Expectation value of a product of an Ito integral and a function of a Brownian motion this problem has come up in my research and is confusing me immensely, any light you can shed would be deeply appreciated.
Let $B(t)$ denote a standard Brownian motion (Wiener process), such that the difference $B(t)-B(s)$ has a normal distribution with zero mean and variance $t-s$.
I am seeking an expression for
$$E\left[ \cos(B(t))\int\limits_0^t \sin(B(s))\,\textrm{d}B(s) \right],$$
where the integral is a stochastic It$\hat{\textrm{o}}$ integral.  My first thought was that the expectation of the integral alone is zero, and that the two terms are statistically independent, hence the whole thing gives zero.  However, I can't prove this.
To give you a little background: this expression arises as one of several terms in a calculation of the second moment of the integral
$$\int\limits_{0}^{t}\cos(B(s))\,\textrm{d}s,$$
after applying It$\hat{\textrm{o}}$'s lemma and squaring.  I can simulate this numerically, so I should know when I get the right final expression!
Thanks.
 A: This addresses the question cited as a motivation.
For every $t\geqslant0$, introduce $X_t=\int\limits_{0}^{t}\cos(B_s)\,\textrm{d}s$ and $m(t)=\mathrm E(\cos(B_t))=\mathrm E(\cos(\sqrt{t}Z))$, where $Z$ is standard normal. 
Then $\mathrm E(X_t)=\int\limits_{0}^{t}m(s)\,\textrm{d}s$ and $\mathrm E(X_t^2)=\int\limits_{0}^{t}\int\limits_{u}^{t}2\mathrm E(\cos(B_s)\cos(B_u))\,\textrm{d}s\textrm{d}u$. 
For every $s\geqslant u\geqslant0$, one has $2\cos(B_s)\cos(B_u)=\cos(B_s+B_u)+\cos(B_s-B_u)$. Furthermore, $B_s+B_u=2B_u+(B_s-B_u)$ is normal with variance $4u+(s-u)=s+3u$ and $B_s-B_u$ is normal with variance $s-u$. Hence, $2\mathrm E(\cos(B_s)\cos(B_u))=m(s+3u)+m(s-u)$, which implies
$$
\mathrm E(X_t^2)=\int\limits_{0}^{t}\int\limits_{u}^{t}(m(s+3u)+m(s-u))\,\textrm{d}s\textrm{d}u.
$$
Since $m(t)=\mathrm e^{-t/2}$, this yields after some standard computations,
$\mathrm E(X_t)=2(1-\mathrm e^{-t/2})$ and 
$$
\mathrm E(X_t^2)=2t-\frac13(1-\mathrm e^{-2t})-\frac83(1-\mathrm e^{-t/2}).
$$
Sanity check: When $t\to0^+$, $\mathrm E(X_t^2)=t^2+o(t^2)$.

To compute the integral $J_t=\mathrm E\left[ \cos(B_t)\int\limits_{0}^{t} \sin(B_s)\,\textrm{d}B_s \right]$, one can start with Itô's formula
$$
\cos(B_t)=1-\int\limits_{0}^{t} \sin(B_s)\,\textrm{d}B_s-\frac12\int\limits_{0}^{t} \cos(B_s)\,\textrm{d}s,
$$
hence 
$$
J_t=\mathrm E(\cos(B_t))-\mathrm E(\cos^2(B_t))-\frac12\int\limits_{0}^{t} \mathrm E(\cos(B_t)\cos(B_s))\,\textrm{d}s,
$$
and it seems each term can be computed easily.
