## Calculus (3rd Edition)

$\int sec(x)tan(x)dx=sec(x)+C$
This is an identity: $\int sec(x)tan(x)dx=sec(x)+C$ However, here it is arithmetically: $\int sec(x)tan(x)dx$ $=\int\frac{sin(x)}{cos^2(x)}dx;$ $u=cos(x);$ $du=-sin(x)dx$ $=-\int\frac{du}{u^2}$ $=\frac{1}{u}+C$ $=sec(x)+C$