Answer
$$0$$
Work Step by Step
$$\eqalign{
& \int_0^\pi {{2^{\sin x}}\cos xdx} \cr
& {\text{substitute }}u = \sin x,{\text{ }}du = \cos xdx \cr
& {\text{express the limits in terms of }}u \cr
& x = \pi {\text{ implies }}u = \sin \left( \pi \right) = 0 \cr
& x = 0{\text{ implies }}u = \sin \left( 0 \right) = 0 \cr
& {\text{the entire integration is carried out as follows}} \cr
& \int_0^\pi {{2^{\sin x}}\cos xdx} = \int_0^0 {{2^u}du} \cr
& {\text{by the integral properties}} \cr
& \int_0^0 {{2^u}du} = 0 \cr} $$
