Solution
= Solution
{c}
For $f(z)=\sin z$, $f'(z)=\cos z$ and $f''(z)=-\sin z$. <Itô's formula> gives
$$
d\sin(B_t)=\cos(B_t)\,dB_t-\tfrac12\sin(B_t)\,dt.
$$
Since $B_0=0$, this is exactly
$$
\boxed{dX_t=\cos B_t\,dB_t-\tfrac12X_t\,dt,\qquad X_0=0.}
$$
The integrand retains $\cos B_t$; replacing it by a fixed sign of $\sqrt{1-X_t^2}$ would lose information about the phase of $B_t$.