Question #180269

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

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1
Expert's answer
2021-05-07T09:06:14-0400

Given f(x) = y = log(s\because ec(x)) on [0,π4\frac{\pi}{4}]


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

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

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


\therefore L = 0π4(tan(x))2+1dx\int_{0}^{\frac{\pi}{4}} \sqrt{(tan(x))^2 +1 }dx


= 0π4tan2(x)+1dx\int_{0}^{\frac{\pi}{4}} \sqrt{tan^2(x) +1 }dx


= 0π4sec2(x)dx\int_{0}^{\frac{\pi}{4}} \sqrt{sec^2(x) }dx ( \because sec2(x) - tan2(x) = 1)

= 0π4sec(x)dx\int_{0}^{\frac{\pi}{4}} sec(x) dx

= [ log( sec(x) +tan(x) )]0π4|^\frac{\pi}{4}_0


= [log(tan(π4\frac{\pi}{4}) + sec(π4\frac{\pi}{4})] - [log(tan(0) + sec(0)]


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


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




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