Question

Integration question regarding inverse hyperbolic trigonometric functions, math posted below in a minute. Just a curiosity.

Answer 1

If I had an integral such that the integrand eventually becomes the value necessary to integrate for either the inverse hyperbolic tangent or inverse hyperbolic cotangent,
(might be a little while, gimme a sec to write all this)

Answer 2

\[\int\limits_{}^{}\frac{ 1 }{ a^{2}-x^{2} }dx = \tanh^{-1}(\frac {x}{a}) + C, |x| < 1\]\[\int\limits_{}^{}\frac{ 1 }{ a^{2}-x^{2} }dx = \cot^{-1}(\frac {x}{a}) + C, |x| > 1\]
If it was a definite integral where my upper and lower bounds were respectively greater than and less than one, would when I evaluate it, I have the upper bound integral be the inverse arccot and the lower bound integral be the inverse arctan?

Answer 3

e.g. \[\int\limits_{c}^{d} \frac{1}{a^{2}-x^{2}}, d > 1 , c < 1\]
\[\int\limits_{c}^{d} \frac {1}{a^{2}-x^{2}}dx = \coth^{-1}(\frac{d}{a}) - \tanh^{-1}(\frac{c}{a})\]

Answer 4

Whoops, forgot, those first two formulae, the inverse arctan and inverse arccot have a 1/a rright before them. But no matter, the question still stands.

Answer 5

?

Answer 6

Well I think part of it comes from the fact that when you are on the unit hyperbola, hyperbolic cosine is not defined for values less than 1.

Not sure, I'm just sort of playing with the integral right now to see if there's a better reason.