Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.3 Exercises - Page 530: 53

$$\int \frac{1}{\sec x \tan x} d x =\ln |\csc t+ \cot t| +\cos x+c$$

$$ \int \frac{1}{\sec x \tan x} d x $$ Since \begin{align*} \int \frac{1}{\sec x \tan x} d x&=\int \frac{1}{\frac{1}{\cos x} \frac{\sin x}{\cos x} } d x\\ &=\int \frac{\cos^2x}{ \sin x } d x\\ &=\int \frac{1-\sin^2x}{ \sin x } d x\\ &=\int (\csc x- \sin x)dx\\ &=\ln |\csc t+ \cot t| +\cos x+c \end{align*}