Question

show that the length of the curve y=logsecx between the points x=0 and x=pi/3 is log(2+underoot3)

Answer 1

Answer on Question #37498 – Math - Integral Calculus

If a function is defined as a function of $x$ by $y = f(x)$ then the arc length is given by:

$$s = \int_{a}^{b} \sqrt{1 + (y')^{2}} \, dx$$

We have $y(x) = \ln \sec x$:

$$y'(x) = (\ln \sec x)' = \frac{1}{\sec x} \cdot (\sec x)' = \frac{1}{\sec x} \cdot \sec x \tan x = \tan x$$

Then take the integral:

$$\begin{aligned} s &= \int_{0}^{\pi/3} \sqrt{1 + \tan^{2} x} \, dx = \int_{0}^{\pi/3} \sqrt{\frac{1}{\cos^{2} x}} \, dx = \int_{0}^{\pi/3} \sqrt{\frac{1}{\cos^{2} x}} \, dx = \left| \cos x > 0 \text{ for } x \in \left[0, \frac{\pi}{3}\right] \right| = \\ &= \int_{0}^{\pi/3} \frac{1}{\cos x} \, dx = \int_{0}^{\pi/3} \frac{\cos x}{\cos^{2} x} \, dx = \int_{0}^{\pi/3} \frac{d \sin x}{1 - \sin^{2} x} \end{aligned}$$

Substitute $\sin x = t$. Then $0 \to \sin 0 = 0, \frac{\pi}{3} \to \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$:

$$\begin{aligned} &\int_{0}^{\frac{\sqrt{3}}{2}} \frac{dt}{1 - t^{2}} = \int_{0}^{\frac{\sqrt{3}}{2}} \frac{dt}{(1 - t)(1 + t)} = \frac{1}{2} \int_{0}^{\frac{\sqrt{3}}{2}} \frac{dt}{(t + 1)} - \frac{1}{2} \int_{0}^{\frac{\sqrt{3}}{2}} \frac{dt}{(t - 1)} = \frac{1}{2} \left[ \ln (t + 1) - \ln (t - 1) \right] \Bigg|_{0}^{\frac{\sqrt{3}}{2}} \\ &= \frac{1}{2} \ln \left. \frac{t + 1}{t - 1} \right|_{0}^{\frac{\sqrt{3}}{2}} \\ &= \frac{1}{2} \left[ \ln \frac{\frac{\sqrt{3}}{2} + 1}{\frac{\sqrt{3}}{2} - 1} - \ln 1 \right] \\ &= \frac{1}{2} \ln \left( \frac{(\sqrt{3} + 2)}{\sqrt{3} - 2} \right) \\ &= \frac{1}{2} \ln \left( \frac{(\sqrt{3} + 2)^{2}}{3 - 2} \right) \\ &= \frac{1}{2} \ln \left( \frac{(\sqrt{3} + 2)^{3}}{3 - 2} \right) = \ln (2 + \sqrt{3}) \\ \end{aligned}$$

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