Chọn phương án B.

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

Khi đó $$\begin{aligned}\displaystyle\int\limits_{0}^{\tfrac{\pi}{2}}x\cos x\mathrm{\,d}x=&x\sin x\bigg|_0^{\tfrac{\pi}{2}}-\displaystyle\int\limits_{0}^{\tfrac{\pi}{2}}\sin x\mathrm{\,d}x\\ =&\dfrac{\pi}{2}+\cos x\bigg|_0^{\tfrac{\pi}{2}}=\dfrac{\pi}{2}-1.\end{aligned}$$