Suppose we have $$ u_y = 0 $$
where $u = u(x,y)$. If we integrate this wrt $x$ on $a \leq x \leq b$, we have
$$ \int_a^b u_y dx = \frac{d}{dy} \int_a^b u(x,y) dx = 0 $$
are we allowed to do this? My book writes this in a step, but Im having trouble seeing why this is true.
If $u_y = 0$, then $u$ is constant with respect to $y$ (and so $u$ can just be considered a function of $x$). Therefore, $\int_a^b u(x,y) dx$ is a constant (not involving $y$), and so its $y$ derivative is zero.
More generally, suppose $u(x,y)$ is continuous, and differentiable with respect to $y$. Then $u$ has a primitive with respect to $x$, say, $U(x,y)$. This is to say, $U_x = u$. Then $$\frac{\partial}{\partial y}\int_a^b u(x,y) dx = \frac{\partial}{\partial y} (U(b,y) - U(a,y)) =$$$$ U_y(b,y) - U_y(a,y) = \int_a^b u_y(x,y) dx$$ The last equality holds because $\frac{\partial}{\partial x}U_y = U_{xy} = (U_x)_y = u_y$.