Answer
$$
\int \frac{\cos x}{\sin ^{2} x-9} d x=\frac{1}{6} \ln \left|\frac{\sin x-3}{\sin x+3}\right|+C
$$
Work Step by Step
$$
\int \frac{\cos x}{\sin ^{2} x-9} d x
$$
Let
$$
u=\sin x\quad d u=\cos x d x
$$
If we look at the Table of Integrals , we see that the closest entry is number $20$ with $a=3:$
$$
\int \frac{\cos x}{\sin ^{2} x-9} d x=\int \frac{1}{u^{2}-9} d u \quad\left[\begin{array}{c}
u=\sin x \\
d u=\cos x d x
\end{array}\right] \\
\quad \stackrel{20}{=} \frac{1}{2(3)} \ln \left|\frac{u-3}{u+3}\right|+C \\
\quad =\frac{1}{6} \ln \left|\frac{\sin x-3}{\sin x+3}\right|+C
$$
