Chọn phương án C.

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

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