## Problem Statement

Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.

$$
y=1-\ln \sin x, \frac{\pi}{3} \leq x \leq \frac{\pi}{2}
$$
## Solution

The length of the arc of a curve given by the equation $y=f(x) ; a \leq x \leq b$ is determined by the formula

$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$

Let's find the derivative of the given function:

$$
f^{\prime}(x)=(1-\ln \sin x)^{\prime}=0-\frac{1}{\sin x} \cdot(\sin x)^{\prime}=-\frac{\cos x}{\sin x}
$$

Then, using the formula above, we get:

$$
\begin{aligned}
L & =\int_{\pi / 3}^{\pi / 2} \sqrt{1+\left(-\frac{\cos x}{\sin x}\right)^{2}} d x= \\
& =\int_{\pi / 3}^{\pi / 2} \sqrt{1+\frac{\cos ^{2} x}{\sin ^{2} x}} d x= \\
& =\int_{\pi / 3}^{\pi / 2} \sqrt{\frac{\sin ^{2} x+\cos ^{2} x}{\sin ^{2} x}} d x= \\
& =\int_{\pi / 3}^{\pi / 2} \sqrt{\frac{1}{\sin ^{2} x}} d x=\left|\begin{array}{c}
\pi / 3 \leq x \leq \pi / 2 ; \\
\sin (x) \geq 0
\end{array}\right|=\int_{\pi / 3}^{\pi / 2} \frac{1}{\sin x} d x= \\
& =\left.\ln \left|\operatorname{tg}\left(\frac{x}{2}\right)\right|\right|_{\pi / 3} ^{\pi / 2}=\ln \left|\operatorname{tg} \frac{\pi}{4}\right|-\ln \left|\operatorname{tg} \frac{\pi}{6}\right|=\frac{\ln 3}{2}
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+17-17$ »

Categories: Kuznetsov Problem Book Integrals Problem 17| Integrals

Ukrainian Banner Network

- Last modified: 10:56, 26 May 2010.
- Content is available under CC-BY-SA 3.0.

Created by Gee Teatoo

## Problem Kuznetsov Integrals 17-18

## Material from PlusPi