I have the following process $X_t = b(X_t)dW_t$ with W a Brownian Motion. I am interested in calculating $Y_t = \int_0^t X_sdX_s = \int_0^t X_sb(X_s)dW_s$ which shows this process is a Martingale as $dY_t = X_tb(X_t)dW_t$ which has no drift term. I am told that $Y_t = \frac{1}{2}X_t^2 + Q_t$ where $Q_t$ involves $b$.
I tried finding $Q_t$ by doing: $$dY_t = (X_t + \frac{d}{dx}Q_t)dW_t + (\frac{1}{2}(1+\frac{d^2}{dx^2}Q_t) + \frac{d}{dt}Q_t)dt$$ and tried solving for $Q_t$ using the previous expression for $dY_t$. How should I approach this problem.
(This is part of my revisions for exams) thank you!
Notice that $dQ_t = dY_t - d(\frac{1}{2}X_t^2)$. Since we know $dY_t = X_t b(X_t) dW_t$ we should try to find an expression for $d(\frac{1}{2}X_t^2)$ of this form. By Ito's lemma we know that $$d(\frac{1}{2}X_t^2) = X_t dX_t + \frac{1}{2} d\langle X \rangle_t = X_t b(X_t) dW_t + \frac{1}{2} b(X_t)^2 dt$$ where the second equality is deduced from the fact that $dX_t = b(X_t)dW_t$. In particular, $dQ_t = -\frac12 b(X_t)^2 dt$ which implies that $Q_t = Q_0 + \int_0^t b(X_s)^2 ds$.