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