## Thomas' Calculus 13th Edition

$$\frac{1}{3}{\left( {\tan \theta } \right)^3} + \tan \theta + C$$
\eqalign{ & \int {{{\sec }^4}\theta } d\theta \cr & {\text{We write }}{\sec ^4}\theta {\text{ as }}{\sec ^2}\theta {\sec ^2}\theta {\text{ }} \cr & = \int {{{\sec }^2}\theta } {\sec ^2}\theta d\theta \cr & {\text{Use the fundamental identiy ta}}{{\text{n}}^2}\theta = {\sec ^2}\theta - 1 \cr & = \int {\left( {{{\tan }^2}\theta + 1} \right)} {\sec ^2}\theta d\theta \cr & {\text{Integrate by the substitution method}} \cr & {\text{Let }}u = \tan \theta,\,\,\,\,du = {\sec ^2}\theta d\theta,\,\,\,\,d\theta = \frac{{du}}{{{{\sec }^2}\theta }} \cr & {\text{Then}}{\text{,}} \cr & = \int {\left( {{u^2} + 1} \right)} {\sec ^2}\theta \left( {\frac{{du}}{{{{\sec }^2}\theta }}} \right) \cr & {\text{Cancel common factor }}{\sec ^2}\theta \cr & = \int {\left( {{u^2} + 1} \right)} du \cr & {\text{Integrate}} \cr & = \frac{1}{3}{u^3} + u + C \cr & {\text{Write in terms of }}\theta,\,\,\,u = \tan \theta \cr & = \frac{1}{3}{\left( {\tan \theta } \right)^3} + \tan \theta + C \cr}