Answer
$$ - \frac{1}{{10}}\ln \left| {\cos 10x} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\tan 10x} dx \cr
& {\text{identity tan}}\theta = \frac{{\sin \theta }}{{\cos \theta }} \cr
& = \int {\frac{{\sin 10x}}{{\cos 10x}}} dx \cr
& {\text{substitute }}u = \cos 10x,{\text{ }}du = - 10\sin 10xdx \cr
& = \int {\frac{{ - 1/10}}{u}} du \cr
& = - \frac{1}{{10}}\int {\frac{1}{u}} du \cr
& {\text{find the antiderivative}} \cr
& = - \frac{1}{{10}}\ln \left| u \right| + C \cr
& {\text{with }}u = \cos 10x \cr
& = - \frac{1}{{10}}\ln \left| {\cos 10x} \right| + C \cr} $$
