## Problem Statement

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

$$
y=\ln \cos x+2, \quad 0 \leq x \leq \frac{\pi}{6}
$$
## 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:

$$
f^{\prime}(x)=(\ln \cos x+2)^{\prime}=\frac{1}{\cos x} \cdot(\cos x)^{\prime}=\frac{-\sin x}{\cos x}=-\operatorname{tg} x
$$

Then, using the above formula, we get:

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

Notice that $\frac{d x}{\cos ^{2} \alpha}=d(\operatorname{tg} \alpha)$ :

$$
\begin{aligned}
L & =\frac{1}{2} \int_{0}^{\pi / 6} \frac{1}{\operatorname{tg}\left(\frac{\pi}{4}+\frac{x}{2}\right)} \cdot \frac{d x}{\cos ^{2}\left(\frac{\pi}{4}+\frac{x}{2}\right)}= \\
& =2 \cdot \frac{1}{2} \int_{0}^{\pi / 6} \frac{1}{\operatorname{tg}\left(\frac{\pi}{4}+\frac{x}{2}\right)} \cdot d\left(\operatorname{tg}\left(\frac{\pi}{4}+\frac{x}{2}\right)\right)= \\
& =\left.\ln \left(\operatorname{tg}\left(\frac{\pi}{4}+\frac{x}{2}\right)\right)\right|_{0} ^{\pi / 6}= \\
& =\ln \operatorname{tg}\left(\frac{\pi}{4}+\frac{\pi}{12}\right)-\ln \operatorname{tg}\left(\frac{\pi}{4}+0\right)= \\
& =\ln \operatorname{tg} \frac{\pi}{3}-\ln \operatorname{tg} \frac{\pi}{4}=\ln \sqrt{3}-\ln 1=\ln \sqrt{3}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+17-25$ »

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

Ukrainian Banner Network

- Last modified: 12:04, 26 May 2010.
- Content is available under CC-BY-SA 3.0.

Created by GeeTeatoo

## Problem Kuznetsov Integrals 17-26

## Material from PlusPi