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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]
Please indicate which specific integral gave you the wrong result.
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.
(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 Wink
At integrating an equivalent to tanh(x)' the result is not as expected. There is no problem with sinh and cosh.

[Image: 46923421524_872517f26b_o.png]

Observation, using an equivalent expression returns the same result.

[Image: 46923421474_c4177df590_o.png]
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.
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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