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Integral with hyperbolic: cosh x

  1. Oct 12, 2013 #1
    I cannot reach the answer for this integral which is part of a bigger question related to discounting investments. I know what the answer to the integral is and have tried all the substitutions and tricks I know. Any pointer would be great!!

    ∫(1/(1+cosh(x))) = tanh(x) + C

    Thanks, Felix
     
  2. jcsd
  3. Oct 12, 2013 #2

    arildno

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    Well, does:
    [tex]\frac{d}{dx}tanh(x)=\frac{1}{\cosh^{2}(x)}[/tex] equal your integrand?
     
  4. Oct 12, 2013 #3
    No, the integrand is 1/(1+cosh(x))
     
  5. Oct 12, 2013 #4

    arildno

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    The integral is readily evaluated by using:
    [tex]\int\frac{dx}{1+\cosh(x)}dx=\int\frac{dx}{1+\cosh(x)}\frac{1-\cosh(x)}{1-\cosh(x)}dx=\int\frac{1-\cosh(x)}{-\sinh^{2}(x)}dx[/tex]
     
  6. Oct 12, 2013 #5

    arildno

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    Correct!
    So, what must the purported answer be called?
     
  7. Oct 12, 2013 #6
    I see,thats a good trick i hadnt thought of using. But how would you manipulate the integral you end up with?
     
  8. Oct 12, 2013 #7
    ?? I'm not sure??
     
  9. Oct 12, 2013 #8

    arildno

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    Maybe what is called TOTALLY WRONG!!

    To give you a hint:
    The correct answer is tanh(x/2)+C, not tanh(x)+C
     
  10. Oct 12, 2013 #9
    yes a typo indeed - apologies. What substitution do u use to get there though - i dont see it!
     
  11. Oct 12, 2013 #10

    arildno

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    First off:
    You can easily do the integral the way I indicated.
    THEN, find the appropriate hyperbolic identities for half-arguments (very analogous to the more well-known half-angle formulas for the trig functions). The result will simplify to tanh(x/2)
     
  12. Oct 12, 2013 #11

    arildno

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    We have:
    [tex]\sinh(x)=2\cosh(\frac{x}{2})\sinh(\frac{x}{2}),\cosh(x)=\cosh^{2}(\frac{x}{2})+\sinh^{2}(\frac{x}{2})[/tex]
    Note the similarity to the trig identities!
    You'll probably also need the fundamental identity:
    [tex]1==\cosh^{2}(\frac{x}{2})-\sinh^{2}(\frac{x}{2})[/tex]
     
  13. Oct 12, 2013 #12
    Ill take a look in the morning - thanks for your help!!!
     
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