Answer
$$\frac{1}{2}{\sin ^{ - 1}}{e^x} + \frac{1}{2}{e^x}\sqrt {1 - {e^{2x}}} + C$$
Work Step by Step
$$\eqalign{
& \int {{e^x}\sqrt {1 - {e^{2x}}} } dx \cr
& {\text{substitute }}{e^x} = \sin \theta ,{\text{ }}{e^x}dx = \cos \theta d\theta \cr
& = \int {\sqrt {1 - {{\sin }^2}\theta } } \cos \theta d\theta \cr
& {\text{identity co}}{{\text{s}}^2}\theta + {\sin ^2}\theta = 1 \cr
& = \int {\sqrt {{{\cos }^2}\theta } } \cos \theta d\theta \cr
& = \int {{{\cos }^2}\theta } d\theta \cr
& {\text{identity co}}{{\text{s}}^2}\theta = \frac{{1 + \cos 2\theta }}{2} \cr
& = \int {\frac{{1 + \cos 2\theta }}{2}} d\theta \cr
& = \int {\left( {\frac{1}{2} + \frac{{\cos 2\theta }}{2}} \right)} d\theta \cr
& {\text{find antiderivative}} \cr
& = \left( {\frac{\theta }{2} + \frac{{\sin 2\theta }}{4}} \right) + C \cr
& = \frac{\theta }{2} + \frac{{\sin 2\theta }}{4} + C \cr
& = \frac{\theta }{2} + \frac{{2\sin \theta \cos \theta }}{4} + C \cr
& = \frac{\theta }{2} + \frac{{\sin \theta \cos \theta }}{2} + C \cr
& {\text{write in terms of }}x \cr
& = \frac{1}{2}{\sin ^{ - 1}}{e^x} + \frac{{{e^x}\left( {\sqrt {1 - {e^{2x}}} } \right)}}{2} + C \cr
& = \frac{1}{2}{\sin ^{ - 1}}{e^x} + \frac{1}{2}{e^x}\sqrt {1 - {e^{2x}}} + C \cr} $$
