<s>[INST] What is $ \dfrac{i}{4-π} \int_{|z|=4} \dfrac{dz}{z \cos z}$? 
How to evaluate the integral   $$ \dfrac{i}{4-π} \int_{|z|=4} \dfrac{dz}{z \cos z}?$$ 

Here $f(z)=\dfrac{1}{z \cos z}$ has poles at $0$ and $\frac{\pm π}{2}$ .
Residue at $z=0 $ is 1 and residues at remaining poles add up to give 0. So the integral using Cauchy integral formula is $2π(4-π)$. 
I think I am wrong. How to get the integral?
 [/INST] It's not true that residues add up to $0$. We have
$$ \cos z = \sin(\frac{\pi}{2}-z) = \sin(\frac{\pi}{2}+z)$$
so
$$ {\rm Res}_{z=0} \frac{1}{z\cos z} = \lim_{z\rightarrow 0}\frac{1}{\cos z} = 1$$
$$ {\rm Res}_{z=\frac{\pi}{2}} \frac{1}{z\cos z} = \lim_{z\rightarrow \frac{\pi}{2}}\frac{(z-\frac{\pi}{2})}{z \sin(\frac{\pi}{2}-z)} = -\frac2\pi$$
$$ {\rm Res}_{z=-\frac{\pi}{2}} \frac{1}{z\cos z} = \lim_{z\rightarrow -\frac{\pi}{2}}\frac{(z+\frac{\pi}{2})}{z \sin(\frac{\pi}{2}+z)} = -\frac2\pi$$
and $$ \sum {\rm Res} = 1 - \frac{4}{\pi} = \frac{\pi-4}{\pi}$$
so the final result is $$ \frac{i}{4-\pi}\cdot 2\pi i\frac{\pi-4}{\pi} = 2$$
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