= Solution
Use <integration by substitution> $u=\cos x$, $du=-\sin x\,dx$. The denominator factorizes as $(u-2)(u-3)$, and <partial fraction decomposition> gives
$$
-\int\frac{du}{(u-2)(u-3)}=\int\left(\frac1{u-2}-\frac1{u-3}\right)\,du.
$$
Integrating and restoring $x$ yields
$$
\boxed{\int\frac{\sin x}{\cos^2x-5\cos x+6}\,dx=\ln\frac{2-\cos x}{3-\cos x}+C.}
$$
For real $x$, both $2-\cos x$ and $3-\cos x$ are positive, so no absolute-value signs are needed in this real <logarithm>. The denominator never vanishes.
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