Answer
$$\ln \frac{1}{\sqrt{2}-1}=\ln (\sqrt{2}+1)$$
Work Step by Step
Since
$$y=\ln \sin x\to y'=\frac{\cos x}{\sin x} $$
Then
\begin{aligned}
s&= \int_{a}^{b}\sqrt{1+y'^2}dx\\
&=\int_{\pi / 4}^{\pi / 2} \frac{\sqrt{\sin ^{2} x+\cos ^{2} x}}{\sin x} d x\\
&=\int_{\pi / 4}^{\pi / 2} \csc x d x\\
&=\left.\ln (\csc x-\cot x)\right|_{\pi / 4} ^{\pi / 2} \\
&=\ln 1-\ln (\sqrt{2}-1)\\
&=\ln \frac{1}{\sqrt{2}-1}=\ln (\sqrt{2}+1)
\end{aligned}
