Answer
$$\frac{1}{2}\tan 2x - x + C$$
Work Step by Step
$$\eqalign{
& \int {{{\tan }^2}2x} dx \cr
& {\text{write in terms of }}u.{\text{ let }}u = 2x,\,\,\,du = 2dx \cr
& \int {{{\tan }^2}2x} dx = \int {{{\tan }^2}u\left( {\frac{1}{2}du} \right)} \cr
& = \frac{1}{2}\int {{{\tan }^2}udu} \cr
& \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral by the formula 28}} \cr
& \left( {28} \right)\int {{{\tan }^2}u} du = \tan u - u + C \cr
& \frac{1}{2}\int {{{\tan }^2}udu} = \frac{1}{2}\tan u - \frac{1}{2}u + C \cr
& \cr
& {\text{Write in terms of }}x{\text{, and replace 2}}x{\text{ for }}u \cr
& = \frac{1}{2}\tan 2x - \frac{1}{2}\left( {2x} \right) + C \cr
& = \frac{1}{2}\tan 2x - x + C \cr} $$
