Hi,

I've pretty much forgotten all my trig, so heres a really basic question if anyone can help.

I have to find the following integral:

upper bound = 1, lower bound = 0

integral: 16 ((sec(x))^2/x^3/2)

I have to use the comparison theorem to do this and so far this is what i did.

I used the identity sec(x) = 1/cos(x)

so subbing that in it would be (1/cos(x))^2/x^3/2

now heres the really stupid question

if i squared cos(x), what would be the result? cos^2(x) or cos(x^2) ??