Tính tích phân I = tích phân 0->pi x^2 cos^2 2x dx bằng cách đặt

A. $I=\frac{1}{2}{x}^2\sin 2x\left|\begin{array}{l}{}^{\pi }\\ {}_0\end{array}\right.-\underset{0}{\overset{\pi }{\int }}x\sin 2xdx.$

B. $I=\frac{1}{2}{x}^2\sin 2x\left|\begin{array}{l}{}^{\pi }\\ {}_0\end{array}\right.-2\underset{0}{\overset{\pi }{\int }}x\sin 2xdx.$

C. $I=\frac{1}{2}{x}^2\sin 2x\left|\begin{array}{l}{}^{\pi }\\ {}_0\end{array}\right.+2\underset{0}{\overset{\pi }{\int }}x\sin 2xdx.$

D. $I=\frac{1}{2}{x}^2\sin 2x\left|\begin{array}{l}{}^{\pi }\\ {}_0\end{array}\right.+\underset{0}{\overset{\pi }{\int }}x\sin 2xdx.$