Register to reply

Integrating tan(x)

by Omid
Tags: integrating, tanx
Share this thread:
Omid
#1
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 ?
Thanks
Phys.Org News Partner Science news on Phys.org
Scientists develop 'electronic nose' for rapid detection of C. diff infection
Why plants in the office make us more productive
Tesla Motors dealing as states play factory poker
Zurtex
#2
Sep7-04, 11:39 AM
Sci Advisor
HW Helper
P: 1,123
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
#3
Sep8-04, 01:05 PM
Sci Advisor
HW Helper
PF Gold
P: 12,016
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|


Register to reply

Related Discussions
Integrating x^x Calculus 11
Need help, soon please Integrating Calculus 12
Integrating (e^4x)/x Calculus 12
Integrating (4x-3)^2 Calculus & Beyond Homework 10
Integrating csc(x) General Math 9