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$$\int \tan x \, dx \equiv \ln|\sec x|$$

What's the significance of the modulus?

I remember doing a question using this formula, and had a negative answer inside the log, so I gave up - because I was using the log with normal brackets.

I know you can't log a negative, but that doesn't help me understand why this mod is here.

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    Have you learned about the anti derivative of $1/x$? – 2017-01-25
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    Yes: $\int{\frac{1}{x} = \ln{x}}$ – 2017-01-25
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    You're wrong: $\;\displaystyle\int\frac1x=\ln\lvert x\rvert$. – 2017-01-25
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    The essence of the answer must be in the significance of the "modulus" in the equality $$ \int \frac 1 x\,dx = \log |x| + C $$ where $C$ is _piecewise_ constant. Maybe I'll post an answer later. $\qquad$ – 2017-01-25
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    @Tobi You need the absolute values round that $x$ because otherwise you cannot integrate for negative values. Also note that the antiderivative of $-1/(-x)$ is is $ln(-x)$ which is then perfectly defined for negative values of $x$. Now you can see where the absolute value comes from – 2017-01-25
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    side question - if ln|x| is the area under a 1/x graph, why is ln|1| = 0? Also, I see people using log, and ln for trig equations, isn't it ln for log base e? How could it be interchangeable? – 2017-01-25
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    @Tobi First part: $ln1$ is zero because according to the definition of ln, you are integrating under the curve $y=1/x$ from $x=1$ to $x=1$ and thus the area of a line segment is zero. Second part (in a nut shell), in higher math (like complex analysis) they use log to denote ln because all the other "logs" are not used. Moreover, if Log is capitalized with $L$, then that strictly indicates a certain interval for the complex number's argument. – 2017-01-26

2 Answers 2

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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 $\sec x < 0$. Hope it helps.

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Two points I'd like to make:

Notice that $\sec x$ comes in intervals of being positive and negative.

Notice that $\tan x$ diverges at the endpoints of the above intervals, so whatever you do, you can't integrate over those points.

Now notice that these are trig functions, and thus periodic, so every integral you take reduces down to one interval (since every interval is the same), and so to preserve this, we take the absolute value of $\sec x$ to get a positive number to put into the integrand.