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it's the Integral sin^2(x)/(3-2cos(x)) dx from x=0 to x=2pi

I tried the substitution z=e^(itheta) plugging in sin and cos as functions of z and I

ultimately get:

i * integral (z^4+2z^2+1)/(z^2-3z+1)dz over |z|=1

Then the poles inside the unit circle are @ z=0 and (3-sqrt(5))/2

so I find the residue @ z=0 to be 3 and @ (3-sqrt5)/2 something like

(looking in my mess of notes) 4(15-6sqrt(5))^2/((3-sqrt(5)^2(-sqrt(5))

then I plug in 2pi X Sum residues X i

but it doesn't agree with the answer I get on my calculator which is around 1.19

any help is really appreciated thanks!