Cho \(I = \int\limits_0^{\frac{\pi }{4}} {\left( {x - 1} \right)\sin 2xdx} \). Tìm đẳng thức đúng?

Cho \(I = \int\limits_0^{\frac{\pi }{4}} {\left( {x - 1} \right)\sin 2xdx} \). Tìm đẳng thức đúng?
A. \(I = - \left( {x - 1} \right)\cos 2x\left| {\begin{array}{*{20}{c}}{\frac{\pi }{4}}\\0\end{array}} \right. + \int\limits_0^{\frac{\pi }{4}} {\cos 2x} dx\)
B. \(I = - \left( {x - 1} \right)\cos 2x\left| {\begin{array}{*{20}{c}}{\frac{\pi }{4}}\\0\end{array}} \right. - \int\limits_0^{\frac{\pi }{4}} {\cos 2x} dx\)
C. \(I = - \frac{1}{2}\left( {x - 1} \right)\cos 2x\left| {\begin{array}{*{20}{c}}{\frac{\pi }{4}}\\0\end{array}} \right. + \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\cos 2x} dx\)
D. \(I = - \frac{1}{2}\left( {x - 1} \right)\cos 2x\left| {\begin{array}{*{20}{c}}{\frac{\pi }{4}}\\0\end{array}} \right. - \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\cos 2x} dx\)

Đặt \(\left\{ {\begin{array}{*{20}{c}}{u = x - 1}\\{dv = \sin 2xdx}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{du = dx}\\{v = - \frac{1}{2}\cos 2x}\end{array} \Rightarrow I = } \right.\left. { - \frac{1}{2}\left( {x - 1} \right)\cos 2x} \right|_0^{\frac{\pi }{4}} + \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\cos 2xdx} .\)