Answer
$$ - \frac{1}{4}{e^{ - 2x}} - \frac{1}{8}\sin \left( {2{e^{ - 2x}}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {{e^{ - 2x}}{{\cos }^2}\left( {{e^{ - 2x}}} \right)} dx,\,\,\,\,u = {e^{ - 2x}} \cr
& {\text{Using the given substitution}} \cr
& u = {e^{ - 2x}},\,\,\,\,\,\,\,du = - 2{e^{ - 2x}}dx,\,\,\,\,\,dx = - \frac{{du}}{{2{e^{ - 2x}}}}\, \cr
& {\text{write in terms of }}u \cr
& \int {{e^{ - 2x}}{{\cos }^2}\left( {{e^{ - 2x}}} \right)} dx = \int {{e^{ - 2x}}{{\cos }^2}u} \left( { - \frac{{du}}{{2{e^{ - 2x}}}}\,} \right) \cr
& = \int {{{\cos }^2}u\left( { - \frac{1}{2}du} \right)} \cr
& = - \frac{1}{2}\int {{{\cos }^2}u} du \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{By formula 27}} \cr
& \left( {27} \right):\,\,\,\,\,\,\int {{{\cos }^2}udu} = \frac{1}{2}u + \frac{1}{4}\sin 2u + C \cr
& - \frac{1}{2}\int {{{\cos }^2}u} du = - \frac{1}{4}u - \frac{1}{8}\sin 2u + C \cr
& {\text{write in terms of }}x{\text{, and replace }}{e^{ - 2x}}{\text{ for }}u \cr
& = - \frac{1}{4}{e^{ - 2x}} - \frac{1}{8}\sin \left( {2{e^{ - 2x}}} \right) + C \cr} $$
