Answer
$$\displaystyle{\int{\frac{\sec \theta \tan \theta}{\sec ^2\theta -\sec \theta}}d\theta=\ln \left| 1-\cos \theta \right|+C}$$
Work Step by Step
$\displaystyle{I=\int{\frac{\sec \theta \tan \theta}{\sec ^2\theta -\sec \theta}d\theta}\\
I=\int{\frac{\sec \theta \tan \theta}{\sec \theta \left( \sec \theta -1 \right)}d\theta}\\
I=\int{\frac{\frac{\sin \theta}{\cos \theta}}{\frac{1}{\cos \theta}-1}d\theta}\\
I=\int{\frac{\frac{\sin \theta}{\cos \theta}}{\frac{1-\cos \theta}{\cos \theta}}d\theta}\\
I=\int{\frac{\sin \theta}{1-\cos \theta}d\theta}}$
$\displaystyle{\left[\begin{matrix}
u=1-\cos \theta& d\theta =\frac{du}{\sin \theta}\\
\end{matrix}\right]}$
$\displaystyle{I=\int{\frac{\sin \theta}{u}\times \frac{du}{\sin \theta}}\\
I=\int{\frac{1}{u}du}\\
I=\ln \left| u \right|+C\\
I=\ln \left| 1-\cos \theta \right|+C}
$
