Mr. Arc's Integral

Integrate by parts to get : $\int_0^1 \frac{\arcsin x}{x} \ \mathrm{d}x = \left[\arcsin x \ln x\right]_0^1 - \int_0^1 \frac{\ln x}{\sqrt{1-x^2}} \ \mathrm{d}x$ Now substitute $x=\sin t$ , to get : $- \int_0^1 \frac{\ln x}{\sqrt{1-x^2}} \ \mathrm{d}x=-\int_0^{\pi/2} \ln(\sin t)\ \mathrm{d}t$ Clearly the integral converges (since the first integral converges), then we call it $I$ and by the substitution $y=\pi/2-t$ we get : $I= -\int_0^{\pi/2} \ln(\sin t)\ \mathrm{d}t = -\int_0^{\pi/2} \ln(\cos y) \mathrm{d}y$ sum them to get : $I=-\int_0^{\pi/2} \ln(\sin(2t)) -\ln 2\ \mathrm{d}t \overset{x=2t}{=} \frac{-\pi \ln 2}{2} - \left(I+\int_{\pi/2}^{\pi} \ln(\sin x) \ \mathrm{d}x\right)$ using the sub $x=\pi/2+t$ on the last integral we get that the last integral is $=I$ , then : $I =\frac{-\pi \ln 2}{2} +2I\Longrightarrow I= \boxed{\frac{\pi\ln 2}{2} \approx 1.08879}$