Decent Integral (2)

$\begin{aligned} I & = \int_0^1 \frac {\sin^{-1}x}x dx & \small \color{#3D99F6} \text{Let }\sin \theta = x, \ \cos \theta \ d\theta = dx \\ & = \int_0^\frac \pi 2 \theta \cot \theta \ d \theta & \small \color{#3D99F6} \text{By integration by parts} \\ & = \theta \ln (\sin \theta) \bigg|_0^\frac \pi 2 - \int_0^ \frac \pi 2 \ln (\sin \theta) \ d\theta \\ & = \int^0_\frac \pi 2 \ln (\sin \theta) \ d\theta & \small \color{#3D99F6} \text{By }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 12 \int^0_\frac \pi 2 \big(\ln (\sin \theta) + \ln(\cos \theta)\big) \ d\theta \\ & = \frac 12 \int^0_\frac \pi 2 \ln (\sin \theta \cos \theta) \ d\theta \\ & = \frac 12 \int^0_\frac \pi 2 \ln \left(\frac {\sin 2 \theta}2\right) \ d\theta \\ & = \frac 12 \int^0_\frac \pi 2 \big(\ln (\sin {\color{#3D99F6}2 \theta}) - \ln 2 \big) \ d\theta & \small \color{#3D99F6} \text{Let }\phi = 2\theta, \ d\phi = 2 \ d\theta \\ & = \frac 1{\color{#3D99F6}4} {\color{#D61F06}\int^0_{\color{#3D99F6}\pi} \ln (\sin {\color{#3D99F6}\phi}) \ d{\color{#3D99F6}\phi}} + \frac 12 \int_0^\frac \pi 2 \ln 2 \ d\theta & \small \color{#D61F06} \text{Note that } \ln(\sin x) \text{ is symmetrical at }\frac \pi 2 \\ & = \frac 14 {\color{#D61F06}(2I)} + \frac \pi 4 \ln 2 \\ & = \frac {\pi \ln 2}2 \end{aligned}$

$\implies a+b+c = 1+2+2 = \boxed{5}$