Answer
$$\int \frac{\sec \theta \tan \theta}{\sec ^{2} \theta - \sec \theta}d\theta=ln\left | 1-\cos \theta \right |+C$$
Work Step by Step
$$\int \frac{\sec \theta \tan \theta}{\sec ^{2} \theta - \sec \theta}d\theta=\int \frac{\tan \theta}{\sec \theta - 1}d\theta$$
$$=\int \frac{\tan \theta}{\sec \theta - 1}\frac{\cos \theta}{\cos \theta}d\theta=\int \frac{\sin\theta}{1-\cos \theta}d\theta$$
$$=ln\left | 1-\cos \theta \right |+C$$
