Correct PDF sampling math check

This also corrects a MathJax problem with the text "solid_angle".

Resolves #1280

Author: hollasch

books/RayTracingTheRestOfYourLife.html

@@ -3493,15 +3493,15 @@
   $$ \cos(\theta_{max}) = \sqrt{1 - \frac{R^2}{length^2(\mathbf{c} - \mathbf{p})}} $$
 
 We also need to evaluate the PDF of directions. For a uniform distribution toward the sphere the PDF
-is $1/\mathit{solid_angle}$. What is the solid angle of the sphere? It has something to do with the
-$C$ above. It is -- by definition -- the area on the unit sphere, so the integral is
+is $1/\text{solid_angle}$. What is the solid angle of the sphere? It has something to do with
+the $C$ above. It is -- by definition -- the area on the unit sphere, so the integral is
 
-  $$ \mathit{solid angle} = \int_{0}^{2\pi} \int_{0}^{\theta_{max}} \sin(\theta)
+  $$ \text{solid_angle} = \int_{0}^{2\pi} \int_{0}^{\theta_{max}} \sin(\theta)
        = 2 \pi \cdot (1-\cos(\theta_{max})) $$
 
 It’s good to check the math on all such calculations. I usually plug in the extreme cases (thank you
 for that concept, Mr. Horton -- my high school physics teacher). For a zero radius sphere
-$\cos(\theta_{max}) = 0$, and that works. For a sphere tangent at $\mathbf{p}$,
+$\cos(\theta_{max}) = 1$, and that works. For a sphere tangent at $\mathbf{p}$,
 $\cos(\theta_{max}) = 0$, and $2\pi$ is the area of a hemisphere, so that works too.