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Trigo Integral

  1. Aug 22, 2009 #1

    Mentallic

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    1. The problem statement, all variables and given/known data
    Find [tex]\int{tan^2xsec^2xdx}[/tex]


    2. Relevant equations
    [tex]tan^2x=sec^2x-1[/tex] (1)


    3. The attempt at a solution
    Using (1): [tex]\int{(sec^4x-sec^2x)dx}[/tex]

    Now, [tex]\int{sec^4xdx}-\int{sec^2xdx}=\int{sec^4xdx}-tanx+c[/tex]

    I can't figure out how to solve [tex]\int{sec^4xdx}[/tex] though.
     
  2. jcsd
  3. Aug 22, 2009 #2

    Dick

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    The derivative of tan(x) is sec^2(x). Mentallic, this is a simple substitution. u=tan(x).
     
    Last edited: Aug 22, 2009
  4. Aug 22, 2009 #3

    Mentallic

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    Oh yeah... ugh I feel like such an idiot.

    [tex]u=tanx[/tex]

    [tex]\int{sec^4xdx}=\int{(1+u^2)du}[/tex]

    Thanks Dick.
     
  5. Aug 22, 2009 #4

    Dick

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    Sure, but why don't you use that substitution in the original integral?
     
  6. Aug 22, 2009 #5

    Mentallic

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    Oh, you mean like [tex]\int{u^2du}[/tex] ? Yeah because I love to make things harder for myself :yuck:
     
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