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Can not do this integral

  1. Jul 20, 2005 #1
    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: Jul 20, 2005
  2. jcsd
  3. Jul 20, 2005 #2


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    [tex]\sec (x) = \frac{1}{\cos (x)}[/tex]

    [tex]\tan (x) = \frac{\sin (x)}{\cos (x)}[/tex]

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

    [tex]\tan (x) \leq \sec (x)[/tex]
  4. Jul 20, 2005 #3
    im such a moron.

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