I want to show that this is true:
$${ \mathbb{E}\big[\sin X_t \big]} = \sum_{n=0}^{\infty} \frac{(-1)^{n}{ \mathbb{E}\big[ X_t^{2n+1} \big]}}{(2n+1)!}$$
($X_t$ is a Brownian Motion).
By linearity I can pass the expectation to each term in the series. Since each term contains only one variable, $X_t$, I can pull out the constants from the expectation.