Cho \[\mathop \smallint \limits_0^1 f\left( x \right)dx = 1.\]Tính \[I = \mathop \smallint \limits_0^{\frac{\pi }{4}} \left( {2{{\sin }^2}x - 1} \right)f\left( {\sin 2x} \right)dx\]

Đặt\[t = \sin 2x \Rightarrow dt = 2\cos 2xdx \Rightarrow - \frac{1}{2}dt = \left( {2{{\sin }^2}x - 1} \right)dx\]

Đổi cận: \(\left\{ {\begin{array}{*{20}{c}}{x = 0 \Rightarrow t = 0}\\{x = \frac{\pi }{4} \Rightarrow t = 1}\end{array}} \right.\)

\[ \Rightarrow I = \mathop \smallint \limits_0^{\frac{\pi }{4}} \left( {2{{\sin }^2}x - 1} \right)f\left( {\sin 2x} \right)dx = - \frac{1}{2}\mathop \smallint \limits_0^1 f\left( t \right)dt = - \frac{1}{2}.1 = - \frac{1}{2}.\]

Đáp án cần chọn là: B