Quick one regarding Integration

Discussion in 'Physics & Math' started by kennedaymr, Dec 27, 2009.

  1. kennedaymr

    kennedaymr Registered Member

    Messages:
    12
    Gday,

    Just a quick one regarding integration.
    The rules at the back of my calculus book states that the Integral of
    tan(u) du = In(|sec u|) + C

    However, my calculators and Wolfram all give me the answer as being
    tan(u) du = -In(|cos u|) + C

    Which one is correct, which do I believe?

    Regards
     
    kennedaymr, Dec 27, 2009
    #1

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  3. Pete

    Pete It's not rocket surgery Registered Senior Member

    Messages:
    10,167
    They are the same, because:
    \(ln \frac{1}{x} = -ln x\)
     
    Pete, Dec 27, 2009
    #2

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  5. James R

    James R Just this guy, you know? Staff Member

    Messages:
    39,426
    That's ln (with a small "L"), not "In". It's the symbol for natural logarithm, which means a logarithm to base e.

    Also:

    \(\sec x = \frac{1}{\cos x}\)

    and

    \(\ln\left(\frac{1}{x}\right) = -\ln x\)

    (as Pete said).
     
    James R, Dec 28, 2009
    #3

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