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Dido525 Group Title

Hyperbolic intergal

  • 2 years ago
  • 2 years ago

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  1. Dido525 Group Title
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    |dw:1353111163841:dw|

    • 2 years ago
  2. Dido525 Group Title
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    No matter how many times I do it I keep getting the wrong answer.

    • 2 years ago
  3. Dido525 Group Title
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    @zepdrix

    • 2 years ago
  4. Dido525 Group Title
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    I believe the anti-derivative is: |dw:1353111289980:dw|

    • 2 years ago
  5. zepdrix Group Title
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    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

    • 2 years ago
  6. Dido525 Group Title
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    Well according to Wikipedia:

    • 2 years ago
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  7. Dido525 Group Title
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    In this case a is 1.

    • 2 years ago
  8. zepdrix Group Title
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    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.

    • 2 years ago
  9. Dido525 Group Title
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    No No. It's a written :) .

    • 2 years ago
  10. zepdrix Group Title
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    Do you have the answer we're SUPPOSE to get, like in a book or anything? :D

    • 2 years ago
  11. Dido525 Group Title
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    No...

    • 2 years ago
  12. zepdrix Group Title
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    Then why do you think you're getting the wrong answer? XD

    • 2 years ago
  13. zepdrix Group Title
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    Oh, so you GAVE me the answer, you're just not sure how to get there? :O

    • 2 years ago
  14. Dido525 Group Title
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    Wolfram.

    • 2 years ago
  15. Dido525 Group Title
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    Ohh wait...

    • 2 years ago
  16. Dido525 Group Title
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    No I didn't give you the answer. I just keep getting different answers every time.

    • 2 years ago
  17. zepdrix Group Title
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    \[\huge \int\limits_{0}^{1}\left(\frac{ 1 }{ 4 }\right)(e^{2x}-e^{-2x}-2) dx\] \[\huge =\left(\frac{ 1 }{ 4 }\right)\left(\left(\frac{ 1 }{ 2 }\right)e^{2x}+\left(\frac{ 1 }{ 2 }\right)e^{-2x}-2x\right)_0^1\]

    • 2 years ago
  18. zepdrix Group Title
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    maybe the problem is... are you forgetting, its a DEFINITE integral.. or you're having trouble before that part? :D

    • 2 years ago
  19. Dido525 Group Title
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    I got it!!!

    • 2 years ago
  20. zepdrix Group Title
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    oh good :3 lol

    • 2 years ago
  21. Dido525 Group Title
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    If you were wondering it is: |dw:1353112228539:dw|

    • 2 years ago
  22. zepdrix Group Title
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    oh cool c:

    • 2 years ago
  23. Dido525 Group Title
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    No constant right? It's a definite integral.

    • 2 years ago
  24. Dido525 Group Title
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    @zepdrix

    • 2 years ago
  25. zepdrix Group Title
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    Yah any constants would fall out c: no constant.

    • 2 years ago
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