Question

exact arc length of y=ln(cos(x)) from x=0 to pi/4
(very important!)

Answer 1

Well, do you understand how arc length is computed?

Answer 2

Yes for the most part...this question just has me stuck

Answer 3

i believe the answer im supposed to get is 1/2ln(3+2sqrt2) but i dont know how

Answer 4

so get that the derivative would be -sinx/cosx and you plug it into the formula and get sqrt(1+(-sinx/cosx)^2dx from 0 to pi/4 but that is where i am stuck

Answer 5

do you know your pythagorean identities?

Answer 6

We can parametrize this such that \(x=t\) and \(y=\ln(\cos(t))\).
Then we can use \[ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}dt\]
We are just trying to find: \[
\int ds = \int \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}dt
\]

Answer 7

@freckles yes i do know my path identities
@wio that is still for arc length?

Answer 8

so you know 1+tan^2(x)=?

Answer 9

sex^2

Answer 10

yep :)

Answer 11

sry sec^2 lol

Answer 12

I know what you meant

Answer 13

For cases where \(y=f(x)\) we let \(x=t\) and our formula simplifies:\[
\int ds = \int \sqrt{1 + \left(\frac{dy}{dx}\right)^2}dx
\]

Answer 14

ohh so since its (-sin^2x/cos^2x) that means its -tan^2? thus being the identity
does the negative matter?

Answer 15

\[
[-\tan(x)]^2 = \tan^2(x)
\]

Answer 16

It should simplify to \(|\sec x|\).

Answer 17

ohh lmfao didnt cath that thanks therefore i get sqrtsecx^2 then giving me the absolute value

Answer 18

However, on the \((0,\pi/4)\) interval, I believe that \(cos\) remains positive, so \(\sec\) should do so as well.

Answer 19

k so then i just do the anti derivative and plug in 0 and pi/4 correct?

Answer 20

I don't think the answer is 1/2ln(3+2sqrt2)

Answer 21

I get
\[\ln (1+\sqrt{2})\]

Answer 22

yes thank you i appreciate the help @alekos after he @wio and @freckles explained the steps i got the same answe...Wish i could at least give them both medals but i cannot i gresstly appreciate everyones help :D