Answer
$$ - \frac{1}{7}\ln \left| {\cos 7x} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\tan 7x} dx \cr
& {\text{set }}u = 7x{\text{ then }}\frac{{du}}{{dx}} = 7,\,\,\,\,\,\,\,\,\frac{{du}}{7} = dx \cr
& {\text{write the integrand in terms of }}u \cr
& \int {\tan 7x} dx = \int {\tan u\left( {\frac{{du}}{7}} \right)} \cr
& {\text{use multiple constant rule}} \cr
& = \frac{1}{7}\int {\tan udu} \cr
& {\text{ using the Basic Trigonometric integral }}\int {\tan x} dx = - \ln \left| {\cos x} \right| + C{\text{ }}\left( {{\text{see page 694}}} \right) \cr
& = \frac{1}{7}\left( { - \ln \left| {\cos u} \right|} \right) + C \cr
& = - \frac{1}{7}\ln \left| {\cos u} \right| + C \cr
& {\text{write in terms of }}x \cr
& = - \frac{1}{7}\ln \left| {\cos 7x} \right| + C \cr} $$
