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

>