Cho hàm số f(x) có f(0)=0 và

Ta có\[f\left( x \right) = \smallint f'\left( x \right)dx = \smallint {\sin ^4}xdx\]

\[\begin{array}{*{20}{l}}{ = \smallint {{\left( {\frac{{1 - \cos 2x}}{2}} \right)}^2}dx}\\{ = \frac{1}{4}\smallint \left( {1 - 2\cos 2x + {{\cos }^2}2x} \right)dx}\\{ = \frac{1}{4}\smallint \left( {1 - 2\cos 2x + \frac{{1 + \cos 4x}}{2}} \right)dx}\\{ = \frac{1}{4}\left( {x - \sin 2x + \frac{1}{2}x + \frac{1}{2}.\frac{{\sin 4x}}{4}} \right) + C}\\{ = \frac{{3x}}{8} - \frac{{\sin 2x}}{4} + \frac{{\sin 4x}}{{32}} + C}\end{array}\]

Theo bài ra ta có\[f\left( 0 \right) = 0 \Leftrightarrow C = 0 \Rightarrow f\left( x \right) = \frac{{3x}}{8} - \frac{{\sin 2x}}{4} + \frac{{\sin 4x}}{{32}}\]

Vậy\[\mathop \smallint \limits_0^{\frac{\pi }{2}} f\left( x \right)dx = \mathop \smallint \limits_0^{\frac{\pi }{2}} \left( {\frac{{3x}}{8} - \frac{{\sin 2x}}{4} + \frac{{\sin 4x}}{{32}}} \right)dx = \frac{{3{\pi ^2} - 16}}{{64}}\] (sử dụng MTCT).