Using Ito's lemma to compute a SDE 
This is the version of Ito's lemma that we are given in our notes.

Now I'm just not able to understand how to begin this problem and arrive at  the given solution. The g(x) integral function that they have given is just confusing me and using Ito's lemma I don't understand what we have to differentiate.
Any help would be much appreciated.
 A: We need to apply Ito's lemma to the function $g(X_t)$, where $g(x) = \int_0^x\frac{1}{\sigma(u)}\ du$, and where $dX_t = \sigma(X_t)dW_t$. Notice that there is no $t$-dependence in the function $g$, so the first term $f_t(X_t,t)dt$ of Ito's lemma vanishes, and the lemma reduces to
$$dg(X_t)  = g_x(X_t)dX_t + \frac{1}{2}g_{xx}(X_t)\sigma(X_t)^2dt.$$
Consider the first term $g_x(X_t)dX_t$. Here $g_x$ is just the derivative of $g$, and by the fundamental theorem of calculus, we have
$$g_x(x) = \frac{d}{dx}g(x) = \frac{d}{dx}\int_0^x\frac{1}{\sigma(u)}\ du = \frac{1}{\sigma(x)},$$
and thus
$$g_x(X_t)dX_t = \frac{1}{\sigma(X_t)}dX_t = \frac{1}{\sigma(X_t)}\sigma(X_t)dW_t = dW_t.$$
Consider then the second term $\frac{1}{2}g_{xx}(X_t)\sigma(X_t)^2dt$. The expression $g_{xx}$ is nothing more but the second derivative of $g$, and we have by the chain rule
$$g_{xx}(x) = \frac{d}{dx}\frac{1}{\sigma(x)} = -\frac{1}{\sigma(x)^2}\sigma'(x).$$
We then obtain
$$\frac{1}{2}g_{xx}(X_t)\sigma(X_t)^2dt = \frac{1}{2}\left(-\frac{1}{\sigma(X_t)^2}\sigma'(X_t)\right)\sigma(X_t)^2dt = -\frac{1}{2}\sigma'(X_t)dt.$$
In total, we have
$$dg(X_t) = dW_t - \frac{1}{2}\sigma'(X_t)dt.$$
