If ${\int }_0^{2\pi }|x\sin x|dx=k\pi$, then the value of $k$ is equal to:

In the interval $[0,2\pi ]$, $x$ is always positive. The sign of $x\sin x$ depends on $\sin x$.For $x\in [0,\pi ]$, $\sin x\ge 0$, so $|x\sin x|=x\sin x$.For $x\in [\pi ,2\pi ]$, $\sin x\le 0$, so $|x\sin x|=-x\sin x$.
$${\int }_0^{2\pi }|x\sin x|dx={\int }_0^{\pi }x\sin xdx-{\int }_{\pi }^{2\pi }x\sin xdx$$
First, find $\int x\sin xdx=-x\cos x+\sin x$.${\int }_0^{\pi }x\sin xdx=[-x\cos x+\sin x{]}_0^{\pi }=(-\pi (-1)+0)-(0)=\pi$.${\int }_{\pi }^{2\pi }x\sin xdx=[-x\cos x+\sin x{]}_{\pi }^{2\pi }=(-2\pi (1)+0)-(-\pi (-1)+0)=-2\pi -\pi =-3\pi$.Total integral = $\pi -(-3\pi )=4\pi$.Comparing with $k\pi$, we get $k=4$.