there is another way... mass reduction but thats not really recommended since its not a fun thing.
S -> integral sign
In = S sec^n x dx
= S (sec^2 x)(sec^(n-2) x) dx
= tan x sec^(n-2) x - (n-2)S tan^2 x sec^(n-2) x dx
= tan x sec^(n-2) x - (n-2)S (sec^2 x - 1) (sec^(n-2) x) dx
= tan x sec^(n-2) x - (n-2)S sec^n x - sec^(n-2) x dx
= tan x sec^(n-2) x - (n-2)(In - I(n-2))
(n-2)In +In = tanx sec^(n-2) x + (n-2)I(n-2)
In = (tanx sec^(n-2) x + (n-2)I(n-2) )/ (n-1)
in which u just plug in numbers to that formula and ull get urself an answer... which in this case n = 3 so
I3 = tanxsecx + I1 / 2
I1 = ln| tanx + secx| + C thus:
I3 = (tanxsecx + ln|tanx + secx| )/ 2 + c
of course this is a long way to do things if u dunno the reduction formula but... its just another way
the thing with this is u cna use it for any power of sec so thats useful =]