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Re: x-ArcSin[Sin[x]]

  • To: mathgroup at
  • Subject: [mg50063] Re: x-ArcSin[Sin[x]]
  • From: Paul Abbott <paul at>
  • Date: Fri, 13 Aug 2004 05:56:33 -0400 (EDT)
  • Organization: The University of Western Australia
  • References: <cfcqmu$446$>
  • Sender: owner-wri-mathgroup at

In article <cfcqmu$446$1 at>,
 Paul Muller < at> 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 

  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.


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 

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