Answer
$$\frac{3}{{\ln 4}}$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /2} {{4^{\sin x}}\cos x} dx \cr
& {\text{Let }}u = \sin x,{\text{ }}du = \cos xdx \cr
& {\text{The new limits of integration are:}} \cr
& x = 0 \Rightarrow u = \sin 0 = 0 \cr
& x = \pi /2 \Rightarrow u = \sin 0 = 1 \cr
& {\text{Applying the substitution we obtain}} \cr
& \int_0^{\pi /2} {{4^{\sin x}}\cos x} dx = \int_0^1 {{4^u}du} \cr
& {\text{Integrating}} \cr
& \int_0^1 {{4^u}du} = \left[ {\frac{{{4^u}}}{{\ln 4}}} \right]_0^1 \cr
& {\text{ }} = \frac{1}{{\ln 4}}\left[ {{4^1} - {4^0}} \right] \cr
& {\text{ }} = \frac{3}{{\ln 4}} \cr} $$