AngouriMath
SolveByWritingAnArcothAsAnArtanhPlusAConstant(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)
Method (no overloads)
Summary
An inverse hyperbolic cotangent of a linear L as its tangent plus a constant:
arcoth(L) - artanh(L) is 1/2 (ln(-q) - ln(q)) for q = (1 + L)/(1 - L) ,
which isi pi/2 where |L| < 1 and -i pi/2 where |L| > 1 --
constant on each interval where both are continuous, and not the same constant on each.
With a symbolK for it the integrand is one in artanh(L) + K , which the
rules for the inverse hyperbolic tangent read; the answer is written back with
K = arcoth(L) - artanh(L) , whose derivative is zero wherever it has one, so that
it is right on every interval without a case for each.
https://github.com/asc-community/AngouriMath/issues/233
which is
constant on each interval where both are continuous, and not the same constant on each.
With a symbol
rules for the inverse hyperbolic tangent read; the answer is written back with
it is right on every interval without a case for each.
https://github.com/asc-community/AngouriMath/issues/233
Remarks
The parser writes arcoth(L) as 1/2 ln((L + 1)/(L - 1)) , which the
substitution underL = tanh(u) does not read as an inverse function: Rubi's 7.4.1
acoth(x)^2/(1 - x^2)^2 was declined in half a second where
(atanh(x) + k)^2/(1 - x^2)^2 is answered in a quarter of one. Last, after by parts,
so that every arcoth answered as written keeps its answer; at the top only, since it asks
the whole question again.
substitution under
so that every arcoth answered as written keeps its answer; at the top only, since it asks
the whole question again.
Angouri © 2019-2023 · Project's repo · Site's repo · Octicons · Transparency · 4862 pages online