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Integrating hyperbolic functions

  1. Nov 9, 2012 #1
    Hi,
    I am trying to integrate (tanh(x)+coth(x))/((cosh(x))^2)
    I am substituting u=tanh(x), du=dx/((cosh(x))^2)
    and end up with 1/2(tanh(x))^2 + ln |tanh(x)| + C
    which is incorrect. What am I doing wrong??
     
  2. jcsd
  3. Nov 9, 2012 #2
    Does someone have an idea what is stymying my answer?
     
  4. Nov 9, 2012 #3

    SteamKing

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    Why did you chose u = tanh(x)? What happens if you expand (tanh(x) + coth (x))?
     
  5. Nov 9, 2012 #4
    I used u=tanhx, as 1/(coshx)^2 is its derivative.
     
  6. Nov 9, 2012 #5

    Dick

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    There is nothing wrong with your answer. If the book is giving one that looks different it may differ from yours by a constant.
     
  7. Nov 9, 2012 #6
    Online calculators claim the integral to be -1/2*(coshx)^2 + ln |tanhx| + c.
    1/2*(tanhx)^2 (which is the first term in my answer) is not equal to -1/2*(coshx)^2, is it?
     
  8. Nov 9, 2012 #7

    Dick

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    Wolfram Alpha gives the first term as -(sechx)^2/2 and that does differ from (tanhx)^2/2 by a constant. What's the constant? Are you sure the online calculator isn't saying -1/(2*(coshx)^2)? You should use more parentheses when you write something like -1/2*(coshx)^2. It's ambiguous.
     
    Last edited: Nov 9, 2012
  9. Nov 9, 2012 #8
    I am not following your argument. Is the answer which Wolfram's calculator yields equal to mine?
    My answer is: (0.5)(tanh(x))^2 + ln |tanh(x)| + C
    Wolfram's calculator's answer: (-0.5)(sech(x)^2) + ln [tanh(x)] + C
     
  10. Nov 9, 2012 #9

    Dick

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    They are only 'equal' if you consider the '+C' part. (tanhx)^2+(sechx)^2=1. Use that identity.
     
  11. Nov 9, 2012 #10
    I see. Thanks a lot!
     
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