Question

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

Answer 1

I know that the derivative of ln(secx) = 1/secx thats where I'm at I dont really know how to evaluate any further

Answer 2

I really just need a shove in the right direction

Answer 3

The derivative of ln(sec) is secxtanx/secx which simplifies to tanx

Answer 4

really.... I thought anything in a ln was just 1/??? like ln(x) = 1/x

Answer 5

The formula for finding the arc length is:
\[\int\limits_{0}^{\pi/4}\sqrt{1+(f'(x))^{2}}dx\]

Answer 6

the derivative of ln(f(x))=f'(x)/f(x)
the derivative of the inner function divided by the original function

Answer 7

Plugging everything in we find that the arc length of of y= ln(secx), 0<x<pi/4 is found by evaluating:
\[\int\limits_{0}^{\pi/4}\sqrt{1+\tan ^{2}x} \ dx\]

Answer 8

Ok, I just worked that out with you now I;m stuck again. So can I get both of those terms out of the sqrt?

Answer 9

if you could just help me through the next step i should be able to take it from there

Answer 10

normally I'd say trig-sub, but because 1+tan^2=sec^2 I would just go from that. sqrt(sec^2)=secx, so just
\[\int\limits_{0}^{\pi/4}\sec x \ dx\]

Answer 11

duh thanks lol

Answer 12

np

Answer 13

Ok, man I got ln(sqrt(2)+1)-ln(1)
but the right answer would have been sqrt(2)+ln(sqrt2+1)

Answer 14

I'm getting the same thing. I'll check with wolfram

Answer 15

Wolfram's saying you're correct as well http://www.wolframalpha.com/input/?i=arc+length+of+y%3D+ln%28secx%29%2C+0%3Cx%3Cpi%2F4 could be a mistake in your answer key