Hướng dẫn giải

Ta có: $\begin{array}{l}{\cos ^4}2x = {\left( {\frac{{1 + \cos 4x}}{2}} \right)^2} = \frac{1}{4}\left( {1 + 2\cos 4x + {{\cos }^2}4x} \right)\\\;\;\;\;\;\;\;\;\;\;\;\; = \frac{1}{4}\left( {1 + 2\cos 4x + \frac{{1 + \cos 8x}}{2}} \right) = \frac{1}{8}\left( {3 + 4\cos 4x + \cos 8x} \right)\end{array}$

Do đó $F\left( x \right) = \frac{1}{8}\int {\left( {3 + 4\cos 4x + \cos 8x} \right)dx} = \frac{1}{8}\left( {3x + \sin 4x + \frac{1}{8}\sin 8x} \right) + C$

Mà $F\left( 0 \right) = 2019$ nên ta có $C = 2019$.

Vậy $F\left( x \right) = \frac{1}{8}\left( {3x + \sin 4x + \frac{1}{8}\sin 8x} \right) + 2019$.

Do đó $F\left( {\frac{\pi }{8}} \right) = \frac{{3\pi + 129224}}{{64}}$

Chọn C.