Re: Indefinite integration problem on ArcTanh(f(Cos))

*To*: mathgroup at smc.vnet.net*Subject*: [mg69982] Re: Indefinite integration problem on ArcTanh(f(Cos))*From*: "David W.Cantrell" <DWCantrell at sigmaxi.net>*Date*: Thu, 28 Sep 2006 06:17:54 -0400 (EDT)*References*: <efdks2$12a$1@smc.vnet.net>

"ANDREW PALFREYMAN" <andrew_ppp at hotmail.com> wrote: > Hi. > I'm integrating Sin[t] / (1 + k Cos[2t]) , over 0 < k < 1, using > Mathematica 4.1. I'm plotting spot values k=0.1, 0.8, 0.95 over 0 < t < > 10. > > This represents the acceleration of a charged particle in an alternating > electric field, where k represents a particular modulation of its > effective inertia. Although the velocity plots (the integral dt) look > fine (symmetrical about zero), the plots for position (integrating > velocity dt) are clearly wrong; I get a massive expression (about 20 > lines!) which will not differentiate successfully back to the original > velocity expression, and they are decidedly asymmetrical about zero. The > position plot ought to look like a sinewave as k->0, but there's no > resemblance. I don't know what answer version 4.1 gives, but the current version's answer seems to be correct. Did you think of trying The Wolfram Integrator <http://integrals.wolfram.com/index.jsp> ? It's free and gives the same result for the indefinite integral as the current version of Mathematica. Anyway, if 0 < k < 1, perhaps the nicest way to write the antiderivative is - ArcTan[Sqrt[2k/(1-k)] Cos[t]] / Sqrt[2k(1-k)]. David