Date: Oct 10, 2013 5:17 AM
Author: Nasser Abbasi
Subject: Re: The A. F. Timofeev symbolic integration test suite

On 10/10/2013 1:21 AM, Albert Rich wrote:

>So I was delighted to discover the following identity that makes
>it easy to transform discontinuous antiderivatives into continuous ones:
>
> If g(x) = arctan(a*tan f(x)) and
>h(x) = f(x) - arctan(cos f(x)*sin f(x) / (a/(1-a) + cos f(x)^2), <-- extra ")" missing
>
>then g?(x) = h?(x).
>


Fyi, verified it:

Mathematica:

g[x_] := ArcTan[a*Tan [f[x]]]
h[x_] := f[x]-ArcTan[ Cos[f[x]]* Sin[f[x]] /(a/(1-a) + Cos[f[x]]^2)]
r1 = D[g[x], x];
r2 = D[h[x], x];
Simplify[r1 - r2]
(* 0 *)

Maple:

g(x) := arctan(a*tan(f(x)));
h(x) := f(x)-arctan(cos(f(x))*sin(f(x)) /(a/(1-a) + cos(f(x))^2));
simplify( diff(g(x),x)-diff(h(x),x));
0

>
> The next release of Rubi will take advantage of these identities to produce
>continuous trig antiderivatives, thereby making the computation of definite integrals easy and reliable.
>


Looking forward for Rubi 4.3 !

> Albert
>