Cho hàm số \[f(x)\]có \[f'(x) = \sin (2x).co{s^2}(4x)\]và \[f(0) = 0\]. Tính \[\int\limits_0^{\frac{\pi }{2}} {f(x)dx} \] bằng:

Ta có

\[\begin{array}{l}f'\left( x \right) = \sin \left( {2x} \right).co{s^2}\left( {4x} \right) \Rightarrow \int {f'\left( x \right)dx = \int {\sin \left( {2x} \right).co{s^2}\left( {4x} \right)dx} } \\ \Leftrightarrow f\left( x \right) = \int {\sin \left( {2x} \right).\frac{2}} dx\\ \Leftrightarrow f\left( x \right) = \frac{{ - 1}}{4}\cos \left( {2x} \right) + \frac{1}{4}\int {\left( {\sin 10x - \sin 6x} \right)} dx\\ \Leftrightarrow f(x) = \frac{{ - 1}}{4}\cos 2x - \frac{1}\cos 10x + \frac{1}\cos 6x + C\end{array}\]

\[f(0) = 0 \Leftrightarrow c = \frac{7}\]

Vậy \[f(x) = \frac{{ - 1}}{4}\cos \left( {2x} \right) - \frac{1}\cos \left( {10x} \right) + \frac{1}\cos \left( {6x} \right) + \frac{7}\]

Do đó:

\[\int\limits_0^{\frac{\pi }{2}} {f(x)dx} = \left. {\left( { - \frac{1}{8}\sin 2x - \frac{1}\sin 10x + \frac{1}\sin 6x + \frac{7}x} \right)} \right|_0^{\frac{\pi }{2}} = \frac{{7\pi }}\]

Chọn đáp án A