= Reciprocal affine sine integral
{title2=$\int_0^{2\pi}(a+b\sin\theta)^{-1}d\theta=2\pi/\sqrt{a^2-b^2}$}
For real $a>b>0$, put $d=\sqrt{a^2-b^2}$. The <change of variables> $z=e^{i\theta}$ changes the integral to $\oint_{|z|=1}(2/b)\,dz/[z^2+2i(a/b)z-1]$. The roots are $-i(a-d)/b$ and $-i(a+d)/b$, with only the first inside the circle because $a-d=b^2/(a+d)$. The <residue> there is $1/(id)$, and the <residue theorem> proves the displayed result. A shift of the angular variable replaces sine by cosine. More generally for real $a>|b|$, the same answer follows by shifting by $\pi$ when $b<0$, and the case $b=0$ follows directly.
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