Tính tích phân $\displaystyle \int \limits _{0}^{\pi} x^2 \cos 2x \mathrm{\,d}x $ bằng cách đặt $\begin{cases} u=x^2\\ dv = \cos 2x \mathrm{\,d}x\end{cases}$. Mệnh đề nào dưới đây đúng?
$\displaystyle I= \dfrac{1}{2} x^2 \sin 2x \bigg|_0 ^ {\pi} - \displaystyle \int \limits _{0}^{\pi} x \sin 2x \mathrm{\,d}x$
$\displaystyle I= \dfrac{1}{2} x^2 \sin 2x \bigg|_0 ^ {\pi} - \displaystyle 2 \int \limits _{0}^{\pi} x \sin 2x \mathrm{\,d}x$
$\displaystyle I= \dfrac{1}{2} x^2 \sin 2x \bigg|_0 ^ {\pi} + \displaystyle 2 \int \limits _{0}^{\pi} x \sin 2x \mathrm{\,d}x$
$\displaystyle I= \dfrac{1}{2} x^2 \sin 2x \bigg|_0 ^ {\pi} + \displaystyle \int \limits _{0}^{\pi} x \sin 2x \mathrm{\,d}x$