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

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.

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. :-)