Answer
$$\ln \left| {\cos {e^{ - x}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {{e^{ - x}}\tan \left( {{e^{ - x}}} \right)} dx \cr
& = \int {\frac{{\sin {e^{ - x}}}}{{\cos {e^{ - x}}}}} \left( {{e^{ - x}}} \right)dx \cr
& {\text{substitute }}u = \cos {e^{ - x}},{\text{ }}du = {e^{ - x}}\sin {e^{ - x}}dx \cr
& = \int {\frac{{du}}{u}} \cr
& {\text{find the antiderivative }} \cr
& = \ln \left| u \right| + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = \cos {e^{ - x}} \cr
& \ln \left| {\cos {e^{ - x}}} \right| + C \cr} $$
