If the value of ${\int }_{-2}^2 |x\cos \pi x|\mathrm{d}x=k/\pi$ then the value of $k$ is

Since the integrand is an even function so required integral is equal to $2{\int }_0^2 |x\cos \pi x|dx$

$\begin{array}{r}=2\left[{\int }_0^{1/2} x\cos \pi xdx-{\int }_{1/2}^{3/2} x\cos \pi xdx+{\int }_{3/2}^2 x\cos \pi xdx\right]\\ =2\left[x\frac{\sin \pi x}{\pi }+{\left.\frac{\cos \pi x}{{\pi }^2}\right|}_0^{1/2}-{\left.\left(x\frac{\sin \pi x}{\pi }+\frac{\cos \pi x}{{\pi }^2}\right)\right|}_{1/2}^{3/2}\right.\\ \left.+{\left.\left(\frac{x\sin \pi x}{\pi }+\frac{\cos \pi x}{{\pi }^2}\right)\right|}_{y_2}^2\right]\end{array}$

$=2\left[\frac{1}{2\pi }-\frac{1}{{\pi }^2}-\left(-\frac{3}{2\pi }-\frac{1}{2\pi }\right)+\frac{1}{{\pi }^2}-\left(-\frac{3}{2\pi }\right)\right]=\frac{8}{\pi }$