Let $B_t$ be standard Brownian Motion. Could someone please help me to show that $$E[(\int_{0}^{t}B_sdB_s)^2] = \int_{0}^{t}E[B_s^2]ds$$
I am sure that it has something to do with Ito's formula but I am pretty new to this stuff so some help would be greatly appreciated.
The RHS is direct to evaluate. The variance of $B_s$ is $s$ so: $$\int_0^t E[B_s^2] ds = \int_0^t s ds = \frac{t^2}{2}$$
You are right about use Ito's Lemma for the LHS. By Ito: $$B_t^2 = t + 2\int_0^t B_s dB_s$$ So \begin{eqnarray*} E[(\int_0^t B_s dB_s)^2] &=& E\left[\left(\frac{B_t^2 - t}{2}\right)^2 \right] \\ &=& \frac{1}{4}(E[B_t^4] - 2tE[B_t^2] + t^2)\\ &=& \frac{1}{4}(3t^2 - 2t\cdot t + t^2)\\ \end{eqnarray*}