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Integrating tan(x) 
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#1
Sep704, 11:34 AM

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Today I was reading my favorite calculus textbook, that saw the integration formula for tan(x).
It was : Integral of tan(x) = lncosx + 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
Sep704, 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]? 


#3
Sep804, 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 antiderivative 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 antiderivative valid for both x greater and less than zero is lnx 


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