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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
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