## Calculus: Early Transcendentals (2nd Edition)

$$0$$
\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}