I'll take a guess at this. It seems like the same idea as "geometric form factors" in radiative heat transfer.
The telescope is "seeing" a piece of space that you can take as a flat disk (diameter 0.116 degrees)
But the surface of Mars that is radiatiing is not a flat disk, it is a hemisphere. The "brightness" as seen by the telescope will be maximum in the center and less at the edges.
That will give you another factor of ##\sin\theta## or ##\cos\theta## in the integral for Mars. Probably the "correction factor" of ##\pi/4## is a well known result, since most astronomical objects are approximately spherical.
