Integral of Wiener Process and Central Limit Theorem I am trying to solve the following exercise:

(1) Given $W$ is a Wiener process, find a constant $M$ such that
  $\lim\limits_{t\to\infty} \frac{1}{t}\int_{0}^{t}\sin^2W_s ds=M$
(2) Then show that
  $\frac{1}{\sqrt{t}}\int_{0}^{t}(\sin^2W_s-M) ds$ converges to $N(0,\sigma^2)$ and compute $\sigma^2$.

So far I have only started the first part of the question before getting stumped.  I started the following:
I simply integrated by parts with respect to s and evaluated at the bounds 0 and t
$$\lim_{t\to\infty} \frac{1}{t}\int_{0}^{t}\sin^2W_s ds= \lim_{t\to\infty} \frac{1}{t}\left(\frac{1}{2}\Big((W_t-\sin(W_t)\cos(W_t)-(W_0-\sin(W_0)\cos(W_0)\Big)-\int_{0}^{t}sd(\sin^2W_s)\right)$$
Given $W_0=0$ We have the following:
$$\lim_{t\to\infty} \frac{1}{t}\left(\frac{1}{2}(W_t-\sin(W_t)\cos(W_t))-\int_{0}^{t}sd(\sin^2W_s)\right)$$
I am assuming I have to apply something like Ito's formula to the stochastic integral term on the RHS. However, I am stumped how to show this limit equals a constant M, and how to begin the second part of the problem statement.
 A: Firstly, by the strong law of large number, the limit should exist. 
Consider $$\theta_k = \inf\{t\geq \theta_{k-1}, |W_t- W_{\theta_{k-1}}| = \pi\}, \theta_0 = 0$$
Then the integral can be written as the sum of $\int_{\theta_{k}}^{\theta_{k+1}}\sin^2(W_s)ds$, we have 
\begin{align}
\frac{1}{t}\int_0^t \sin^2(W_s)ds = \dfrac{\sum_{k=1}^{+\infty}1_{\theta_k < t}}{t} \dfrac{\sum_{k=0}^{+\infty}\int_{\theta_{k}}^{\theta_{k+1}}\sin^2(W_s)ds1_{\theta_{k+1} < t} + \int_{\theta_{k+1}}^t \sin^2(W_s)ds1_{\theta_{k+1} < t \leq  \theta_{k+2}} }{\sum_{k=1}^{+\infty}1_{\theta_k < t}}
\end{align}
Remark that $$\dfrac{\sum_{k=0}^{+\infty}\int_{\theta_{k}}^{\theta_{k+1}}\sin^2(W_s)ds1_{\theta_{k+1} < t} + \int_{\theta_{k+1}}^t \sin^2(W_s)ds1_{\theta_{k+1} < t \leq  \theta_{k+2}} }{\sum_{k=1}^{+\infty}1_{\theta_k < t}}\to E\int_0^{\theta_1}\sin^2(W_s)ds$$ alomst surely by the strong law of large number because $\int_{\theta_{k}}^{\theta_{k+1}}\sin^2(W_s)ds$'s are i.i.d.
And that $$\dfrac{\sum_{k=1}^{+\infty}1_{\theta_k < t}}{t} \to \dfrac{1}{E(\theta_1)}$$ alomost sutrly(Imagine a Poisson process which jumps once a new $\theta_k$ is reached) 
So the limit is equal to $\dfrac{E\int_0^{\theta_1}\sin^2(W_s)ds}{E(\theta_1)}$. To compute it, let $$I_1 = \frac{1}{t}\int_0^t \sin^2(W_s)ds$$ $$I_2 = \frac{1}{t}\int_0^t \cos^2(W_s)ds$$ then we have $I_1 + I_2 = 1$.
Intuitively, when $t\to \infty$, the limit of $I_1$ and $I_2$ should be the same, since $\sin^2 x$  is just $\cos^2 x$ delayed by $\frac{\pi}{2}$, so we get $M = \frac{1}{2}$.
To prove it rigorously, let $\tau = \inf\{t \geq 0, W_t = \frac{\pi}{2}\}$, then 
\begin{align}
I_2 &= \frac{1}{t} \left(\int_0^\tau \cos^2(W_s)ds + \int_{\tau}^t\cos^2(W_s)ds\right) \\
& = \frac{1}{t} \left(\int_0^\tau \cos^2(W_s)ds + \int_{\tau}^t\sin^2(W_s - \frac{\pi}{2})ds\right)
\end{align}
Since $\tau$ is finite almost surely: $$\lim I_2 = \lim \frac{1}{t} \int_{\tau}^t\sin^2(W_s - \frac{\pi}{2})ds = \lim \frac{1}{t} \int_{0}^{t-\tau}\sin^2(W_s')ds = \lim I_1$$
where $W_s' = W_{s+ \tau} - \frac{\pi}{2}$
So the limit of $I_1$ and $I_2$ are both equal to $\frac{1}{2}$
Once the first part is finished, we can realize that we can treat the problem as a renewal process, which jumps once a new $\theta_k$ is reached and with the jump size $\int_{\theta_{k-1}}^{\theta_k}\sin^2(W_s)ds$. Therefore we can apply the central limit theorem for renewal process to resolve the second question.
