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Integration result ln || - confusing and apparent contradictions!

  1. Jun 24, 2012 #1
    The integral of sec x is ln|sec x + tan x| + c. That's what the internet is telling me.

    Now, I am having trouble understanding the modulus sign. I understand that the ln operator gives real numbers for positive values of the argument and complex numbers for negative values of the argument. So, if the ln operator is defined for negative values, why incorporate the modulus sign?

    Also, I differentiated ln |sec x + tan x| + c and I get |sec x|. Doesn't that mean that the integral of sec x is only defined for positive values of sec x? But that contradicts common sense when we have a look at the graph of sec x (google 'graph sec x' for a display of the graph). It's clear the function does have an integral for negative values of sec x, right?
     
  2. jcsd
  3. Jun 24, 2012 #2
    Real integrals can't take complex values, to be precise.
    Differentiating the expression you gave, we get this: We apply the chain rule once to get the derivative is
    [tex]\frac{1}{\sec x + \tan x}\frac{d}{dx}|\sec x + \tan x|[/tex]
    Now, we apply the chain rule once more. The derivative of the absolute value function is the signum function, defined everywhere but at 0. Hence this yields [itex]\mathrm{sgn}(\sec x + \tan x)\sec(x)[/itex] as the derivative. Now, the expression inside the signum function is positive or zero because of this trigonometric simplification:

    [tex]\sec x + \tan x = \frac{1+\sin x}{\cos x} = \frac{1+\sin x}{\sqrt{1-\sin^2 x}} \geq 1+\sin x \geq 0[/tex]

    Hence we are allowed to discard the signum function (for it will be 1) and rewrite the derivative as [itex]\sec(x)[/itex].

    I am curious how you obtained the result you did. To obtain a more general answer, try differentiating and integrating a general reciprocal.
     
  4. Jun 24, 2012 #3
    One problem. [itex]\cos x = \sqrt{1-\sin^2 x}[/itex] is only true when [itex]\cos x[/itex] is positive.
     
  5. Jun 25, 2012 #4
    I'd like to remind you that if cos x is negative, due to the identity [itex]|\sec x|\geq |\tan x|[/itex], the expression [itex]\sec x + \tan x[/itex] is negative or zero: values where the real logarithm is not defined.
     
  6. Jun 26, 2012 #5
    How about going back to ∫1/x dx = ln|x| + C?
    My calc textbook said that since 1/x is odd, ∫1/x dx is even, so you can take the antiderivative to be ln|x| + C with absolute values to make it even. Then you can apply it to the problem with ∫secx dx.
     
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