Let \(x=\arcsin y\), so that \(\sin x=y\) and \(\mathrm dx=\dfrac{\mathrm dy}{\sqrt{1-y^2}}\). Then
\(\displaystyle\int_0^{\pi/2}\log\sin x\,\mathrm dx=\int_0^1\dfrac{\log y}{\sqrt{1-y^2}}\,\mathrm dy\)
Integrate by parts, setting
\(u=\log y\implies \mathrm du=\dfrac{\mathrm dy}y\)
\(\mathrm dv=\dfrac{\mathrm dy}{\sqrt{1-y^2}}\implies v=\arcsin y\)
so the integral is equivalent to
\(\displaystyle\int_0^1\dfrac{\log y}{\sqrt{1-y^2}}\,\mathrm dy=uv-\int v\,\mathrm du\)
\(=\log y\arcsin y\bigg|_{y=0}^{y=1}-\displaystyle\int_0^1\frac{\arcsin y}y\,\mathrm dy\)
The first term vanishes, since
\(\displaystyle\lim_{y\to0^+}\log y\arcsin y=0\)
Now for comparison, consider the function \(f(y)=\dfrac1{\sqrt{1-y}}\). We have that the integral of this function over the same interval is convergent:
\(\displaystyle\int_0^1\frac{\mathrm dy}{\sqrt{1-y}}=-2\sqrt{1-y}\bigg|_{y=0}^{y=1}=2\)
so if we could show that \(\dfrac{\arcsin y}y\le f(y)\) at least for \(0, then we're done. Replacing \(y\) with \(\sin x\) (which means \(0), the inequality can be written as
\(\dfrac1{\sqrt{1-y}}\ge\dfrac{\arcsin y}y\implies \dfrac y{\sqrt{1-y}}\ge\arcsin y\implies\dfrac{\sin x}{\sqrt{1-\sin x}}\ge x\)
This is certainly true, because as \(x\to0^+\), both sides approach 0, while as \(x\to\dfrac\pi2^-\), you have the right side approach \(\dfrac\pi2\) while the left side approaches \(+\infty\). You can compare these two functions' derivatives to see that the left side is monotonic over the interval.
So we've established that
\(\displaystyle\frac{\arcsin y}y\le\frac1{\sqrt{1-y}}\implies\int_0^1\frac{\arcsin y}y\,\mathrm dy\le\int_0^1\frac{\mathrm dy}{\sqrt{1-y}}=2\)
so the first integral must converge, and so must
\(\displaystyle\int_0^{\pi/2}\log\sin x\,\mathrm dx\)
to some value between -2 and 0.
