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Integral of ln(secx + tanx)?

  1. Mar 12, 2008 #1
    I was taking the integral of the secant function. Twice...
    The first one is simple, but what is the integral of
    ln(secx + tanx)dx?

    I've tried a few things, the first being integration by parts with u = ln(secx + tanx+) and dv = dx
    This just cancels in the end to 0 = 0
    I also rewrote it as int[ln(1+sinx) - ln(cosx)]dx but that doesn't seem to be any easier.

    Any suggestion would be greatly appreciated.
     
    Last edited: Mar 12, 2008
  2. jcsd
  3. Mar 12, 2008 #2

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    It doesn't have a closed form answer in terms of elementary functions.
     
  4. Mar 12, 2008 #3
    sure it does, unless you consider the polylog erudite:

    [tex]
    \frac{1}{2} \left(\pi \left(i x+\text{Log}\left[-\text{Sin}\left[\frac{1}{4} (\pi -2 x)\right]\right]-\text{Log}[1+i \text{Cos}[x]-\text{Sin}[x]]-\text{Log}[1-i
    \text{Cos}[x]+\text{Sin}[x]]+\text{Log}\left[\text{Sin}\left[\frac{1}{4} (\pi +2 x)\right]\right]\right)+2 x \left(2 i \text{ArcTan}\left[e^{i x}\right]+\text{Log}[\text{Sec}[x]+\text{Tan}[x]]\right)+2
    i \text{PolyLog}\left[2,i e^{i x}\right]-2 i \text{PolyLog}[2,-i \text{Cos}[x]+\text{Sin}[x]]\right)[/tex]

    lol off screen. whatever it's in the body of the post, just click quote or something
     
    Last edited: Mar 12, 2008
  5. Mar 12, 2008 #4

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    Elementary functions are finite combinations and compositions of algebraic, logarithmic, and exponential functions. Polylog obviously doesn't fit that description.
     
  6. Mar 12, 2008 #5
    yea you're right
     
  7. Jan 12, 2011 #6
    @ice109: How do you get that formula? (The one that involves PolyLog)
     
  8. Jan 12, 2011 #7
  9. Jan 13, 2011 #8
    Thank you!!!
     
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