= Solution
Put $z=e^{i\theta}$, so $\sin\theta=(z-z^{-1})/(2i)$ and $d\theta=dz/(iz)$. The <integral> becomes
$$
\oint_{|z|=1}\frac{dz}{6z^2+13iz-6}.
$$
The poles are $-2i/3$ and $-3i/2$; only the first is inside the unit circle. The <derivative> of the denominator there is $12(-2i/3)+13i=5i$. The <residue theorem> therefore gives
$$
\boxed{\int_0^{2\pi}\frac{d\theta}{13+12\sin\theta}=\frac{2\pi}{5}.}
$$
There are no poles on the contour because the real denominator is at least one.
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