Double Integration in Polar and its Applications

Still a little bit confused as to what you would use for the u-sub (or v, in your case), but I guess that won't have any relevance since it's wrong to begin with.

Oops, misspoke. I mean integration by parts. sec^3x dx, u=secx, dv=sec^2x dx. That has du = secxtanx dx, v = tanx. So you get int udv = uv - int vdu = secxtanx - int (tan^2x secx)
Which by sec^2x = 1 + tan^2x is int (sec^3 x - sec x)
so 2 int (sec^3 x dx) = secxtanx + int(secx)
The integral of secant x is not particularly easy to figure out so I'd just expect to memorize it or look it up.