I am trying to evaluate the integral $\int_0^{\pi/2}\frac{1}{\sin^2 t+(\sin t)^{-2}}dt$. The first step I took was using symmetry to get $$\int_{0}^{\pi/2}\frac{1}{\sin^2 t+(\sin t)^{-2}}dt=\frac{1}{4}\int_{0}^{2\pi}\frac{1}{\sin^2t+(\sin t)^{-2}}dt.$$ Then if $z=e^{it}$, then we have $dz=ie^{it}dt=izdt$, and moreover $$\frac{1}{\sin^2t+(\sin t)^{-2}}=\frac{1}{(\frac{z-z^{-1}}{2i})^2+(\frac{2i}{z-z^{-1}})^2}=\cdots =\frac{-4z^2(z^2-1)^2}{z^8-4z^6+22z^4-4z^2+1}.$$ Hence the integral becomes $$\frac{1}{4}\oint_{|z|=1}\frac{-4z^2(z^2-1)^2}{z^8-4z^6+22z^4-4z^2+1}\frac{dz}{iz}.$$ My question is, how do we proceed from here? Using the residue theorem requires first computing the roots of $z^8-4z^6+22z^4-4z^2+1$ lying inside the unit circle, but I wasn't able to do that by hand. What is a more efficient way?
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You need not to decompose the denominator polynomial. Your integral can be written as
$${1\over 4}\oint_{|z|=1}{1\over iz}{({z-z^{-1}\over 2i})^2\over ({z-z^{-1}\over 2i})^4+1}dz$$with singularities lying in $z=0$ and the roots of $({z-z^{-1}\over 2i})^4=-1$ or $${z-z^{-1}\over 2i}={\pm 1\pm i\over \sqrt2}$$I leave to you finding the residue of the function at the singularities.
Mostafa Ayaz
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$$ f(z)=\frac1{z^2+z^{-2}}=\frac{z^2}{z^4+1} $$
and $h(z)=\frac1{2\mathrm i}\left(z-z^{-1}\right)=\sin\left(\frac{\log z}{\mathrm i}\right)$, but I think in view of the simple solution already provided by Mostafa Ayaz it's more trouble than it's worth.
– joriki Apr 23 '20 at 06:59