Since $d\cos(\theta)=-\sin(\theta)d\theta$,
\[ d^3r = -r^2 dr d\cos(\theta) d\phi ,\]
let $\mu = \cos(\theta)$,
\[ \int \int \int_0^{\pi} d^3r = \int \int \int_0^{\pi} -r^2 dr d\cos(\theta) d\phi \\ = \int \int \int_{\cos(0)}^{\cos(2\pi)} -r^2 dr d\mu d\phi \\ =\int \int \int_{\cos(0)}^{\cos(2\pi)} -r^2 dr d\mu d\phi \\ = \int \int \int_{\cos(2\pi)}^{\cos(0)} r^2 dr d\mu d\phi \\ = \int \int \int_{-1}^{1} r^2 dr d\mu d\phi \]
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