integrales de funciones trigonometricas hiperbolicas
04-16-2019, 09:13 PM
Post: #1
 eduardo_MO0@hotmail.com Junior Member Posts: 5 Joined: Apr 2019
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]
04-17-2019, 04:18 AM
Post: #2
 Wes Loewer Member Posts: 162 Joined: Jan 2014
RE: integrales de funciones trigonometricas hiperbolicas
Please indicate which specific integral gave you the wrong result.
04-17-2019, 04:28 AM
Post: #3 Tim Wessman Senior Member Posts: 2,108 Joined: Dec 2013
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.
04-17-2019, 06:17 AM
Post: #4
 Aries Member Posts: 131 Joined: Oct 2014
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 04-20-2019, 02:02 AM
Post: #5 Carlos295pz Member Posts: 296 Joined: Sep 2015
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
04-20-2019, 04:21 AM (This post was last modified: 04-20-2019 04:24 AM by Wes Loewer.)
Post: #6
 Wes Loewer Member Posts: 162 Joined: Jan 2014
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.
04-22-2019, 06:43 AM
Post: #7
 Wes Loewer Member Posts: 162 Joined: Jan 2014
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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