I tried to solve the following integral using Residue Theorem but I am not sure about its singular values. Any other way without using Residue Theorem is also appreciated.
\begin{equation} \int_{-\pi}^{\pi}\frac{\sin^2 \theta}{(\frac{1}{2}+iw\cos \theta)^4} d\theta \end{equation}
Here $i$ is the imaginary unit and w is an arbitrary real number.
Let$$R(x,y)=\frac{y^2}{\left(\frac12+iwx\right)^4}.$$Then the integral that you wish to compute is $\int_{-\pi}^\pi R(\cos\theta,\sin\theta)\,\mathrm d\theta$. Let\begin{align}f(z)&=\frac1zR\left(\frac{z+1/z}2,\frac{z-1/z}{2i}\right)\\&=-\frac{4z\left(z^2-1\right)^2}{\left(w z^2+w-i z\right)^4}.\end{align}Then the poles of $f$ are located at$$\frac{1\pm\sqrt{1+4w^2}}{2w}i$$and the only pole inside the unit circle is the one with the $-$ sign. The residue of $f$ at that point is $\frac8{\left(4w^2+1\right)^{5/2}}$ and therefore your integral is equal to$$\frac{16\pi}{\left(4w^2+1\right)^{5/2}}.$$