I'm stuck with the computation of $$\int_0^t E[\cos(B_u)|F_s] du$$
where $F_s$ is the natural filtration and $0<s<t$.
I tried to separate the cases when $u \in [0,s]$ and $u \in[s,t]$, but I'm not sure how to go on.
How should I move?
Edit
Now I write $$\int_0^t E[\cos(B_u)|F_s]du = \int_0^t E[\cos(B_u) \mathcal{1}_{[0,s]} + \cos(B_u) \mathcal{1}_{[s,t]}|F_s]du= \\ \int_0^s E[\cos(B_u)]du + \int_s^t E[\cos(B_u)|F_s]du$$
Now $E[\cos(B_u)|F_s] = E[\cos(B_u - B_s + B_s)|F_s] = E[\cos(B_u - B_s) \cos(B_s) - \sin(B_s) \sin(B_u - B_s)|F_s] = \cos(B_s) E[\cos(B_u - B_s)]$
The expectation with the sine is $0$ since $B_u - B_s$ is normally distributed and since the integrand function $\sin(x) \exp(\ldots)$ is odd on a symmetric interval.
Now, $E[\cos(B_u - B_s)] = e^{-\frac{u-s}{2}}$, by looking at the characteristic function.
Therefore
$$\int_s^t E[\cos(B_u)|F_s]du = \int_s^t \cos(B_s) e^{-\frac{u-s}{2}} du = \cos(B_s) \int_s^t e^{-\frac{u-s}{2}} du = \cos(B_s) (2-2 e^{\frac{s-t}{2}})$$
Summing up everything I get:
$$ \int_0^t E[\cos(B_u)|F_s] du = \int_0^s \cos(B_u)du + \cos(B_s) (2-2 e^{\frac{s-t}{2}})$$
Note
$$\int_0^t E[\cos B_u|F_s] du=\int_0^s \cos B_u du+\int_s^t E[\cos B_u|F_s] du\tag 1$$
where the first integral is deterministic and known given the filtration $F_s$. To calculate the second integral, apply Ito's law to $\cos B_t$
$$d(\cos B_t) = -\sin B_tdB_t - \frac12 \cos B_t dt$$
Integrate over $(s,t)$ to get
$$\cos B_t -\cos B_s = -\int_s^t \sin B_tdB_t - \frac12 \int_s^t \cos B_u du$$
Therefore,
$$\int_s^t E[\cos B_u|F_s]du = 2\cos B_s - 2E[\cos B_t| F_s]$$
where the expectation on the RHS can be computed with the normal density function,
$$E[\cos B_t| F_s] = \frac1{\sqrt{2\pi (t-s)}}\int_{-\infty}^{-\infty}\cos x \>e^{-\frac{(x- B_s)^2}{2(t-s)}} = \cos B_se^{-\frac {t-s}2}$$
Thus, plug the result into (1) to obtain
$$\int_0^t E[\cos B_u|F_s] du = \int_0^s \cos B_u du + 2\cos B_s (1-{e^{-\frac {t-s}2}})$$