$x = x^2$ ?

First, let's do integration by parts for the integral on the right. Let $u = \sin ^2 x, du = 2 \sin x \cos x = \sin 2x, dv = x^{-2}, v = - x^{-1}$ . Then the integral to the right is equal to $- \frac{\sin^2 x}{x} \big|_{-\infty}^{\infty} + \int_{-\infty}^{\infty} \frac{ \sin 2x}{x} dx$ . Because $\sin^2(x) = o(x)$ the term outside the integral vanishes so we are left with just $\int_{-\infty}^{\infty} \frac{ \sin 2x}{x} dx$ . Now, let $u = 2x, du = 2dx$ . Then the integral is equal to $\int_{-\infty}^{\infty} \frac{ \sin u}{\frac{u}{2}} \frac{du}{2} = \frac{2}{2} \int_{-\infty}^{\infty} \frac{ \sin u}{u} du = \int_{-\infty}^{\infty} \frac{ \sin u}{u} du$ .