# Integrating hyperbolic functions

1. Nov 9, 2012

### peripatein

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. Nov 9, 2012

### peripatein

Does someone have an idea what is stymying my answer?

3. Nov 9, 2012

### SteamKing

Staff Emeritus
Why did you chose u = tanh(x)? What happens if you expand (tanh(x) + coth (x))?

4. Nov 9, 2012

### peripatein

I used u=tanhx, as 1/(coshx)^2 is its derivative.

5. Nov 9, 2012

### Dick

There is nothing wrong with your answer. If the book is giving one that looks different it may differ from yours by a constant.

6. Nov 9, 2012

### peripatein

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?

7. Nov 9, 2012

### Dick

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
8. Nov 9, 2012

### peripatein

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

9. Nov 9, 2012

### Dick

They are only 'equal' if you consider the '+C' part. (tanhx)^2+(sechx)^2=1. Use that identity.

10. Nov 9, 2012

### peripatein

I see. Thanks a lot!