Đặt \(\begin{cases}
u=x^2\\ v'=\sin x
\end{cases}\Rightarrow\begin{cases}
u'=2x\\ v=-\cos x
\end{cases}\)

Khi đó $$\begin{aligned}I&=-x^2\cos x\bigg|_0^{\pi}+2\displaystyle\int\limits_{0}^{\pi}x\cos x\mathrm{\,d}x\\ &=\pi^2+2\displaystyle\int\limits_{0}^{\pi}x\cos x\mathrm{\,d}x.\end{aligned}$$

Đặt \(\begin{cases}
u=x\\ v'=\cos x
\end{cases}\Rightarrow\begin{cases}
u'=1\\ v=\sin x
\end{cases}\)

Khi đó $$\begin{aligned}I&=\pi^2+2\left(x\sin x\bigg|_0^{\pi}-\displaystyle\int\limits_{0}^{\pi}\sin x\mathrm{\,d}x\right)\\ &=\pi^2+2\cos x\bigg|_0^{\pi}=\pi^2-4.\end{aligned}$$