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Let $B_t$ be a standard Brownian motion, $t≤T$. How could we show that the expectations of the following equality hold true: $$\mathbb{E}[sin^2(B_t)]=\int_{0}^{t}\mathbb{E}[cos(2B_s)]ds.$$

Since $B_t$ is normally distributed with mean $0$ and variance $t$, using density function $$\mathbb E(\sin^2(B_t))=\frac{1}{\sqrt{2\pi t}}\int_{-\infty}^{\infty}\sin^2(x)e^{-x^2/2t}dx,$$ and $$\int_{0}^{t}\mathbb{E}[cos(2B_s)]ds=\int_{0}^{t}\frac{1}{\sqrt{2\pi t}}\int_{-\infty}^{\infty}[\cos(2x)]e^{-x^2/2t}dxds.$$

But, I'm stuck to proceed. I am truly having trouble in simplifying the two. Any suggestions would be helpful!

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    Do you know Itô's formula? – 2017-02-23
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    @saz yes I got it. Thanks. However, I used Fubini's Theorem as well. – 2017-02-23

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Just treating $B_t$ as a normal random variable we know from its characteristic function that \begin{align*} \varphi_X(s) &= \mathbb{E}\left[ e^{isX} \right] = \mathbb{E}\left[ \cos(sX) \right] + i \mathbb{E}\left[ \sin(s X) \right] \\ &= e^{i \mu s - \frac{1}{2}\sigma^2 s^2 }, \end{align*} now combine this with what you found in your previous question for the expected value of the odd function $\sin (sX)$ and using $\mu = 0$ and $\sigma^2 = t$ and see how you get on from there.

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    I should make it clear that my answer is only addressing how one can go about easily calculating the expectation of some trigonometric functions as appear in your second and third equations - not whether the first identity you have set out to prove is in fact true. – 2017-02-23