Answer
The exact length of the curve
$$
y=2\ln (\sin \frac{1}{2}x), \quad\quad \frac{\pi}{3} \leq x \leq \pi.
$$
is equal to $ (-2 \ln (2-\sqrt{3}) \approx 2.63 ). $
Work Step by Step
$$
y=2\ln (\sin \frac{1}{2}x), \quad\quad \frac{\pi}{3} \leq x \leq \pi.
$$
We have
$$
y=2\ln (\sin \frac{1}{2}x) \Rightarrow \frac{d y}{d x}=2 \cdot \frac{1}{\sin \left(\frac{1}{2} x\right)} \cdot \cos \left(\frac{1}{2} x\right) \cdot \frac{1}{2}=\cot \left(\frac{1}{2} x\right)
$$
$$
\Rightarrow 1+\left(\frac{d y}{d x}\right)^{2}=1+\cot ^{2}\left(\frac{1}{2} x\right)=\csc ^{2}\left(\frac{1}{2} x\right)
$$
Thus, the arc length is
$$
\begin{aligned} L &=\int_{\pi / 3}^{\pi} \sqrt{\csc ^{2}\left(\frac{1}{2} x\right)} d x \\
&=\int_{\pi / 3}^{\pi}\left|\csc \left(\frac{1}{2} x\right)\right| d x \\
&=\int_{\pi / 3}^{\pi} \csc \left(\frac{1}{2} x\right) d x \\
& \quad \quad\left[\begin{array}{c} { \text { Let } u=\frac{1}{2} x} \\ { \text { then }d u=\frac{1}{2} d x}\end{array}\right] , so \\
&=\int_{\pi / 6}^{\pi / 2} \csc u(2 d u) \\
& =2[\ln |\csc u-\cot u|]_{\pi / 6}^{\pi / 2} \\
&= 2\left[\ln \left|\csc \frac{\pi}{2}-\cot \frac{\pi}{2}\right|-\ln \left|\csc \frac{\pi}{6}-\cot \frac{\pi}{6}\right|\right] \\ &=2[\ln |1-0|-\ln |2-\sqrt{3}|]=-2 \ln (2-\sqrt{3}) \approx 2.63 \end{aligned}
$$
Hence, the exact length of the curve
$$
y=2\ln (\sin \frac{1}{2}x), \quad\quad \frac{\pi}{3} \leq x \leq \pi.
$$
is equal to $(-2 \ln (2-\sqrt{3}) \approx 2.63 ). $
