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: $$ dY_t=X_t\,dt = X_t\,dt +\ 0\,dB_t. $$