For some point $x \in \mathbb{R},$ we consider the SDE: $$ \forall t \geq 0, \quad X_{t}=x+\int_{0}^{t} \sin \left(2 X_{s}\right) d s+\int_{0}^{t}\left(1+\cos ^{2}\left(X_{s}\right)\right)^{1 / 2} d B_{s} $$ 1) Show that the SDE admits a unique solution.
2) Show that $\left(\phi\left(X_{t}\right)\right)_{t \geq 0}$ is a martingale, where $$ \phi(x)=\int_{0}^{x}\left(1+\cos ^{2}(u)\right)^{2} d u $$
for the first one, one must show that : $x \mapsto\sin(2x)$ and $x \to \sqrt{1 + \cos^2(x)}$ satisfy both the lipschitz and growth condition.
which they do, since they are both differentiable and bounded functions.
however I'm lost at 2), the drift of $\phi(X)$ is :
$$-(2\sin X_t \cos X_t)(1 + \cos^2 X_t))(1 + \cos^2 X_t)$$
unless I'm missing something, since the drift isn't $0$ then $\phi(X)$ shouldn't be a martingale ?
what did I do wrong ?
I found your mistake there is a missing $(2\sin X_t \cos X_t).(1+\cos(X_t)^2)^2$ term in your drift calculation which offsets your result.
It comes from the $dt$ embedded term in the $((1+\cos(X_t)^2)^2.dX_t$ term that you forgot. I remind you that :
$$dF(X_t)= F'(X_t)dX_t+1/2F''(X_t)d\langle X_t\rangle$$ and that $$2\sin X_t \cos X_t=\sin(2X_t).$$