I previously asked whether $X_t=\sin B_t$ is a martingale, where $B_t$ is a standard Brownian motion and the underlined filtration is the canonical one.
As it can be seen in the answer provided, this turned out to be false. The answer consisted of an application of Ito's Lemma, and as I understood from the fact that an Ito Process with a drift can not be a martingale.
I would like now to see if I can check whether $X_t=\sqrt{t}\sin(B_t)$ is a martingale. I have been given the hint that it is, but I am not sure why is this the case. Following the notation of this page: we have that our function $f(w,t)=\sqrt{t} \sin(w)$ so that $\frac{\partial f }{\partial t}=\frac{1}{2 \sqrt{t}}\sin(w)$, $f'(w,t) = \sqrt{t} \cos (w)$ and $f''(w,t) = -\sqrt{t}\sin w$.
Since $dW(t)=dB(t)$ and $dB(t)^2=dt$ we would then have that:
$$dX_t = \frac{\partial f }{\partial t}(B_t,t)dt + f'(B_t,t) dB_t + \frac{1}{2}f''(B_t,t)dt = \sqrt{t}\cos (B_t)dB_t + \\ \sin(B_t)dt\bigg(\frac{1}{2 \sqrt{t}} - \sqrt{t}\frac{1}{2}\bigg)$$
Here it seems that again we have a drift, which would indicate that the process is not a martingale. Where is my mistake?