Answer
$$\sqrt 2 $$
Work Step by Step
$$\eqalign{
& \int_{ - \pi /4}^{\pi /4} {\sqrt {1 + \cos 4x} dx} \cr
& = \int_{ - \pi /4}^{\pi /4} {\sqrt {1 + \cos 2\left( {2x} \right)} dx} \cr
& {\text{Use the identity }}\cos 2\theta = 2{\cos ^2}\left( {2\theta } \right) - 1 \cr
& \int_{ - \pi /4}^{\pi /4} {\sqrt {1 + \cos 2\left( {2x} \right)} dx} = \int_{ - \pi /4}^{\pi /4} {\sqrt {1 + 2{{\cos }^2}\left( {2x} \right) - 1} dx} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int_{ - \pi /4}^{\pi /4} {\sqrt {2{{\cos }^2}\left( {2x} \right)} dx} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int_{ - \pi /4}^{\pi /4} {\sqrt 2 \cos 2xdx} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt 2 \int_{ - \pi /4}^{\pi /4} {\cos 2xdx} \cr
& {\text{Integrate}} \cr
& \sqrt 2 \int_{ - \pi /4}^{\pi /4} {\cos 2xdx} = \sqrt 2 \left[ {\frac{1}{2}\sin 2x} \right]_{ - \pi /4}^{\pi /4} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\sqrt 2 }}{2}\left[ {\sin \left( {\frac{\pi }{2}} \right) - \sin \left( { - \frac{\pi }{2}} \right)} \right] \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\sqrt 2 }}{2}\left[ {1 + 1} \right] \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt 2 \cr} $$