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
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]?
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|