Cho hàm số f(x) có f(0)=0 và \[f\prime (x) = si{n^4}x\forall x \in \mathbb{R}\]. Tích phân \[\mathop \smallint \limits_0^{\frac{\pi }{2}} f\left( x \right)dx\] bằng:

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

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

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

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

Đáp án cần chọn là: C