I have that $\sigma$ is a piecewise continuous function on $[0,t]$, $W$ is Brownian motion, $X(t)=\int_0^t\sigma(s)dW(s)$, and $Z(t)= e^{iuX(t)},$ for some fixed $u\in\mathbb{R}$. It is then stated that with Ito's rule it is easy to see that $$Z(t)=1+ \int_0^tiuZ(s)dW(s)- \frac{1}{2}u^2\int_0^t\sigma(s)^2Z(s)ds.$$
I'm having difficulties verifying this equation. I started by using Ito's rule for differentiation, and if I haven't made any mistakes, I end up with $$dZ(t)=iue^{iuX(t)}dX(t)-\frac{1}{2}u^2e^{iuX(t)}\sigma(t)^2dt,$$ which gets me close to my goal. I have a feeling all that is left is just to integrate both sides, but I'm not sure how, especially when dealing with the term $dX(t).$
Your calculation is almost good. However, your application of Ito's lemma is not entirely correct. I will outline the details for you:
Since $X(t) = \int_{0}^{t} \sigma(s) dW_s$, it follows that $dX(t) = \sigma(t)dW_t$. Define $Z(t) = \exp(iuX(t)) = f(t,X(t))$. Now, an easy calculation shows $$\begin{cases} \displaystyle\frac{\partial f}{\partial t} = 0, \\ \displaystyle \frac{\partial f}{\partial X_t} = \exp(iuX(t))iu = iuZ(t), \\ \displaystyle \frac{\partial f}{\partial X_t^2} = -u^2 Z(t). \end{cases}$$ Hence, applying Itô's lemma gives (note that the process $X_t$ does not have a drift term) \begin{align} dZ(t) &= \left(\frac{\partial f}{\partial t}+\frac{\partial f}{\partial X_t} \times 0 + \frac{1}{2} \sigma(t)^2 \frac{\partial^2 f}{\partial X_t^2} \right)dt + \sigma(t) \frac{\partial f}{\partial X_t} dW_t \\ \\& = \left(-\frac{1}{2} \sigma(t)^2 u^2 Z_t\right)dt + iu Z_t \sigma(t) dW_t \\ \end{align} In differential form (by integrating both sides): $$Z(t) = \underbrace{Z(0)}_{=1} -\frac{1}{2} u^2 \int_{0}^{t} \sigma(s)^2 Z_s ds + i u \int_{0}^{t} \sigma(s) Z_s dW_s.$$ Hope this helps. Final remark: in my notation I vary between the notations $Z(t)$ and $Z_t$ but they are the same.