Ito's formula and sin(Brownian motion) I would like to compute:
$d(e^{\frac12t}sinB_t)$
using the integration by parts of Ito I come up with the following:$$\frac12e^{\frac12t}sinB_tdt +e^{\frac12t}cosB_tdB_t + 0$$ however I the solution should be only $e^{\frac12t}cosB_tdB_t$. where is the mistake?
moreover what is the result of $E[e^{\frac12t}sinB_t]$?
 A: You are computing $d(\sin B_t)$ like if it was a differentiable function.
If we apply Ito's formula with the function $f(x)= \sin x,$ we get 
$$d(\sin B_t) = \cos B_t dB_t - \frac{1}{2} \sin B_t dt.$$
Now applying the integration by parts formula, you will get the result
\begin{align*}
d(e^ {\frac{1}{2}t} \sin B_t) &= e^{\frac{1}{2}t} d(\sin B_t) + \frac{1}{2}e^ {\frac{1}{2}t} \sin B_t dt \\
&= ...
\end{align*} 
For the other question, you can use the fact that $B_t \sim N(0, \sqrt{t})$ in conjunction with this: Compute $E(\sin X)$ if $X$ is normally distributed
A: The correct formula in this case is
$$
d f(B_t,t)=f_1(B_t,t)\,dB_t+f_2(B_t,t)\,dt+\frac{1}{2}f_{11}(B_t,t)\,d[B_t].
$$
Applying this to $f(x,t)=e^{t/2}\sin(x)$, one gets
\begin{align}\label{1}\tag{1}
df(B_t,t)&=e^{t/2}\cos(B_t)\,dB_t+\frac{1}{2}e^{t/2}\sin(B_t)\,dt-\frac{1}{2}e^{t/2}\sin(B_t)\,dt \\
&=e^{t/2}\cos(B_t)\,dB_t.
\end{align}
It follows that $\mathsf{E}f(B_t,t)=0$ ($\because$ the RHS of $\eqref{1}$ is a martingale starting at $0$).
