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: <email@example.com>
- 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 ("Â²" 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!