Dido525
Hyperbolic intergal
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Dido525
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|dw:1353111163841:dw|
Dido525
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No matter how many times I do it I keep getting the wrong answer.
Dido525
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@zepdrix
Dido525
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I believe the anti-derivative is:
|dw:1353111289980:dw|
zepdrix
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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
Dido525
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Well according to Wikipedia:
Dido525
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In this case a is 1.
zepdrix
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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.
Dido525
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No No. It's a written :) .
zepdrix
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Do you have the answer we're SUPPOSE to get, like in a book or anything? :D
Dido525
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No...
zepdrix
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Then why do you think you're getting the wrong answer? XD
zepdrix
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Oh, so you GAVE me the answer, you're just not sure how to get there? :O
Dido525
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Wolfram.
Dido525
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Ohh wait...
Dido525
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No I didn't give you the answer. I just keep getting different answers every time.
zepdrix
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\[\huge \int\limits_{0}^{1}\left(\frac{ 1 }{ 4 }\right)(e^{2x}-e^{-2x}-2) dx\]
\[\huge =\left(\frac{ 1 }{ 4 }\right)\left(\left(\frac{ 1 }{ 2 }\right)e^{2x}+\left(\frac{ 1 }{ 2 }\right)e^{-2x}-2x\right)_0^1\]
zepdrix
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maybe the problem is... are you forgetting, its a DEFINITE integral.. or you're having trouble before that part? :D
Dido525
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I got it!!!
zepdrix
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oh good :3 lol
Dido525
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If you were wondering it is:
|dw:1353112228539:dw|
zepdrix
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oh cool c:
Dido525
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No constant right? It's a definite integral.
Dido525
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@zepdrix
zepdrix
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Yah any constants would fall out c: no constant.