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MathGroup Archive 2003

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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



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