I have following issue on my hands:
Use Ito's formula to compute the semimartingale decomposition and the quadratic variation $[X_i]_t$ for $X_t= sin^2(B_t) + cos^2(B_t)$
This is quite strange, since we all know $sin^2(x) + cos^2(x)=1$. I think that this should apply also for Brownian motion. Hence, Ito's integral is:
$$X_t = 1 + \int_0^t0dB_t + 0.5\int_0^t0dt = 1$$
And the quadratic variation is then also $0$. I just want to make sure, that this makes sense, as I find this result quite odd. Thanks!