Answer
$$ \int \frac{\sin x}{\cos ^{2} x-3 \cos x} d x=\frac{1}{3} \ln \left|\frac{\cos x}{\cos x-3}\right|+C$$
Work Step by Step
Given$$ \int \frac{\sin x}{\cos ^{2} x-3 \cos x} d x$$
Let
\begin{array}{l}{\cos x=u} \\ {\sin x\ d x=d u}\end{array}
So, we get
\begin{aligned} I&= \int \frac{\sin x}{\cos ^{2} x-3 \cos x} d x\\
&=\int \frac{-1}{u^{2}-3 u} d u \\ &=\int \frac{-1}{u(u-3)} d u \\ &=\frac{1}{3} \int \frac{-3}{u(u-3)} d u \\
&= \frac{1}{3} \int \frac{(u-3)-u}{u(u-3)} d u \\
&= \frac{1}{3} \int \frac{(u-3)}{u(u-3)}-\frac{u}{u(u-3)} d u \\
&=\frac{1}{3} \int \frac{1}{u}-\frac{1}{(u-3)} d u\\
&=\frac{1}{3}[\ln |u|-\ln |u-3|]+C\\
& =\frac{1}{3} \ln \left|\frac{u}{u-3}\right|+C \\
& =\frac{1}{3} \ln \left|\frac{\cos x}{\cos x-3}\right|+C\end{aligned}
