s, 6 orders of magnitude longer. So clearly averaging is completely justified in this scenario.
A: No, the integral you wrote is not what we mean by the average over a long time. 
Here's a plot of $cos^2 t$

The average is just the average height on that plot. 1/2 looks quite reasonable. You can define it by
$$\lim_{T\to\infty}\frac{1}{T}\int_0^T\cos^2(\omega t)dt$$
The reason it comes to 1/2 is that $cos^2 \omega t + \sin^2 \omega t = 1$. Since they have the same shape, they have the same average value, so they are both 1/2 on average.
It is hard to say why you are justified in taking the average value, since you did not include any context. We don't know what $W$ is or really any of what's going on. 
