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MathGroup Archive 2006

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Re: Solving integrals

  • To: mathgroup at smc.vnet.net
  • Subject: [mg68663] Re: Solving integrals
  • From: Jens Benecke <jens-news at spamfreemail.de>
  • Date: Mon, 14 Aug 2006 06:44:47 -0400 (EDT)
  • References: <ebhi41$11p$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Jens Benecke wrote:

> Hello everybody,
> 
> I am trying to solve this integral:
> 
> f :== int ( cos(x) / sqrt[A=C2=B2=C2=B7sin=C2=B2(x/2) + (Bx)=C2=B2 =
> ] dx, x==0..N*2*pi

Hello,

sorry, this post apparently got mangled ("²" aka "^2" wasn't recognized).
Here is the integral again:

f := int ( cos(x) / sqrt[ A*A*sin(x/2)*sin(x/2) + (Bx)^2] dx, x=0..N*2*pi


> where N==(int)0..20, and A and B are small fixed values (around
> 0.0001..0.05).
 
> The problem is the additional summand in the sqare root of the denominator
> (ie., (Bx)=C2=B2). Without this part the integral would be an elliptic one
> or could be represented with a combination of E(x) and F(x).
> 
> I have tried developing the sine function into rows (x-x^3/3!+...) but to
> be precise enough I would have to go up to x^150/150=C2=B0 which is hardly
> practical. I have also tried seperating the denominator into expressions
> where either part of the sum is << or >> the other part so that suitable
> approximations can be made, without much luck.
> 
> 
> I would really appreciate any kind of hint or idea on how to tackle this
> problem.
> 
> Thank you! :)

-- 
Jens Benecke
Please do not carbon copy me, I read the lists I post in!


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