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 $E[e^{\frac12t}sinB_t]$?

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
