I am fairly new to these exercises so please bear with me.
I am given that $X_t = 2t + 4B^2_t$ where $B_t$ is a standard Brownian motion. And I am asked to calculate $d X_t$.
This is confusing to me since standard notation is $d X_t:= X_t - X_0$, but I have a feeling the exercise wants me to apply Ito's formula (Indeed this is what some fellow students of mine did).
To apply Ito's formula I need to somehow write $X_t$ as an Ito process, it is tempting to write
$$X_t = 2\int_0^t \, ds+ 8\int_0^tB_s\, dBs$$ But I don't think $8\int_0^tB_s\, dBs = 4B^2_t$. Is this equality true? if so why? (I know the technical definition of Ito integral).
EDIT: Ito's formula says that (in the formulation I know of) that Given an Ito process $X$ and a function $f(t,x) \in C^{1,2}(R^2)$. Then the stochastic process
$$Y_t = f(t,X_t)$$ is an Ito process and we have that
$$df(t,X_t) = \partial_tf(t,X_t) dt + \partial_xf(t,X_t) dX_t +\frac{1}{2} \partial_{xx} f(t,X_t) d [X]_t.$$
EDIT2: If $B_t$ is an Ito process then we can apply Ito's formula and obtain
$$d X_t= 2dt + 8B_t dB_t + 4dt = 6dt + 8 B_t dB_t $$
Use linearity to get $\mathrm{d}X_t=2\, \mathrm{d}t+4\, \mathrm{d}B^2_t$
Then apply Ito formula to $B^2$:
$$B^2_t=B^2_0+\int_{0}^{t} 2B_s\, \mathrm{d}B_s+\int_{0}^{t} 1 \, \mathrm{d}s$$
$$\mathrm{d}B^2_t=2B_t\, \mathrm{d}B_t+\mathrm{d}t$$
Substitute this to $\mathrm{d} X_t$ to get the final form.
$$\mathrm{d}X_t=2\, \mathrm{d}t+4\, \mathrm{d}B^2_t=2\, \mathrm{d}t+4(2B_t\, \mathrm{d}B_t+\mathrm{d}t)=8B_t\, \mathrm{d}B_t+6\mathrm{d}t.$$