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) ??