Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 48

Determine the integral $\displaystyle \int \frac{dx}{\cos x - 1}x$


$\begin{equation} \begin{aligned} \int \frac{1}{\cos x - 1} \cdot \frac{\cos x + 1}{\cos x + 1} dx &= \int \frac{\cos x + 1}{\cos^2 x -1} dx \qquad \text{Apply Trigonometric Identity } \cos^2 x - \sin^2 x =1\\ \\ \int \frac{\cos x + 1}{\sin^2 x } dx &= \int \left( - \frac{\cos x}{\sin^2 x} - \frac{1}{\sin^2 x} \right) dx\\ \\ \int \frac{\cos x + 1}{\sin^2 x } dx &= - \int (\cot x \csc x + \csc^2 x) dx \\ \\ \int \frac{\cos x + 1}{\sin^2 x } dx &= -(-\csc x + (-\cot x) + c)\\ \\ \int \frac{\cos x + 1}{\sin^2 x } dx &= \csc x + \cot x + c \end{aligned} \end{equation}$