transformation between Ito and Stratonovich calculus Let's define the Stratonovich integral as
$$
\int_0^T f(w,t)\circ dW_t \sim \sum_j \frac{f(t_j)+f(t_{j+1})}{2}(W_{t_{j+1}}-W_{t_j})
$$
for later short hand notation, let $\delta W_{t_j} = W_{t_{j+1}}-W_{t_j}, \delta t_j = t_{t+1}-t_j$.
The transformation between Ito and Stratonovich is shown as, $b=b(t,X_t),\sigma=\sigma(t,X_t)$
Stratonovich
$$
dX_t = b dt+\sigma \circ dW_t 
$$
Ito
$$
dX_t = (b+\frac{1}{2}\sigma'\sigma) dt+\sigma dW_t 
$$
The proof in the textbook follows the definition of two integrals. When it concludes the convergence, I am confused why the $\delta t_j \delta W_{t_j}$ goes out. Here's the major calculation in the proof:

Assume the associated Ito's process is
$$
dX_t = \alpha dt+\beta dW_t 
$$
$$
\sum \sigma(X_{t_{j+1}},t_{j+1}) = \sigma(X_{t_{j}},t_{j})\delta W_{t_j} + \sigma_t(X_{t_{j}},t_{j})\delta W_{t_j}\delta t_j + \sigma_x(X_{t_{j}},t_{j})\delta W_{t_j} (dX_t) 
$$
$$
\sum \sigma(X_{t_{j+1}},t_{j+1}) =\sigma(X_{t_{j}},t_{j})\delta W_{t_j} + \sigma_t(X_{t_{j}},t_{j})\delta W_{t_j}\delta t_j + \sigma_x(X_{t_{j}},t_{j})\delta W_{t_j} (\alpha \delta t_j + \beta \delta W_j) \\
\int \sum \sigma(X_{t_{j+1}},t_{j+1}) \rightarrow \sigma(t,X_t)dW_t + \sigma_x\beta dt
$$

Basically $\delta W_{t_j}\delta t_j $ cancel out. I thought it was the mean zero property of Brownian. but why did the first $\delta W_{t_j}$ survived?
----Solution:------
I went back to read a few pages before and turns out that
$$
\delta t\delta W_t = 0, \delta t^2=0, \delta W_t^2 = \delta t
$$
by a Taylor expansion.
 A: I find that proof using those $\delta W\,$s and $\delta t\,$s not being my cup of tea. Consider this:
From the definition of the Stratonovich integral you can see easily that
$$
\int_0^tf(s)\circ\,dW_s
$$
is the limit of
$$
\underbrace{\sum_{j}\frac{f(t_{j+1})\color{red}{-}f(t_j)}{2}\Big(W_{t_{j+1}}-W_{t_j}\Big)}_{(*)}+\underbrace{\sum_j
f(t_j)\Big(W_{t_{j+1}}-W_{t_j}\Big)}_{(**)}\,.
$$
The term $(*)$ converges to half of the covariation $\langle W,f\rangle_t$. The term $(**)$ converges to the Ito integral $\int_0^tf(s)\,dW_s\,.$
This proves easily the familiar relationship
$$\tag{1}
\int_0^tf(s)\circ\,dW_s=\int_0^tf(s)\,dW_s+\frac{1}{2}\langle W,f\rangle_t\,.
$$
To find the covariation of $\sigma(\,.\,X)$ and $W$ we apply the Ito formula to $\sigma(t,X_t)$ which gives
\begin{align}
\sigma(t,X_t)=\sigma(0,X_0)+\int_0^t\sigma_x(s,X_s)\,dX_s+\int_0^t\sigma_t(s,X_s)\,ds+\frac{1}{2}\int_0^t\sigma_{xx}(s,X_s)\,d\langle X\rangle_s\,.
\end{align}
A non zero covariation of $\sigma(\,.\,X)$ and $W$ can only come from a term $\sigma(s,X_s)\,dW_s$ in $dX_s$ and is now
easily seen to be
$$\tag{2}
\Big\langle\sigma(\,.,X),W\Big\rangle_t=\int_0^t\sigma_x(s,X_s)\,\sigma(x,X_s)\,ds\,.
$$
(You wrote this as the integral of $\sigma'\sigma$.)
Finally applying (1) to
$$
X_t=X_0+\int_0^tb(s,X_s)\,ds+\int_0^t\sigma(s,X_s)\circ dW_s
$$
gives
\begin{align}
X_t
&=X_0+\int_0^tb(s,X_s)\,ds+\int_0^t\sigma(s,X_s)\,dW_s+\frac{1}{2}\underbrace{\int_0^t\sigma_x(s,X_s)\sigma(s,X_s)\,ds}_{(2)}\,.
\end{align}
