The arc length of the curve is traced out by the graph of $f\left(x\right)=\ln (csxx)$on the interval $\left[\frac{\pi }{4},\frac{\pi }{2}\right]$.

given,

$f\left(x\right)=\ln (cscx)$

Recalling that the exact value of the arc length of a function $f\left(x\right)$, which is differentiable and has continuous derivate on an interval $[a,b]$, is computed by the definite integral.

$L={\int }_0^b \sqrt{1+{\left(f^{\mathrm{\prime }}\left(x\right)\right)}^2}dx.....\left(1\right)$

Observe that the function $f\left(x\right)=\ln (csxx)$is a differentiable function and has a continuous derivative on the interval $\left[\frac{\pi }{4},\frac{\pi }{2}\right]$.Differentiate the function with respect to $x$by using the chain rules of differentiation.

$\begin{array}{r}{f}^{\mathrm{\prime }}\left(x\right)=\frac{1}{\csc x}\cdot \frac{d}{dx}(\csc x)\\ =\frac{1}{\csc x}\cdot (-\csc x\cot x)\\ =-\cot x\end{array}$

$\begin{array}{r}L={\int }_{\pi /4}^{\pi /2} \sqrt{1+(-\cot x{)}^{2}x}dx\\ ={\int }_{\pi /4}^{\pi /2} \sqrt{1+{\cot }^{2}x}dx\end{array}$

Using trigonometric identity to solve the integral

$L={\int }_{\pi /4}^{\pi /2} \csc xdx$

Using the trigonometric integration

$\int \csc xdx=-\ln (\csc x+\cot x)$

Using the above formula to calculate the arc length

localid="1649316613365" $\begin{array}{r}L=-[\ln (\csc x+\cot x){]}_{\pi /4}^{\pi /2}\\ =-\left[\ln \left(\csc x\frac{\pi }{2}+\cot \frac{\pi }{2}\right)-\ln \left(\csc \frac{\pi }{4}+\cot \frac{\pi }{4}\right)\right]\\ =-[\ln (1+0)-\ln (\sqrt{2}+1\left)\right]\\ =\ln (\sqrt{2}+1)\end{array}$

Therefore the arc length of the given function is localid="1649316644825" $\ln (\sqrt{2}+1)\approx 0.88$