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Integrals of Trig Functions

  1. May 20, 2005 #1
    Hello everyone,

    I am having some trouble finding the integral of tanxsec^2x. I honestly have no idea where to go with this one. I've finished all the others but this one is really screwing me up. I tried doing something with the chain rule but it didn't work out at all. I know that I am supposed to show my work, but at this point I don't have anything to show you guys. Thanks a lot for any help you can give me.
     
  2. jcsd
  3. May 20, 2005 #2

    OlderDan

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    Write everything in terms of sines and cosines. Then you can do a simple substitution.
     
  4. May 20, 2005 #3
    Or just note that the derivative of tangent is the secant squared... u-substitution, anybody?

    --J
     
  5. May 20, 2005 #4
    Ok guys, thanks for the help, but I'm still completely lost. I still don't have a clue of what I should do.
     
  6. May 20, 2005 #5
    Are you familiar with u-substitution? You're going to have to make one to evaluate the integral. You have two choices for what to let u equal, both work. What do you think u might equal?

    --J
     
  7. May 20, 2005 #6
    I am sort of familiar with it. Could you let u equal tanx?
     
  8. May 20, 2005 #7

    OlderDan

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    If you did, what would du be?
     
  9. May 20, 2005 #8
    Ok, actually I might let u = sec^2x, then du should equal tanx+c?
     
  10. May 20, 2005 #9

    OlderDan

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    That is not correct

    You should review the derivatives of all the trig functions.
     
  11. May 20, 2005 #10
    You're integrating u to get du for some reason. You must differentiate u to get du.

    Stick with your original substitution.

    --J
     
  12. May 20, 2005 #11

    dextercioby

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    [tex] \int \sec^{2} x \tan x \ dx=-\int \frac{d {}\cos x}{\cos^{3} x}=\frac{1}{2}\sec^{2}x +\mathcal{C} [/tex]

    Daniel.
     
  13. May 21, 2005 #12
    shouldnt it be:
    [tex]\int sec^2x~tanx~dx=\int\frac{sinx}{cos^3x}dx[/tex]
    then, you could do u-sub and set [itex]u=cosx[/itex] and go from there?

    *Edit*
    dextercioby, you put sinx instead of cosx. i think tanx = sinx/cosx
    *Edit*
     
    Last edited: May 21, 2005
  14. May 21, 2005 #13

    OlderDan

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    and du would be

    [tex]du = d cosx = -sinx dx[/tex]

    giving

    [tex] sinx dx = -d cosx[/tex]

    which is what Dexter did to get his result, and what earlier hints were pointing to. It could also be done using

    [tex] u= tanx [/tex]

    [tex]du= d tanx = sec^2x dx[/tex]

    [tex]sec^2x dx = d tanx[/tex]

    as suggested by Justin early on
     
  15. May 24, 2005 #14
    Oh ok....I think I am starting to get it
     
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