Answer
$$0$$
Work Step by Step
$$\eqalign{
& \int_{ - \pi /2}^{\pi /2} {\left( {\sin 4x + \cos 4x} \right)} dx \cr
& {\text{Write the integral as a sum of integrals }} \cr
& = \int_{ - \pi /2}^{\pi /2} {\sin 4x} dx + \int_{ - \pi /2}^{\pi /2} {\cos 4x} dx \cr
& {\text{Where}} \cr
& \sin 4x:{\text{ Odd function}} \cr
& \cos 4x:{\text{Even function}} \cr
& {\text{Using }}\left( {{\text{THEOREM 4}}.{\text{16}}} \right) \cr
& = 0 + 2\int_0^{\pi /2} {\cos 4x} dx \cr
& {\text{Integrating}} \cr
& = 0 + 2\left[ {\frac{1}{4}\sin 4x} \right]_0^{\pi /2} \cr
& = 0 + 2\left[ {\frac{1}{4}\sin 4\left( {\frac{\pi }{2}} \right) - 0} \right] \cr
& = 0 \cr} $$
