91影视

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.

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

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

$\begin{array}{r}{f}^{\mathrm{鈥�}}\left(x\right)=\frac{1}{\csc 鈦�x}鈰�\frac{d}{dx}(\csc 鈦�x)\\ =\frac{1}{\csc 鈦�x}鈰�(鈭�\csc 鈦�x\cot 鈦�x)\\ =鈭�\cot 鈦�x\end{array}$

$\begin{array}{r}L={鈭�}_{蟺/4}^{蟺/2}鈥�\sqrt{1+(鈭�\cot 鈦�x{)}^{2}x}dx\\ ={鈭�}_{蟺/4}^{蟺/2}鈥�\sqrt{1+{\cot }^2鈦�x}dx\end{array}$

Using trigonometric identity to solve the integral

$L={鈭�}_{蟺/4}^{蟺/2}鈥�\csc 鈦�xdx$

Using the trigonometric integration

$鈭�\csc 鈦�xdx=鈭�\ln 鈦�(\csc 鈦�x+\cot 鈦�x)$

Using the above formula to calculate the arc length

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

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