Chọn phương án B.

Đặt \(\begin{cases}
u=x\\ v'=\sin2x
\end{cases}\Rightarrow\begin{cases}
u'=1\\ v=-\dfrac{1}{2}\cos2x
\end{cases}\)

Khi đó $$\begin{aligned}\displaystyle\int\limits_{0}^{\tfrac{\pi}{4}}x\sin x\mathrm{\,d}x=&-\dfrac{x}{2}\cos2x\bigg|_0^{\tfrac{\pi}{4}}+\dfrac{1}{2}\displaystyle\int\limits_{0}^{\tfrac{\pi}{4}}\cos2x\mathrm{\,d}x\\ =&\dfrac{1}{4}\sin2x\bigg|_0^{\tfrac{\pi}{4}}=\dfrac{1}{4}.\end{aligned}$$