Answer
The arc length is
$$l=\ln(\sqrt{2}+1).$$
Work Step by Step
By definition the arc length between $x=x_1$ and $x=x_2$ is given by the formula
$$l=\int_{x_1}^{x_2}\sqrt{1+(y')^2}dx = \int_{x_1}^{x_2}\sqrt{1+((\ln(\sin x))')^2}dx.$$
By chain rule we have that
$$(\ln(\sin x))'=\frac{1}{\sin x}(\sin x)'=\frac{1}{\sin x}\cos x =\cot x,$$
so we have by substituting $x_1=\pi/4$ and $x_2=\pi/2$
$$l=\int_{\pi/4}^{\pi/2}\sqrt{1+\cot^2 x}dx = \int_{\pi/4}^{\pi/2}\sqrt{\frac{\sin^2 x}{\sin^2 x}+\frac{\cos^2 x}{\sin^2 x}}dx=\int_{\pi/4}^{\pi/2}\sqrt{\frac{\sin^2x+\cos^2x}{\sin^2x}}dx.$$
We use that that $\sin^2x+\cos^2x=1$ so we have
$$l=\int_{\pi/4}^{\pi/2}\sqrt{\frac{1}{\sin^2x}}dx.$$
In the region of integration the sine is positive so $\sqrt{\sin^2 x}=\sin x$ so we get
$$l=\int_{\pi/4}^{\pi/2}\frac{1}{\sin x}dx=\int_{\pi/4}^{\pi/2}\csc x dx=\ln|\csc x-\cot x|\left.\right|_{\pi/4}^{\pi/2}.$$
We evaluate this by substituting integration bounds
$$l=\ln|\csc(\pi/2)-\cot (\pi/2)|-\ln|\csc(\pi/4)-\cot (\pi/4)| \\=\ln 1 - \ln(\sqrt 2-1)= -\ln(\sqrt 2-1) =\\
\ln\frac{1}{\sqrt{2}-1}=\ln(\sqrt{2}+1).$$
