Question

I'm trying to use integrals to find the arc length of y= ln(secx), 0 13 years ago

Answer 1

I understand the formula but Im kind of confused on how to use it on this particular problem

Answer 2

Arc Length = ∫ (√(1 + [f '(x)]^2)) dx {from a to b}

∫ (√(1 + [tan(x)]^2)) dx {from 0 to pi/4}

Use the trig identity sec(x)^2 = 1 + tan(x)^2
= ∫ √(sec(x)^2) dx {from 0 to pi/4}

= ∫ sec(x) dx {from 0 to pi/4}

= ∫ 1/cos(x) dx {from 0 to pi/4}

Multiply the integrand by cos(x)/cos(x).
= ∫ cos(x) / cos(x)^2 dx {from 0 to pi/4}

= ∫ cos(x) / (1 - sin(x)^2) dx {from 0 to pi/4}

Now do a substitution. Let u=sin(x), then du=cos(x) dx.
∫ cos(x) / (1 - sin(x)^2) dx {from 0 to pi/4}

= ∫ du / (1 - u^2) {from 0 to pi/4}

= ∫ du / ((1 - u)*(1 + u)) {from 0 to pi/4}

= (1/2) * ∫ (1 / (1 - u) + 1 / (1 + u)) du {from 0 to pi/4}

= (1/2) * [∫ (1 / (1 - u) du + ∫ 1 / (1 + u)) du] {from 0 to pi/4}

= (1/2) * [ - ln(|1 - u|) + ln(|1 + u|) ] {from 0 to pi/4}

= ln(|1 + u|) / 2 - ln(|1 - u|) / 2 {from 0 to pi/4}

= ln(|1 + sin(x)|) / 2 - ln(|1 - sin(x)|) / 2 {from 0 to pi/4}

= (ln(|1 + sin(pi/4)|) / 2 - ln(|1 - sin(pi/4)|) / 2) - (ln(|1 + sin(0)|) / 2 - ln(|1 - sin(0)|) / 2)

= ln(|1 + sin(pi/4)|) / 2 - ln(|1 - sin(pi/4)|) / 2

= ln(|1 + √2 / 2|) / 2 - ln(|1 - √2 / 2|) / 2

= [ln(|1 + √2 / 2|) - ln(|1 - √2 / 2|)] / 2

= ln(|(1 + √2 / 2) / (1 - √2 / 2)|) / 2

= ln(2*√2 + 3) / 2

= ln((2*√2 + 3)^(1/2))

The √(2*√2 + 3) = 1 + √2, so ln((2*√2 + 3)^(1/2)) = ln(1 + √2)

Answer 3

but how did you get to the first step without getting 1/secx

Answer 4

i have just go it. i cant explain u

Answer 5

well ln(sec(x)) the derivative of this is 1/secx right?

Answer 6

yes

Answer 7

then you square it. and you get 1/sec^2x

Answer 8

how did you get 1+tan^2x in the first step of you integral

Answer 9

cos + x square

Answer 10

where did you get cos + x square?