Integral with hyperbolic: cosh x

  • Thread starter FelixHelix
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  • #1
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Main Question or Discussion Point

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
 

Answers and Replies

  • #2
arildno
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Well, does:
[tex]\frac{d}{dx}tanh(x)=\frac{1}{\cosh^{2}(x)}[/tex] equal your integrand?
 
  • #3
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No, the integrand is 1/(1+cosh(x))
 
  • #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]
 
  • #5
arildno
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No, the integrand is 1/(1+cosh(x))
Correct!
So, what must the purported answer be called?
 
  • #6
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I see,thats a good trick i hadnt thought of using. But how would you manipulate the integral you end up with?
 
  • #7
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Correct!
So, what must the purported answer be called?
?? I'm not sure??
 
  • #8
arildno
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?? I'm not sure??
Maybe what is called TOTALLY WRONG!!

To give you a hint:
The correct answer is tanh(x/2)+C, not tanh(x)+C
 
  • #9
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yes a typo indeed - apologies. What substitution do u use to get there though - i dont see it!
 
  • #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)
 
  • #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]
 
  • #12
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Ill take a look in the morning - thanks for your help!!!
 

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