Integral with hyperbolic: cosh x

  • Thread starter FelixHelix
  • Start date
  • #1
FelixHelix
28
0
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
Science Advisor
Homework Helper
Gold Member
Dearly Missed
10,025
135
Well, does:
[tex]\frac{d}{dx}tanh(x)=\frac{1}{\cosh^{2}(x)}[/tex] equal your integrand?
 
  • #3
FelixHelix
28
0
No, the integrand is 1/(1+cosh(x))
 
  • #4
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
10,025
135
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
Science Advisor
Homework Helper
Gold Member
Dearly Missed
10,025
135
No, the integrand is 1/(1+cosh(x))
Correct!
So, what must the purported answer be called?
 
  • #6
FelixHelix
28
0
I see,thats a good trick i hadnt thought of using. But how would you manipulate the integral you end up with?
 
  • #7
FelixHelix
28
0
Correct!
So, what must the purported answer be called?
?? I'm not sure??
 
  • #8
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
10,025
135
?? 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
FelixHelix
28
0
yes a typo indeed - apologies. What substitution do u use to get there though - i dont see it!
 
  • #10
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
10,025
135
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
Science Advisor
Homework Helper
Gold Member
Dearly Missed
10,025
135
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
FelixHelix
28
0
Ill take a look in the morning - thanks for your help!!!
 

Suggested for: Integral with hyperbolic: cosh x

  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
4
Views
895
  • Last Post
Replies
5
Views
1K
Replies
3
Views
1K
Replies
1
Views
4K
Replies
89
Views
10K
Replies
4
Views
1K
  • Last Post
Replies
11
Views
4K
  • Last Post
Replies
3
Views
943
  • Last Post
Replies
6
Views
1K
Top