Simple Improper Integral from OBM Exam

Let $I$ be the given integral.

$\begin{aligned} I & = \int_0^\pi \log_{10} (\sin x) \ dx & \small \color{#3D99F6} \text{Since }\sin x \text{ is symmetrical about }x = \frac \pi 2 \\ & = 2 \int_0^\frac \pi 2 \log_{10} (\sin x) \ dx & \small \color{#3D99F6} \text{Using }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \int_0^\frac \pi 2 \left(\log_{10} (\sin x) + \log_{10} (\cos x) \right) \ dx \\ & = \int_0^\frac \pi 2 \log_{10} (\sin x \cos x) \ dx \\ & = \int_0^\frac \pi 2 \log_{10} \left(\frac {\sin 2x}2 \right) \ dx \\ & = {\color{#3D99F6}\int_0^\frac \pi 2 \log_{10} (\sin 2x) \ dx} - \int_0^\frac \pi 2 \log_{10} 2 \ dx & \small \color{#3D99F6} \text{Let }u = 2x \implies du = 2\ dx \\ & = {\color{#3D99F6}\frac 12 \int_0^\pi \log_{10} (\sin u) \ du} - x \log_{10} 2\ \bigg|_0^\frac \pi 2 & \small \color{#3D99F6} \text{Note that }I = \int_0^\pi \log_{10} (\sin x) \ dx \\ & = {\color{#3D99F6}\frac 12 I} - \frac {\pi \log_{10} 2}2 \ & \small \color{#3D99F6} \text{Rearranging} \\ & = - \pi \log_{10} 2 \approx \boxed{-0.946} \end{aligned}$