Answer
$$ - \frac{1}{5}\ln \left| {\cos 5x} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\tan 5x} dx \cr
& = \int {\frac{{\sin 5x}}{{\cos 5x}}} dx \cr
& {\text{substitute }}u = \cos 5x,{\text{ }}du = - 5\sin 5xdx \cr
& \int {\frac{{\sin 5x}}{{\cos 5x}}} dx = \int {\frac{{ - 1/5}}{u}} du \cr
& = - \frac{1}{5}\int {\frac{1}{u}} du \cr
& {\text{find the antiderivative }} \cr
& = - \frac{1}{5}\ln \left| u \right| + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = \cos 5x \cr
& = - \frac{1}{5}\ln \left| {\cos 5x} \right| + C \cr} $$
