Tính tích phân $I=\displaystyle\displaystyle\int\limits_{0}^{\pi}x^2\cos2x\mathrm{d}x$ bằng cách đặt $\begin{cases}u=x^2\\ \mathrm{d}v=\cos2x\mathrm{d}x\end{cases}$. Mệnh đề nào dưới đây đúng?

	$I=\dfrac{1}{2}x^2\sin2x\bigg|_{0}^{\pi}-\displaystyle\displaystyle\int\limits_{0}^{\pi}x\sin2x\mathrm{d}x$
	$I=\dfrac{1}{2}x^2\sin2x\bigg|_{0}^{\pi}-2\displaystyle\displaystyle\int\limits_{0}^{\pi}x\sin2x\mathrm{d}x$
	$I=\dfrac{1}{2}x^2\sin2x\bigg|_{0}^{\pi}+2\displaystyle\displaystyle\int\limits_{0}^{\pi}x\sin2x\mathrm{d}x$
	$I=\dfrac{1}{2}x^2\sin2x\bigg|_{0}^{\pi}+\displaystyle\displaystyle\int\limits_{0}^{\pi}x\sin2x\mathrm{d}x$