# Can not do this integral

#### unggio

i need to find the area between these curves

sec(x) & tan(x) between 0 --> Pi/2

i assume that sec(x) increases faster than tan(x) , cuz when i plug in numbers it does, i forgot how to prove it. l'hopitals rule doesnt work .

so far i have integral of secx - tanx from 0 --> Pi/2

that gives me ln(secx + tanx) + ln(cosx) from 0 --> Pi/2

my answer i got is :ln(2)

this is a test problem from my friends 2nd qtr calculus class

thanks if u can help

Last edited:

#### quasar987

Homework Helper
Gold Member
i assume that sec(x) increases faster than tan(x) , cuz when i plug in numbers it does, i forgot how to prove it. l'hopitals rule doesnt work .
$$\sec (x) = \frac{1}{\cos (x)}$$

$$\tan (x) = \frac{\sin (x)}{\cos (x)}$$

But $\forall x \in [0,\pi /2]$, $\sin (x) \leq 1$. So $\forall x \in [0,\pi /2]$,

$$\tan (x) \leq \sec (x)$$

im such a moron.

thanks

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