Answer
$${e^{1/2}} - 1$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /4} {{e^{{{\sin }^2}x}}\sin 2x} dx \cr
& \int_0^{\pi /4} {{e^{{{\sin }^2}x}}\left( {2\sin x\cos x} \right)} dx \cr
& {\text{substitute }}u = {\sin ^2}x,{\text{ }}du = 2\sin x\cos xdx \cr
& {\text{express the limits in terms of }}u \cr
& x = \pi /4{\text{ implies }}u = {\sin ^2}\left( {\pi /4} \right) = 1/2 \cr
& x = 0{\text{ implies }}u = {\sin ^2}0 = 0 \cr
& {\text{the entire integration is carried out as follows}} \cr
& \int_0^{\pi /4} {{e^{{{\sin }^2}x}}\left( {2\sin x\cos x} \right)} dx = \int_0^{1/2} {{e^u}du} \cr
& {\text{find the antiderivative}} \cr
& = \left. {\left( {{e^u}} \right)} \right|_0^{1/2} \cr
& {\text{use the fundamental theorem}} \cr
& = {e^{1/2}} - {e^0} \cr
& {\text{simplify}} \cr
& = {e^{1/2}} - 1 \cr} $$
