Question

Show that the length of the curve y = logSecx between the points x = 0 and x = π/3 is log (2 + √3 ).

Answer 1

Answer on question #34678 – Math – Real Analysis

Show that the length of the curve $y = \log \sec x$ between the points $x = 0$ and $x = \pi/3$ is $\log(2 + \sqrt{3})$

Solution

Using the formula

$$l(f) = \int_{a}^{b} \sqrt{1 + \left(f'(x)\right)^2} dx$$

We obtain

$$l(y) = \int_{0}^{\pi/3} \sqrt{1 + \left(\left(\ln \sec x\right)'\right)^2} dx$$ $$(\ln \sec x)' = \frac{1}{\sec x} (\sec x)' = \cos x \left(\frac{\sin x}{\cos^2 x}\right) = \tan x$$

Substitute this into (*) we get

$$\begin{aligned} l(y) = & \int_{0}^{\frac{\pi}{3}} \sqrt{1 + (\tan x)^2} dx = \int_{0}^{\frac{\pi}{3}} \sqrt{\frac{1}{\cos^2 x}} dx = \int_{0}^{\frac{\pi}{3}} \frac{1}{\cos x} dx = \int_{0}^{\frac{\pi}{3}} \sec x \, dx = \\ & = \ln |\sec x + \tan x| \Big|_{0}^{\frac{\pi}{3}} = \ln(2 + \sqrt{3}). \end{aligned}$$

QED.

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