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Quick one regarding Integration

  1. Dec 27, 2009 #1

    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?

  2. jcsd
  3. Dec 27, 2009 #2


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    Science Advisor

    sec(u)= 1/cos(u) so ln(sec(u))= ln(1/cos(u))= -ln(cos(u)). They are exactly the same thing.

    (And the natural logarithm is represented by ln ("ell en") not In ("eye en").)
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