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Integrating tan(x)

  1. Sep 7, 2004 #1
    Today I was reading my favorite calculus textbook, that saw the integration formula for tan(x).
    It was : Integral of tan(x) = -ln|cosx| + C .

    I know that when we say integral of tanx we mean, what is the function whose derivative is tanx. So started to take the derivative of -ln |cosx|, in order to prove the formula. But what could I do with the absolute value sign ? I just ignored it and took the derivative. It worked and I arrived at the answer, tan(x). Now there are 2 questions. 1. why is the sign there anymore? 2. what is the right approach while taking derivative of functions involving absolute value sign? Do we ignore them always, as I did in this case ?
    Thanks
     
  2. jcsd
  3. Sep 7, 2004 #2

    Zurtex

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    Real values integrate to real values, so if we didn't have the modulus symbol we would be taking the natural logarithm of a negative number when [itex]\cos x < 0[/itex].

    Anyway you do know to integrate [itex]\tan x[/itex] you just just write it as [tex]\frac{\sin x}{\cos x}[/tex]?
     
  4. Sep 8, 2004 #3

    arildno

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    The absolute value sign is needed in order to gain the proper integral value of the function [tex]\frac{1}{x}[/tex] on intervals where x<0
    (Remember, you can't find the natural logarithm of a negative real number among the reals!)
    To illustrate:
    Given x>0, we may show that a proper anti-derivative is ln(x).
    For example,
    [tex]\int_{a}^{b}\frac{1}{x}dx=ln(b)-ln(a)=ln(|b|)-ln(|a|)(a,b>0)[/tex]

    Let's consider:
    [tex]\int_{-b}^{-a}\frac{1}{x}dx[/tex]
    Let us make the substitution t=-x:
    [tex]\int_{-b}^{-a}\frac{1}{x}dx=\int_{b}^{a}\frac{1}{t}dt=-\int_{a}^{b}\frac{1}{t}dt=-\frac{ln(b)}{ln(a)}[/tex]

    Or, further:
    [tex]=-\frac{ln(b)}{ln(a)}=\frac{ln(a)}{ln(b)}=ln(|-a|)-ln(|-b|)[/tex]

    Hence, we see that a proper anti-derivative valid for both x greater and less than zero is ln|x|
     
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