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How to get (secx)(tanx) from (1/cosx)(sinx/cosx)?

  1. Dec 7, 2006 #1
    Quick question:

    h(x) = sinx/cos^(2) x

    = (1/cosx)(sinx/cosx)

    Then you get (secx)(tanx)..

    I do not get how they get secx x tanx?? Anyone?? Thanks
     
  2. jcsd
  3. Dec 7, 2006 #2

    chroot

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    1/cos(x) is also called sec(x).

    sin(x)/cos(x) is also called tan(x).

    - Warren
     
  4. Dec 7, 2006 #3
    so the answer would be

    (secx)(tanx) + c

    Correct??
     
  5. Dec 7, 2006 #4

    chroot

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    All you've done so far is convert the function you gave me into a slightly simpler form.

    sin(x) / cos^2(x) = sec(x) tan(x).

    Since you didn't actually post the problem as it was given to you, I don't know if h(x) is a function of which you need to find the antiderivative, or whether you've already done that step. You probably need to actually perform the antiderivative now.

    - Warren
     
  6. Dec 7, 2006 #5
    sorry how would i go about doing this.. Iam so confused.
     
  7. Dec 7, 2006 #6

    chroot

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    Find the function which has a derivative of sec(x) tan(x). You should have a list of such facts in your book.

    - Warren
     
  8. Dec 7, 2006 #7
    alright.. I get it now.. Sorry.. I dont know why this problem was causing me problems.. Thanks for clearing it up.
     
  9. Dec 8, 2006 #8

    dextercioby

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    Do you know the method of substitution to find antiderivatives ? If so, just plug

    [tex] \cos x = t [/tex]

    and c what u get.

    Daniel.
     
  10. Dec 8, 2006 #9

    HallsofIvy

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    It would have helped if you had told us from the beginning that you were trying to find an anti-derivative! All you said was that you couldn't see how they had gone from Quick question:

    h(x) = sinx/cos^(2) x

    to h(x)= (secx)(tanx).
     
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