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 Post subject: complex analysisPosted: Tue, 24 May 2011 18:15:10 UTC
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Joined: Tue, 24 May 2011 18:01:46 UTC
Posts: 2
hi

i really like to know how to solve this integrate using complex analysis theory:

Integrate[Cos[3 x]^2/(5 - 4 Cos[2 x]), {x, -Pi, Pi}]

i tryed to solve this using numerical methods, but i want to know the complex analysis solvation.

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 Post subject: Re: complex analysisPosted: Tue, 24 May 2011 18:23:31 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12103
Location: Austin, TX
ali_er wrote:
hi

i really like to know how to solve this integrate using complex analysis theory:

Integrate[Cos[3 x]^2/(5 - 4 Cos[2 x]), {x, -Pi, Pi}]

i tryed to solve this using numerical methods, but i want to know the complex analysis solvation.

(by periodicity)

You want to use that on the unit circle to convert that integral into a path integral around , counterclockwise, et cetera. Since you have a you'll need to actually do on the unit circle, but then it will be an easy rational function with no zeros or poles on the boundary.

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 Post subject: Posted: Tue, 24 May 2011 19:04:13 UTC
 S.O.S. Newbie

Joined: Tue, 24 May 2011 18:01:46 UTC
Posts: 2
sorry, but i don't understand!

how can i solve an integrate with complex analysis without zeros or poles on the boundary?

i'm confused...

and 1 other thing, i guess Cos(nx)=( z^n + z^(-n) ) / 2

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 Post subject: Posted: Tue, 24 May 2011 19:08:04 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12103
Location: Austin, TX
ali_er wrote:
sorry, but i don't understand!

how can i solve an integrate with complex analysis without zeros or poles on the boundary?

i'm confused...

and 1 other thing, i guess Cos(nx)=( z^n + z^(-n) ) / 2

Yes, that's right the 1/2 should be there. And why on Earth would you want zeros or poles on the boundary, that would prevent you from using the Residue theorem/Argument principle.

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