Chapter 5 - The Integral - 5.7 Substitution Method - Exercises - Page 276: 61

$$ \frac{1}{2} (\sec \theta-1)^{2}+C$$

Given $$ \int \sec \theta \tan \theta(\sec \theta-1) d \theta $$ Let $$u=\sec \theta-1 \ \ \ \ \Rightarrow \ \ \ du =\sec \theta \tan \theta d\theta $$ \begin{aligned} \int \sec \theta \tan \theta(\sec \theta-1) d \theta &=\int u d u \\ &=\frac{1}{2} u^{2}+C\\ &= \frac{1}{2} (\sec \theta-1)^{2}+C \end{aligned}