I have this proof that I need to do and I don't know how to execute a certain step. The part that I don't get is this; $$-\nabla \times E = \nabla \times (w \times B)$$ The notes just say, "Integrating we get"; $$E = (-w \times B) + (-\nabla \psi)$$ Where $-\nabla \psi$ is the constant of integration.
How do we integrate a curl operator? and what does that even mean?
Here is one interpretation of what's going on. Noting that there is a curl on both sides, we can rewrite the equation as $$ \nabla\times \left[w \times B + E \right] = 0. $$ "Integrating" in this case seems to refer generically to undoing a differential operator, in this case the curl. It is known that if a vector field (over a simply connected domain) has curl zero, then it must be of the form $\nabla \phi$ for some scalar field $\phi$. Thus, we can write $$ w \times B + E = \nabla \phi \implies E = -w \times B + \nabla \phi. $$ Substituting $\psi = -\phi$ gives us the result from your notes.