Let $C_R$ denote the counterclockwise semicircular arc extending from $R$ to $-R$ in the plane and $C_{\rho_1} , C_{\rho_2}$ be the counterclockwise semicircular arcs extending from $-1 - \rho_1$ to $-1 + \rho_1$ and from $1 -\rho_1$ and $1+\rho_1,$ respectively. Note that $\frac{x\sin 4x}{x^2 -1} = \Im \frac{x\exp{4ix}}{x^2 -1}.$
Consider also that $\,\displaystyle{f(z) = \frac{z\exp{4iz}}{z^2 -1}}\, $ has simple poles at $-1$ and $1.$ By Jordan's Lemma, there exists $\theta \in [0, \pi ]$ such that $\displaystyle\int_{C_R} \frac{z\exp{4iz}}{z^2-1} \le \frac{\pi }{4} \frac{1}{R^2 \exp{2i\theta }-1}$ The right hand side of this inequality tends to zero. Using the residue theorem,
$\int_{C_R} \frac{z\exp{4iz}}{z^2-1} + \int_{-C_{\rho_1}} \frac{z\exp{4iz}}{z^2-1} + \int_{-C_{\rho_2}} \frac{z\exp{4iz}}{z^2-1} +\int_{-R}^{-1 -\rho_1} \frac{x\exp{4ix}}{x^2-1} dx +$ $+\int_{-1 +\rho_1}^{1 - \rho_e} \frac{x\exp{4ix}}{x^2-1} dx +\int_{1 + \rho_1}^{R} \frac{x\exp{4ix}}{x^2-1} dx = 0$
Letting $R$ tend to infinity and $\rho_1 \rho_2$ tend to zero, we have $\textrm{pv }\int_{-\infty }^{\infty } \frac{x\exp{4ix}}{x^2 -1 }dx = \pi i [\textrm{ Res } (f, -1) + \textrm{ Res } (f, 1) ] = \pi i \cos 4 $ Hence $\textrm{pv } \int_{-\infty }^{\infty } \frac{x\sin 4x}{x^2 -1 }dx = \pi \cos 4.$