## Problem Statement

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

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

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

Then, using the above formula, we get:

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

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\� \%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5\%D0\%B3 $\% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_17-21$ »

Categories: Kuznetsov's Problem Book Integrals Problem 17| Integrals

- Last modified: 11:48, 26 May 2010.
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## Problem Kuznetsov Integrals 17-22

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