Small mistake, complex calculus

[joris_pixie]

Im trying to solve:
But i always come out:
$$\int^{2PI}_{0}$$ (sin²(x) / (5 - 4cos(x))dx

==> (i/4)$$\oint(z^2-1)^2 / (z^2(z-2)(z-1/2)) dz$$= 2pi i (i/4) (RES (z=0) + RES (z = 1/2)) = (-pi/2) (5/2 - 3/2) = -pi/2

instead of pi/4 please help

 

[Dick]

I can't really tell what you are doing wrong from what you've posted, but I get a factor of (-i/8) in front of your contour integral.

 

[joris_pixie]

I can't really tell what you are doing wrong from what you've posted, but I get a factor of (-i/8) in front of your contour integral.

That would do the trick, but could you please write how you get to -i/8 in front of it ?

 

[Hurkyl]

How did you get your i/4?

 

[joris_pixie]

my mistake

my mistake should be here:

http://www.pixie.be/wrong.jpg

But I can't see it ...
Could some one please help ? :(

 

[Dick]

In going from the second contour to the third you seem to have just moved an i from the denominator to the numerator. You are now off by a sign. 5z-2z^2-2=(-2)(z-2)(z-1/2). There's another sign and a factor of two you made disappear.

 

[joris_pixie]

In going from the second contour to the third you seem to have just moved an i from the denominator to the numerator.

ow yes, that was a typing error !

You are now off by a sign. 5z-2z^2-2=(-2)(z-2)(z-1/2).

Aha, got it;
After studying a lot of math, it's mostly the problem that i forget to do the 'simple' things correct ;) haha
Stupid mistake, but thank you so much !