complex analysis

ali_er · **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.

Shadow · **Posted:** Tue, 24 May 2011 18:23:31 UTC

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.

$$\int_{-\pi}^{\pi}{\cos^2 (3x)\over 5-4\cos(2x)}\,dx=\int_0^{2\pi}{\cos^2 (3x)\over 5-4\cos(2x)}\,dx$ (by periodicity)

You want to use that $\cos z=z+{1\over z}$ on the unit circle to convert that integral into a path integral around |z|=1

, counterclockwise, et cetera. Since you have a $\cos nx$ you'll need to actually do $\cos nz=z^n+z^{-n}$ on the unit circle, but then it will be an easy rational function with no zeros or poles on the boundary.

ali_er · **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

Shadow · **Posted:** Tue, 24 May 2011 19:08:04 UTC

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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