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I know this is a bit late, but if you still haven't figured it out, here goes:

Firstly, let the mass per unit area of the disk be $\sigma = \frac{m}{\pi R^2}$.

Consider a small element of area $dA$ at a distance $x$ from the center of the disk. The mass of this element is $dm = \sigma dA$. The frictional force on this element is $dF = \mu (dm) g = \mu \sigma g \times dA$

The torque on this small element is $d\tau = x \times dF = \mu \sigma g x  \times dA$. (note that by $\times$ I just mean normal multiplication, not the cross product or anything).

Now let us consider a thin circular ring with thickness $dx$ at a distance $x$ from the center (I mean to say a ring of radius $x$ whose center is the center of the disk). The net torque $\tau$ on this ring would be the integral of all the small torques on the small elements of area $dA$ lying on this thin ring. 

$$\tau = \int d\tau = \mu \sigma g x \int dA = \mu \sigma g \times (2 \pi x^2 dx)$$

Note that in this integral $x$ is a constant, since $x$ is the distance from the center for all the small elements on the circular ring we have taken under consideration. 

As $dx \to 0$, we have infinite such rings on the disk. So now we need to integrate over all these rings. The equation above matches the second equation of your post. 

I think you should be able to take it from here. The important thing is to keep in mind what we are doing: first we find the net torque on a thin ring situated at a distance $x$ from the center which I've done above, and then we integrate over all such rings.