Answer
$$ - 2\ln \left| {\cos {{\left( {\frac{x}{4}} \right)}^2}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{x}{4}\tan {{\left( {\frac{x}{4}} \right)}^2}} dx \cr
& or \cr
& \int {\frac{x}{4}\tan \left( {\frac{{{x^2}}}{{16}}} \right)} dx \cr
& {\text{set }}u = \frac{{{x^2}}}{{16}}{\text{ then }}\frac{{du}}{{dx}} = \frac{x}{8},\,\,\,\,\,\,\,\,\frac{{8du}}{x} = dx \cr
& {\text{write the integrand in terms of }}u \cr
& \int {\frac{x}{4}\tan \left( {\frac{{{x^2}}}{{16}}} \right)} dx = \int {\frac{x}{4}\tan u} \left( {\frac{{8du}}{x}} \right) \cr
& {\text{cancel the common factors}} \cr
& = \int {\tan u} \left( {2du} \right) \cr
& {\text{use multiple constant rule}} \cr
& = 2\int {\tan u} du \cr
& {\text{ using the Basic Trigonometric integral }}\int {\tan x} dx = - \ln \left| {\cos x} \right| + C{\text{ }}\left( {{\text{see page 694}}} \right) \cr
& = - 2\ln \left| {\cos u} \right| + C \cr
& {\text{write in terms of }}x \cr
& = - 2\ln \left| {\cos \frac{{{x^2}}}{{16}}} \right| + C \cr
& or \cr
& = - 2\ln \left| {\cos {{\left( {\frac{x}{4}} \right)}^2}} \right| + C \cr} $$
