Answer
$$\int \frac{cos\,x}{sin^{2}x-9}dx=\frac{1}{6}ln\frac{sin\,x-3}{sin\,x+3}+C$$
Work Step by Step
$\int \frac{du}{u^{2}-a^{2}}=\frac{1}{2a}ln\frac{u-a}{u+a}+C$
Thus: $$\int \frac{cos\,x}{sin^{2}x-9}dx=\int \frac{d(sin\,x)}{sin^{2}x-3^{2}}$$
$$=\frac{1}{6}ln\frac{sin\,x-3}{sin\,x+3}+C$$
