The differential and integral forms of <PERSON>'s equations are truly equivalent; they are essentially the same set of equations.
One can convert between the two using two mathematical theorems:
Divergence Theorem (Wikipeda - Divergence Theorem) <PERSON>' Theorem (Wikipedia - Stokes Theorem)
The divergence theorem states that the flux over a closed surface of a vector field is equal to integrating the divergence of that vector field over the volume enclosed by the surface. <PERSON>' theorem states that the line integral of a vector field over a closed path is equal to integrating the curl of that vector field over any surface bounded by the closed path.
Now, what you are seeking to show is a wave equation, which as you have stated is a differential form; which is as follows:
$\nabla^{2} \vec{E} = \mu_0 \epsilon_0 \frac{\partial^{2}E}{\partial t^{2}} $
Now, to get there from the integral form, what you do is the following: Take the third equation, for instance, which states:
$\oint \limits_C \vec{E} \cdot \vec{dl} = -\frac{\partial}{\partial t} \iint\limits_S \vec{B}\cdot \vec{dS} $
Now Stokes Theorem states that:
$\oint \limits_C \vec{E} \cdot \vec{dl} = \iint\limits_S \vec{\nabla} \times \vec{E} \cdot \vec{dS}$
Plug it in, and you get:
$\iint\limits_S \vec{\nabla} \times \vec{E} \cdot \vec{dS} = -\frac{\partial}{\partial t} \iint\limits_S \vec{B}\cdot \vec{dS}$
Change order of time derivative and integral (yes, you can do that) and collect under same integral sign:
$\iint\limits_S (\vec{\nabla} \times \vec{E} + \frac{\partial \vec{B}}{\partial t} )\cdot \vec{dS}$
Now, since this must hold for any surface S, the integrand itself must be zero. 418
So we get:
$\vec{\nabla} \times \vec{E} + \frac{\partial \vec{B}}{\partial t} = 0$
Now, we are on our way to getting the wave equation, but what we have done as the first step is simply converting the integral form of the equation to its differential form! To continue, you would need to do the same to the first two equations, then apply $(\vec{\nabla}\times)$ operator to the equation we just obtained above... Use a vector identity, and get there...
You may say, this is not really using the integral forms to come up with the wave equation! Well, it is starting from the integral forms of the equations, but what choice do we have if what we are trying to reach a second-order differential equation anyways??
(The above procedure will give you the wave equation for $\vec{E}$. To get the one for $\vec{B}$, you will need to start with equation four...) 418