= Solution
{c}
Condition on the Brownian path up to $t\leq T$. The independent normal increment $B_T-B_t$ has $\mathbb E\cos(B_T-B_t)=e^{-(T-t)/2}$ and zero expected sine. Hence
$$
M_t:=\mathbb E(\sin B_T\mid\mathcal F_t)
=e^{-(T-t)/2}\sin B_t.
$$
Apply <Itô's formula> to this expression. Its time derivative and second spatial derivative cancel, leaving
$$
dM_t=e^{-(T-t)/2}\cos B_t\,dB_t,\qquad M_0=0.
$$
Therefore the required representation is
$$
\boxed{\sin B_T=\int_0^T e^{-(T-s)/2}\cos B_s\,dB_s.}
$$
The bounded integrand is square integrable, and uniqueness follows from the <Itô isometry>. The drift-containing decomposition in part (b) alone would not be the required representation as a constant plus one <stochastic integral>.
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