Integrating tan(x)

by Omid
Tags: integrating, tanx
Omid is offline
Sep7-04, 11:34 AM
P: 188
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 ?
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Zurtex is offline
Sep7-04, 11:39 AM
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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]?
arildno is offline
Sep8-04, 01:05 PM
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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,

Let's consider:
Let us make the substitution t=-x:

Or, further:

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

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