Complex analysis $\int^{2\pi}_0 \frac{d \theta}{(2-\sin \theta)^2}$

Question

how do I compute

$$\int^{2\pi}_0 \frac{d \theta}{(2-\sin \theta)^2}$$

I tried substituting $z=e^{i\theta}$ but it just got very messy..

Answer 1

Can't suggest an alternative method, but here are some suggestions for coping with the calculations after substituting $z=e^{i\theta}$.

If my algebra is correct, you should end up having to find the residue of $$\frac{z}{(z^2-4iz-1)^2}$$ at its unique singular point inside the unit circle. You will often find that it is simplest to do this kind of thing in symbolic form and then substitute back actual values at the end. So, write the above as $$\frac{z}{(z-\alpha)^2(z-\beta)^2}\ .$$ It is easy to find $$\alpha=(2-\sqrt3)i\ ,\quad\beta=(2+\sqrt3)i\ ,$$ and the residue is $$\frac{d}{dz}\frac{z}{(z-\beta)^2}\Bigg|_{z=\alpha}\ .$$ You can find this initially in terms of $\alpha$ and $\beta$, then simplify using the facts $$\alpha-\beta=-2\sqrt3i\ ,\quad \alpha+\beta=4i\ .$$ It's not all that hard if you do it this way, just take it slowly and carefully.