Chọn phương án B.

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

Khi đó ta có $$\begin{aligned}
\displaystyle\int\limits_{0}^{\tfrac{\pi}{4}}x\sin x\mathrm{\,d}x&=-x\cos x\bigg|_{0}^{\tfrac{\pi}{4}}-\displaystyle\int\limits_{0}^{\tfrac{\pi}{4}}\left(-\cos x\right)\mathrm{\,d}x\\
&=-\dfrac{\pi}{4}\cdot\dfrac{\sqrt{2}}{2}+\sin x\bigg|_{0}^{\tfrac{\pi}{4}}\\
&=-\dfrac{\pi\sqrt{2}}{8}+\dfrac{\sqrt{2}}{2}\\
&=\dfrac{4-\pi}{4\sqrt{2}}.
\end{aligned}$$