- #1

- 28

- 0

∫(1/(1+cosh(x))) = tanh(x) + C

Thanks, Felix

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter FelixHelix
- Start date

- #1

- 28

- 0

∫(1/(1+cosh(x))) = tanh(x) + C

Thanks, Felix

- #2

arildno

Science Advisor

Homework Helper

Gold Member

Dearly Missed

- 9,970

- 134

Well, does:

[tex]\frac{d}{dx}tanh(x)=\frac{1}{\cosh^{2}(x)}[/tex] equal your integrand?

[tex]\frac{d}{dx}tanh(x)=\frac{1}{\cosh^{2}(x)}[/tex] equal your integrand?

- #3

- 28

- 0

No, the integrand is 1/(1+cosh(x))

- #4

arildno

Science Advisor

Homework Helper

Gold Member

Dearly Missed

- 9,970

- 134

[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

- 9,970

- 134

Correct!No, the integrand is 1/(1+cosh(x))

So, what must the purported answer be called?

- #6

- 28

- 0

- #7

- 28

- 0

?? I'm not sure??Correct!

So, what must the purported answer be called?

- #8

arildno

Science Advisor

Homework Helper

Gold Member

Dearly Missed

- 9,970

- 134

?? 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

- 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

- 9,970

- 134

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

- 9,970

- 134

[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

- 28

- 0

Ill take a look in the morning - thanks for your help!!!

Share: