Example:
$(W_{t},\mathcal{F}_{t})$ is standard Brownian.
$\cos^{2}W(t)=1+ \underbrace{-2\int_{0}^{t}\cos(W_{s})\sin(W_{s})dW_{s}}_{=:M_{t}} +\frac{1}{2}\int_{0}^{t}[2-4\cos^{2}(W_{s})]ds $ by Ito's formula.
$\Rightarrow E[\cos^{2}W(t)]$ contains term $E[M_{t}]\overset{\text{optional sampling}}=E[M_{0}] \overset{?}=1$
First, I do not see how this term is $1$ and not $0$ given we are now evaluating the upper bound of the integral at $0$.
Second, given that it is then the case that optional sampling can be applied, would this then imply that any expectation of this form (expectation of ito integrals which are also martingales) is going to be $0$ given that the upper bound will always be evaluated at $0$?