State if the Ito integral exist for $\int_0^T sin(B_t)dB_t $ calculate its mean and variance.
After applying Ito's lemma this is what I got:
$B_0 = 0$
$sin(B_t) = \int_0^T cos(B_t)dB_t - \frac 12 \int_0^T sin(B_t)dt $
I am a bit stuck on how to show its mean and variance.
The function $\sin(\cdot)$ is bounded and hence $\int_0^T E(|\sin(B_t)|^2)dt<\infty$. This implies that the Itô integral is a martingale and hence it has $0$ mean.
For the variance apply the Itô isometry
$$E\left(\big|\int_0^T \sin(B_t)dB_t\big|^2\right)=E\left(\int_0^T \sin(B_t)^2dt\right)$$
Then you can apply Tonelli lemma to change the order of integration, and then calculate
$$E(\sin^2(B_t)).$$
To do so use the Euler formula $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$ and using the fact that $B_t$ is centered Gaussian with variance $t$ you can solve this expectation using the Characteristic function (I leave you this computations). Then you can simply compute the time integral to obtain the variance.