I've carried out the steps for the time average for $\cos^2x$ for limits $0$ to $T$.
I've gotten : $\frac{1}{T}\left[\frac{1}{2}[T+\frac{1}{4}\sin2T\right]$
I'm trying to find the average over a long period of time. Would it be correct to set $T =0$ ?
The expression for the time average from $0$ to $T$ is $$\frac{1}{T}\left[\frac{1}{2}T+\frac{1}{4}\sin 2T\right].$$
Express this time average as $$\frac{1}{2}+\frac{1}{4T}\sin 2T.$$
The time $T$ becomes very large, it does not approach $0$. As time becomes large, the term $\frac{1}{4T}(\frac{1}{4}\sin 2T$ approaches $0$, since $\sin 2T$ remains between $-1$ and $1$. So the long term time average is very close to $\frac{1}{2}$. In terms of limits, we have $$\lim_{T\to\infty}\frac{1}{T}\left[\frac{1}{2}T+\frac{1}{4}\sin 2T\right]=\frac{1}{2}.$$
Remarks: $1.$ There may be no good reason to assume that the process starts at time $0$. The time average from time $t=T_0$ to time $T$ is $$\frac{1}{T-T_0}\left[\frac{1}{2}(T-T_0)+\frac{1}{4}(\sin^2 2T -\sin^2 2T_0)\right].$$ This expression also approaches $\frac{1}{2}$ as $T$ becomes very large.
$2.$ In your post, you gave an expression for the time average over the interval from $0$ to $T$, and we obtained the long-term time average from that. However, we can do it more simply. If $T$ is very large, the time average is the same as the time average of $\sin^2 x$ over one full period of the sine function. If we look at the graph of $\cos x$, we see that this is the same as the time average of $\cos^2 x$ over a full period. But $\sin^2 x+\cos^2 x=1$ for all $x$, so the time average of $\sin^2 x+\cos^2 x$ over a full period is $1$. It follows that each of $\sin^2 x$ and $\cos^2 x$ has time average $\frac{1}{2}$ over a full period.
