Question

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

Answer 1

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:

The Length of the curve $y = \log \sec x$ between the points $x = 0$ and $x = \pi / 3$ given a formula

$$s = \int_ {0} ^ {\pi / 3} \sqrt {1 + (y ^ {\prime}) ^ {2}}$$ $$y ^ {\prime} = (\log (\sec (x))) ^ {\prime} = \frac {1}{\sec (x)} (\sec (x)) ^ {\prime} = \tan (x)$$ $$\sqrt {1 + \tan^ {2} x} = \frac {1}{\cos (x)}$$ $$s = \int_ {0} ^ {\pi / 3} \frac {d x}{\cos (x)} = \log \left| \tan (\frac {x}{2} + \frac {\pi}{4}) \right| _ {0} ^ {\pi / 3} = \log \left| \tan (\frac {\pi}{6} + \frac {\pi}{4}) \right| - \log \left| \tan \left(\frac {\pi}{4}\right) \right|$$ $$\tan \left(\frac {\pi}{6} + \frac {\pi}{4}\right) = \frac {\tan \left(\frac {\pi}{6}\right) + \tan \left(\frac {\pi}{4}\right)}{1 - \tan \left(\frac {\pi}{6}\right) * \tan \left(\frac {\pi}{4}\right)} = \frac {\frac {\sqrt {3}}{3} + 1}{1 - \frac {\sqrt {3}}{3}} = 2 + \sqrt {3}$$

So

$$s = \int_ {0} ^ {\pi / 3} \frac {d x}{\cos (x)} = = = \log \left| \tan \left(\frac {\pi}{6} + \frac {\pi}{4}\right) \right| - \log \left| \tan \left(\frac {\pi}{4}\right) \right| = \log (2 + \sqrt {3})$$