## anonymous 3 years ago Hyperbolic intergal

1. anonymous

|dw:1353111163841:dw|

2. anonymous

No matter how many times I do it I keep getting the wrong answer.

3. anonymous

@zepdrix

4. anonymous

I believe the anti-derivative is: |dw:1353111289980:dw|

5. zepdrix

Hmm I'm not sure how you're able to deal with the fact that the outside term is a square. Is there a double angle formula for hyperbolic sine? Otherwise I would convert to exponentials and break it down from there. $\large \sinh x=\frac{ e^x-e^{-x} }{ 2 }$ $\large \sinh^2 x=\left(\frac{ e^x-e^{-x} }{ 2 }\right)^2=\left(\frac{ 1 }{ 4 }\right)(e^{2x}-e^{-2x}-2)$ Hopefully I expanded that correctly :O

6. anonymous

Well according to Wikipedia:

7. anonymous

In this case a is 1.

8. zepdrix

Hmm yah it looks good. But if you're trying to enter this into webassign (or something similar), then maybe your teacher is looking for an answer in exponential form. Hmmmmm.

9. anonymous

No No. It's a written :) .

10. zepdrix

Do you have the answer we're SUPPOSE to get, like in a book or anything? :D

11. anonymous

No...

12. zepdrix

Then why do you think you're getting the wrong answer? XD

13. zepdrix

Oh, so you GAVE me the answer, you're just not sure how to get there? :O

14. anonymous

Wolfram.

15. anonymous

Ohh wait...

16. anonymous

No I didn't give you the answer. I just keep getting different answers every time.

17. zepdrix

$\huge \int\limits_{0}^{1}\left(\frac{ 1 }{ 4 }\right)(e^{2x}-e^{-2x}-2) dx$ $\huge =\left(\frac{ 1 }{ 4 }\right)\left(\left(\frac{ 1 }{ 2 }\right)e^{2x}+\left(\frac{ 1 }{ 2 }\right)e^{-2x}-2x\right)_0^1$

18. zepdrix

maybe the problem is... are you forgetting, its a DEFINITE integral.. or you're having trouble before that part? :D

19. anonymous

I got it!!!

20. zepdrix

oh good :3 lol

21. anonymous

If you were wondering it is: |dw:1353112228539:dw|

22. zepdrix

oh cool c:

23. anonymous

No constant right? It's a definite integral.

24. anonymous

@zepdrix

25. zepdrix

Yah any constants would fall out c: no constant.