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