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!
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.