Phương pháp giải:

Ta có: \(f\left( x \right) = \int {f'\left( x \right)dx.} \)

Từ đó ta tính tích phân cần tìm.

Chú ý công thức biến đổi lượng giác \(\cos a\cos b = \dfrac{1}{2}\left( {\cos \left( {a + b} \right) + \cos \left( {a - b} \right)} \right)\)

Lời giải chi tiết:

Ta có: \(f\left( x \right) = \int {f'\left( x \right)dx.} \)

\(\begin{array}{l} \Leftrightarrow f\left( x \right) = \int {\left( {\cos x{{\cos }^2}2x} \right)dx} \\ \Leftrightarrow f\left( x \right) = \int {\left( {\cos x.\dfrac{{1 + \cos 4x}}{2}} \right)dx} \\ \Leftrightarrow f\left( x \right) = \dfrac{1}{2}\int {\left( {\cos x + \cos x\cos 4x} \right)dx} \\ \Leftrightarrow f\left( x \right) = \dfrac{1}{2}\int {\cos xdx}  + \dfrac{1}{2}\int {\dfrac{1}{2}\left( {\cos 5x + \cos 3x} \right)dx} \\ \Leftrightarrow f\left( x \right) = \dfrac{1}{2}\sin x + \dfrac{1}{4}\left( {\dfrac{{\sin 5x}}{5} + \dfrac{{\sin 3x}}{3}} \right) + C\\ \Leftrightarrow f\left( x \right) = \dfrac{1}{2}\sin x + \dfrac{1}{{20}}\sin 5x + \dfrac{1}{{12}}\sin 3x + C\end{array}\)

Suy ra \(f\left( x \right) = \dfrac{1}{2}\sin x + \dfrac{1}{{20}}\sin 5x + \dfrac{1}{{12}}\sin 3x + C\)

Mà \(f\left( 0 \right) = 0 \Rightarrow C = 0\)

Do đó \(f\left( x \right) = \dfrac{1}{2}\sin x + \dfrac{1}{{20}}\sin 5x + \dfrac{1}{{12}}\sin 3x\)

\(\begin{array}{l} \Rightarrow \int\limits_0^\pi  {f\left( x \right)dx}  = \int\limits_0^\pi  {\left( {\dfrac{1}{2}\sin x + \dfrac{1}{{20}}\sin 5x + \dfrac{1}{{12}}\sin 3x} \right)dx} \\ = \left. {\left( { - \dfrac{1}{2}\cos x + \dfrac{1}{{20}}.\dfrac{{ - \cos 5x}}{5} + \dfrac{1}{{12}}.\dfrac{{ - \cos 3x}}{3}} \right)} \right|_0^\pi \\ = \left. {\left( { - \dfrac{1}{2}\cos x - \dfrac{1}{{100}}\cos 5x - \dfrac{1}{{36}}\cos 3x} \right)} \right|_0^\pi \\ =  - \dfrac{1}{2}\left( { - 1 - 1} \right) - \dfrac{1}{{100}}\left( { - 1 - 1} \right) - \dfrac{1}{{36}}\left( { - 1 - 1} \right)\\ = \dfrac{{242}}{{225}}.\end{array}\)

Chọn C.