now here's a trick or shortcut for doing an anti-derivative of any secx function with an odd power...
~secnx = 1/(n-1)[sec(n-2)xtanx + 1/(n-1)[(n-2)~sec(n-2)x]
that n stuff can be confusing looking so here's a few examples
~sec3x
n = 3
(n-1) = 2
(n-2) = 1
so ~sec3x = 1/2sec1xtanx...
you guys are on the right track but not quite there...let me help :)
I don't know how to make those fancy integral symbols on here either, so I'm going to use the tilda for it :)
to solve ~sec5x you actually first need to solve ~sec3x. You'll see why when we get there. So let's do that...