If
$${\rm Cov}[dW_t,dB_t]=\rho \, dt$$
then why
$\mathbb{Cov} \left( \int_0^t \sigma_{1}(s) \mathrm{d} W_s, \int_0^t \sigma_{2}(u) \mathrm{d} B_u \right)$ $\stackrel{\text{Ito isometry}}{=} \rho \int_0^t \sigma_1(s) \sigma_2(s) \, \mathrm{ds}$
where $\sigma_{1s}$ and $\sigma_{2s}$ are two deterministic functions of $t$?
I do not understand why the $\rho$ emerges. I mean from the application of the Ito isometry the answear should not only be $\int_0^t \sigma_1(s) \sigma_2(s) \, \mathrm{d}s$?