RE: Re: Fwd: Antiderivatives and Definite Integrals

*To*: mathgroup at smc.vnet.net*Subject*: [mg39906] RE: [mg39868] Re: Fwd: Antiderivatives and Definite Integrals*From*: "Wolf, Hartmut" <Hartmut.Wolf at t-systems.com>*Date*: Tue, 11 Mar 2003 02:36:38 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

>-----Original Message----- >From: David W. Cantrell [mailto:DWCantrell at sigmaxi.org] To: mathgroup at smc.vnet.net >Sent: Sunday, March 09, 2003 11:29 AM >To: mathgroup at smc.vnet.net >Subject: [mg39906] [mg39868] Re: Fwd: Antiderivatives and Definite Integrals > > >Garry Helzer <gah at math.umd.edu> wrote: >[snip] >> > Redefine: >> > >> > In[6]:= f[x_] /; -Pi < x < Pi = f[x] >> > Out[6]= 2*Sqrt[1 + Cos[x]]*Tan[x/2] >> > >> > In[7]:= f[x_] /; Pi < x < 3 Pi = f[x] + 4*Sqrt[2] >> > Out[7]= 4*Sqrt[2] + 2*Sqrt[1 + Cos[x]]*Tan[x/2] >> >> Or 4Sqrt[2]Round[x/(2Pi)]+ 2*Sqrt[1 + Cos[x]]*Tan[x/2] . But these >> formulas are less than perfect since they are indeterminate at odd >> multiples of Pi. > >But of course there are "perfect" formulas for the antiderivative of >Sqrt[1 + Cos[x]]. I mentioned one such formula in the parent thread of >this one. If we let y denote Floor[(x+Pi)/(2*Pi)], then a >"perfect" (and >maximally neat?) formula for the antiderivative is > > 2*Sqrt[2]*( (-1)^y*Sin[x/2] + 2*y ) > >David Cantrell > Not quite as neat, just to indicate that there are more such formulas, see 2(Sqrt[1 - Cos[#]]*Sign[#] + (x - #)*Sqrt[2]/Pi) & [Mod[x, 2 Pi, -Pi]] or 2*Sqrt[2]*(Sin[#/2] + (x - #)/Pi) & [Mod[x, 2 Pi, -Pi]] which comes quite close to your expression. We reach those formulas when substituting the functions of half angle for tan and (in reverse) for sin in Integrate[Sqrt[1 + Cos[x]], x]. -- Hartmut Wolf