Apply Ito's formula to $d(X_t)$, where $X_t = e^{\int_0^t(X_s-1/2)ds+W_t}$

Question

I'd like to find $dX_t$, where $X_t = e^{\int_0^t(X_s-1/2)ds+W_t}$. I have no idea how to apply Ito's formula here.

Answer 1

Let's denote $Y_t = \int_0^t(X_s-1/2)ds+W_t$, so we have $X_t=e^{Y_t}$.

Applying the Ito's lemma on $X_t=e^{Y_t}$ $$dX_t = e^{Y_t}dY_t+\frac{1}{2}e^{Y_t}<dY_t,dY_t> \tag{1}$$

But from $Y_t = \int_0^t(X_s-1/2)ds+W_t$, we have $$dY_t = (X_t-\frac{1}{2})dt+dW_t \tag{2}$$

Replacing (2) to (1), by knowing that $<dY_t,dY_t> = <dW_t,dW_t> = dt$: \begin{align} dX_t &= e^{Y_t}dY_t+\frac{1}{2}e^{Y_t}<dY_t,dY_t> \\ &= X_t ((X_t-\frac{1}{2})dt+dW_t)+\frac{1}{2}X_tdt \\ &= X_t (X_tdt+dW_t) \\ &= X_t^2 dt + X_tdW_t \end{align}