Q: Find the volume of the solid of revolution obtained by revolving the region bounded by the line $y = 2$ and the curve $y = \sec^2x$, $-90 \lt x \lt 90$, around the x-axis.
Attempted solution: $(1/3) \tan x ( \sec^2x + 2)$
I'm sure I did the integral correctly, but now when I need to sub in for $90, \tan(90)$ does not exist? (The answer is $2(\pi)^2 - 8\pi/3$)