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Re: x-ArcSin[Sin[x]]
*To*: mathgroup at smc.vnet.net
*Subject*: [mg50063] Re: x-ArcSin[Sin[x]]
*From*: Paul Abbott <paul at physics.uwa.edu.au>
*Date*: Fri, 13 Aug 2004 05:56:33 -0400 (EDT)
*Organization*: The University of Western Australia
*References*: <cfcqmu$446$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
In article <cfcqmu$446$1 at smc.vnet.net>,
Paul Muller <paul.muller-at-epfl.ch at sicinfo2.epfl.ch> wrote:
> I am calculating the magnetic field in any point of the plane of a spire (or
> spiral inductor). When I apply this to the center of the spire, I get the
> following integral:
>
> Integrate[Sin[x+ArcSin[Sin[Pi/2-x]]], {x, 0, 2Pi}]
>
> According to textbook results for magnetic fields, the result should be 2Pi.
> Because of the definition of ArcSin[Sin[x]] in Mathematica, the integral
> results in Pi.
> Maybe I'm not careful enough when using the ArcSin function on the angles of
> my geometric problem (I'm an engineer, not a mathematician), but it would
> be helpful if this integral was solved "correctly".
Indeed, you are not being careful enough. If you enter
Plot[Sin[x+ArcSin[Sin[Pi/2-x]]], {x, 0, 2Pi}]
you will see why Mathematica returns Pi for the integral. Also, if you
enter
FullSimplify[Sin[x+ArcSin[Sin[Pi/2-x]]], 0 < x < Pi]
you will see get the reduction you expect over the (reduced) range
0 < x < Pi.
Cheers,
Paul
--
Paul Abbott Phone: +61 8 9380 2734
School of Physics, M013 Fax: +61 8 9380 1014
The University of Western Australia (CRICOS Provider No 00126G)
35 Stirling Highway
Crawley WA 6009 mailto:paul at physics.uwa.edu.au
AUSTRALIA http://physics.uwa.edu.au/~paul
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