How to Find the Area Between the Curves Sec(x) and Tan(x) from 0 to Pi/2?

[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) , because when i plug in numbers it does, i forgot how to prove it. l'hopitals rule doesn't 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

 

[quasar987]

i assume that sec(x) increases faster than tan(x) , because when i plug in numbers it does, i forgot how to prove it. l'hopitals rule doesn't 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)

 

[unggio]

im such a moron.



thanks