integrales de funciones trigonometricas hiperbolicas
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04-16-2019, 09:13 PM
Post: #1
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integrales de funciones trigonometricas hiperbolicas
I recently tried to calculate an integral with hyperbolic trigonometric functions and I got the result wrong, please notify me of this to hp to solve it in the next firmware update, so I think I only think that the tanh (x), the sinh ( x) and the cosh (x) are fine, I do not know if more things fail[/size][/font]
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04-17-2019, 04:18 AM
Post: #2
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RE: integrales de funciones trigonometricas hiperbolicas
Please indicate which specific integral gave you the wrong result.
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04-17-2019, 04:28 AM
Post: #3
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RE: integrales de funciones trigonometricas hiperbolicas
I suspect "wrong result" will simply be "not in the form I think it should be", but we shall see...
Unless you provide your input, and expected output. Nobody can help in any way. TW Although I work for the HP calculator group, the views and opinions I post here are my own. |
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04-17-2019, 06:17 AM
Post: #4
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RE: integrales de funciones trigonometricas hiperbolicas
(04-16-2019 09:13 PM)eduardo_MO0@hotmail.com Wrote: I recently tried to calculate an integral with hyperbolic trigonometric functions and I got the result wrong, please notify me of this to hp to solve it in the next firmware update, so I think I only think that the tanh (x), the sinh ( x) and the cosh (x) are fine, I do not know if more things fail[/size][/font] Could you please show any example ? Thanks and best, Aries ![]() |
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04-20-2019, 02:02 AM
Post: #5
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RE: integrales de funciones trigonometricas hiperbolicas
At integrating an equivalent to tanh(x)' the result is not as expected. There is no problem with sinh and cosh.
![]() Observation, using an equivalent expression returns the same result. ![]() Viga C | TD | FB |
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04-20-2019, 04:21 AM
(This post was last modified: 04-20-2019 04:24 AM by Wes Loewer.)
Post: #6
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RE: integrales de funciones trigonometricas hiperbolicas
If I were doing this integral by hand, I would do the following:
∫(1-tanh(x)^2) dx = ∫sech(x)^2 dx = tanh(x)+C The Prime CAS gives the result as -2/((e^x)^2+1)+C At first glance, these might look different, but since tanh(x) - 1 = -2/((e^x)^2+1), then the two expressions differ only by a constant. This means that the two expressions are both correct, they just have different integration constants. If you look at other CAS's, you'll see different but equivalent results. Maxima: 2/(e^(-2*x)+1) Nspire: −2/(e^(2*x)+1) WolframAlpha: tanh(x) When my students say that they got a different answer than the textbook, I encourage them to graph both results. If they get the same graph but shifted up or down, then they can be reasonable certain that their answers are equivalent. |
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04-22-2019, 06:43 AM
Post: #7
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RE: integrales de funciones trigonometricas hiperbolicas
I was asked privately about the integral ∫tanh(x) dx
The Prime gives ∫(tanh(x),x) --> ln((e^x)^2+1)-x which is correct. By hand I would have done the following: ∫tanh(x) dx = ∫sinh(x)/cosh(x) dx = ln(cosh(x))+C but this can be rewritten as = ln((e^x+e^-x)/2) + C = ln((e^(2x)+1)/(2e^x)) + C = ln(e^2x)+1) - ln(2e^x) + C = ln(e^2x)+1) - ln(2) - ln(e^x) + C = ln(e^2x)+1) - x + (C-ln(2)) = ln(e^2x)+1) - x + D Once again, the correct answers can be rewritten such that they differ by only a constant. I've learned over the years that whenever a CAS's antiderivative looks different than mine, it's usually that they differ by a constant, or my antiderivative is wrong. :-) |
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