Integrating tan(x): Solving the Absolute Value Puzzle

[Omid]

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

 

[Zurtex]

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 $\cos x < 0$.

Anyway you do know to integrate $\tan x$ you just just write it as $$\frac{\sin x}{\cos x}$$?

 

[arildno]

The absolute value sign is needed in order to gain the proper integral value of the function $$\frac{1}{x}$$ 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,
$$\int_{a}^{b}\frac{1}{x}dx=ln(b)-ln(a)=ln(|b|)-ln(|a|)(a,b>0)$$

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

Or, further:
$$=-\frac{ln(b)}{ln(a)}=\frac{ln(a)}{ln(b)}=ln(|-a|)-ln(|-b|)$$

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