# Math Help - Need help with some easy trig + integral

1. ## Need help with some easy trig + integral

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) ??

2. Is this your integral? : $\displaystyle 16 \int_0^1 \frac{(sec(x))^2}{ \left( \frac{x^3}{2} \right) } \, dx$

And you want to evaluate it (if it converges) or just test its convergence?

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Originally Posted by Kuma
if i squared cos(x), what would be the result? cos^2(x) or cos(x^2) ??
It will be : $cos^2(x)=(cos(x))^2 \not= cos\left(x^2\right)$

3. i just want to test if it converges. The denominator is x^1.5 = x^3/2 = x sqrt x. The rest is correct.