Ito formula when g(t,x) is an integral Suppose we have a stochastic process which is written as an Ito process.
$$dX_t=\mu_t\ dt +\sigma_t\ dB_t$$.
If $Y_t$ is defined as a stochastic process as a function of $X_t$, then we can find $dY_t$ using the Ito formula. The key is to have the function $g(t,x)$ which relates $X_t$ to $Y_t$.  Then we can find the derivatives with respect to $t$ and $x$ to plug into the Ito formula.  For instance, if we write 
$Y_t=tX_t$, we use the Ito formula with the function $g(t,x)=tx$.
However, what if we want to define $Y_t$ as a time integral of $X_t$?  That is, $$Y_t=\int^t_0 X_u \ du.$$  Then how do we write $g(t,x)$ in order to find $Y_t$ as an Ito process?  Is it simply
$$\int^t_0 x\ du = tx-0*x=tx?$$  That doesn't feel right.  Or is it maybe $$\int^t_0 x\ dx = \frac{t^2}{2}?$$  I definitely don't think that is right though since we changed $du$ to $dx$.
Edit: To be clear, I want to write $Y_t$ as a function of $\mu_t$ and $\sigma_t$. 
Thanks for the help!

Express $Y$ as an Ito process:
$$
dY_t=X_t\,dt = X_t\,dt +\ 0\,dB_t.
$$
