Chọn phương án C.

Ta biết \(f\left(x\right)\) là một nguyên hàm của \(f'\left(x\right)\). $$\begin{aligned}
\displaystyle\int f'\left(x\right)\mathrm{\,d}x&=\displaystyle\int\cos x\cdot\cos^22x\mathrm{\,d}x\\
&=\displaystyle\int\cos x\cdot\dfrac{1+\cos4x}{2}\mathrm{\,d}x\\
&=\displaystyle\int\dfrac{\cos x}{2}\mathrm{\,d}x+\displaystyle\int\dfrac{\cos x\cdot\cos4x}{2}\mathrm{\,d}x\\
&=\dfrac{1}{2}\displaystyle\int\cos x\mathrm{\,d}x+\dfrac{1}{4}\displaystyle\int\left(\cos5x+\cos3x\right)\mathrm{\,d}x\\
&=\dfrac{1}{2}\sin x+\dfrac{1}{20}\sin5x+\dfrac{1}{12}\sin3x+C.
\end{aligned}$$Suy ra \(f\left(x\right)=\dfrac{1}{2}\sin x+\dfrac{1}{20}\sin5x+\dfrac{1}{12}\sin3x+C\).

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

Vậy \(f\left(x\right)=\dfrac{1}{2}\sin x+\dfrac{1}{20}\sin5x+\dfrac{1}{12}\sin3x\).

Khi đó $$\begin{aligned}
\displaystyle\int\limits_{0}^{\pi}f\left(x\right)\mathrm{\,d}x&=\displaystyle\int\limits_0^{\pi}\left(\dfrac{1}{2}\sin x+\dfrac{1}{20}\sin5x+\dfrac{1}{12}\sin3x\right)\mathrm{\,d}x\\
&=\left(-\dfrac{1}{2}\cos x-\dfrac{1}{100}\cos 5x-\dfrac{1}{36}\cos3x\right)\bigg|_0^{\pi}\\
&=\dfrac{242}{225}.
\end{aligned}$$