Answer to Question #180269 in Calculus for Prathamesh

Question

Determine the length of curve 𝑦 = log(sec 𝑥) between 0 ≤ 𝑥 ≤

𝜋

4

Answer 1

Given f(x) = y = log(s$\because$ ec(x)) on [0,$\frac{\pi}{4}$]

The arc length is given by L = $\int_{a}^{b} \sqrt{(f'(x))^2 +1 } \,dx$

we have f(x) = log(sec(x))

$\implies$ f'(x) = $\frac{1}{sec(x)} \cdot$ (sec(x) $\cdot$ tan(x)) = tan(x)

$\therefore$ L = $\int_{0}^{\frac{\pi}{4}} \sqrt{(tan(x))^2 +1 }dx$

= $\int_{0}^{\frac{\pi}{4}} \sqrt{tan^2(x) +1 }dx$

= $\int_{0}^{\frac{\pi}{4}} \sqrt{sec^2(x) }dx$ ( $\because$ sec²(x) - tan²(x) = 1)

= $\int_{0}^{\frac{\pi}{4}} sec(x) dx$

= [ log( sec(x) +tan(x) )]$|^\frac{\pi}{4}_0$

= [log(tan($\frac{\pi}{4}$) + sec($\frac{\pi}{4}$)] - [log(tan(0) + sec(0)]

= [log(1+ $\sqrt{2}$ )] - [log(1)]

$\therefore$ L = log(1+ $\sqrt{2}$ )