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