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