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Let $W_t$ be a standard Brownian motion, $t\leq T$. Then how can we prove that the expectation

$\mathbb{E}\bigg(\int_{0}^{T}\sin(3W_t)dW_t\bigg)=0.$

I was trying to use zero mean property, but I'm stuck. Anyone could help me?

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Let $Y_T=\int_{0}^{T}{sin(3W_t)dW_t}$ defines an Ito process. One way to prove your result is to show that $Y$ is a martingale, which can be done by proving that

$$E\left(\int_{0}^{T}{sin(3W_t)^2dt}\right)< \infty$$

We know that $sin(3W_t)^2\leq 1$ almost surely therefore

$$E\left(\int_{0}^{T}{sin(3W_t)^2dt}\right) \leq E\left(\int_{0}^{T}{1dt}\right)=T< \infty$$

It proves that $Y_T$ is a martingale and therefore that $$E(Y_T)=Y_0=0$$

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