Find the arc length of the curve y = ln sin(x - 1)

Find the arc length of the curve $y = \ln\sin(x - 1)$ between the points with abscissas $x = 1 + \dfrac{\pi}{3}$ and $x = 1 + \dfrac{2\pi}{3}$.

We differentiate, simplify using a Pythagorean identity, and integrate the cosecant.

Step 1: Differentiate

\[y'=\frac{\cos(x-1)}{\sin(x-1)}=\cot(x-1)\]

Step 2: Simplify $1+(y')^2$

\[1+\cot^2(x-1)=\csc^2(x-1)\]

Step 3: Set up and simplify the integral

For $x\in\left[1+\frac{\pi}{3},\,1+\frac{2\pi}{3}\right]$, i.e. $x-1\in\left[\frac{\pi}{3},\frac{2\pi}{3}\right]$, we have $\sin(x-1)>0$, so $|\csc(x-1)|=\csc(x-1)$. With $u=x-1$:

\[L=\int_{1+\pi/3}^{1+2\pi/3}\csc(x-1)\,dx = \int_{\pi/3}^{2\pi/3}\csc u\,du\]

Step 4: Evaluate

Using $\displaystyle\int\csc u\,du=\ln|\csc u-\cot u|+C$:

\[L=\big[\ln|\csc u-\cot u|\big]_{\pi/3}^{2\pi/3}\]

At $u=\frac{2\pi}{3}$: $\csc\frac{2\pi}{3}=\frac{2\sqrt3}{3}$, $\cot\frac{2\pi}{3}=-\frac{\sqrt3}{3}$, so $\csc u-\cot u=\sqrt3$.

At $u=\frac{\pi}{3}$: $\csc\frac{\pi}{3}=\frac{2\sqrt3}{3}$, $\cot\frac{\pi}{3}=\frac{\sqrt3}{3}$, so $\csc u-\cot u=\frac{\sqrt3}{3}$.

Step 5: Combine

\[L=\ln\sqrt3-\ln\frac{\sqrt3}{3}=\ln\left(\sqrt3\cdot\frac{3}{\sqrt3}\right)=\ln3\]

\[\boxed{L=\ln3\approx1.099}\]