Đáp án:
$\begin{array}{l}
f'\left( x \right) = {\sin ^2}x.cos4x\\
= \dfrac{{1 - \cos 2x}}{2}.\left( {1 - 2{{\sin }^2}2x} \right)\\
= \dfrac{1}{2} - \dfrac{1}{2}.\cos 2x + {\sin ^2}2x.\cos 2x\\
\Rightarrow f\left( x \right) = \int {\left( {\dfrac{1}{2} - \dfrac{1}{2}\cos 2x + {{\sin }^2}2x.\cos 2x} \right)dx} \\
= \dfrac{1}{2}x - \dfrac{1}{4}.\sin 2x + \int {\dfrac{1}{2}.{{\sin }^2}2xd\sin 2x} \\
= \dfrac{1}{2}.x - \dfrac{1}{4}\sin 2x + \dfrac{1}{6}.{\sin ^3}2x\\
\Rightarrow \int\limits_0^{\pi /2} {f\left( x \right)dx} \\
= \int\limits_0^{\pi /2} {\left( {\dfrac{1}{2}x - \dfrac{1}{4}\sin 2x + \dfrac{1}{6}{{\sin }^3}2x} \right)dx} \\
= \left( {\dfrac{1}{4}{x^2} + \dfrac{1}{8}\cos 2x} \right)_0^{\pi /2} + \dfrac{1}{{12}}.\int\limits_0^{\pi /2} {{{\sin }^2}2x.sin2x.d2x} \\
= \dfrac{{{\pi ^2}}}{{16}} - \dfrac{1}{8} - \dfrac{1}{8} + \dfrac{1}{{12}}.\int\limits_0^{\pi /2} {\left( {{{\cos }^2}2x - 1} \right).d\left( {\cos 2x} \right)} \\
= \dfrac{{{\pi ^2}}}{{16}} - \dfrac{1}{4} + \dfrac{1}{{12}}.\left( {\dfrac{1}{3}{{\cos }^3}2x - \cos 2x} \right)_0^{\pi /2}\\
= \dfrac{{{\pi ^2}}}{{16}} - \dfrac{1}{4} + \dfrac{1}{{12}}.\left( {\dfrac{2}{3} + \dfrac{2}{3}} \right)\\
= \dfrac{{{\pi ^2}}}{{16}} - \dfrac{5}{{36}}
\end{array}$
