Let $S(t)$ be the standard stock price expression $S(0)$ $\exp \left[\left(\mu-\frac{1}{2} \sigma^{2}\right) t+\sigma B(t)\right]$. Verify that $\int_{t=0}^{1} \sigma S(t) d B(t)$ is an Ito stochastic integral.
I got the SDE of $S_t$, $$dS_t=\mu S_t\:dt+\sigma S_t\: dB_t$$ Now, we want to get the stochastic integral, $$\int_{t=0}^{1} \sigma S(t) d B(t)$$ We may want to reverse engineer $f(t,x)$ using Ito's formula as follows. Ito's formula gives:
$$ \,d f(t,S_t) = \left( \frac{\partial f}{\partial t}(t,S_t) + \mu S_t \frac{\partial f}{\partial x}(t,S_t) + \frac{1}{2}\sigma^2 S_t^2\frac{\partial^2 f}{\partial x^2}(t,S_t)\right)\,d t + \sigma S_t \frac{\partial f}{\partial x}(t,S_t)\,d B_t$$ or,
$$ \int_0^T\sigma S_t \frac{\partial f}{\partial x}(t,S_t)\,d B_t = f(t,S_t) - f(0,S_0) - \int_0^T\left( \frac{\partial f}{\partial t}(t,S_t) + \mu S_t \frac{\partial f}{\partial x}(t,S_t) + \frac{1}{2}\sigma^2 S_t^2\frac{\partial^2 f}{\partial x^2}(t,S_t)\right)\,d t.$$
So we will be done if we can find $f$ such that $$ \sigma x \frac{\partial f}{\partial x}(t,x) = \sigma x$$ which will lead us to the integral $$ f(t,x) - f(t,0) = \int_0^xdy=x-x_0 $$
But, now I feel lost. From here, how can I conclude something? Like, I need to show that the $M_t\:\left( \int_{t=0}^{1} \sigma S(t) d B(t) \right)$ is driftless. But from here, how can I?