arc length / y=ln(e^x+1)/(e^x-1) / y=ln(cos x)

[Guest]

I have 2 different problems here.
First,

y=ln(e^x+1)/(e^x-1)

Second,

y=ln(cos x)

I take the derivative and then do the integral in the region of sqrt(1+y'(x))

So, the first I get to sqrt(1+[(-2e^x)/(e^x-1)^2]^2

I can't seem to figure out how to reduce this further so I can integrate and solve for arclength with the given region. Can someone help with this?

I have the same issue with the second one. I get sqrt.(1-(sinx/cosx)^2)
where do I go from here?

Thanks

 

[soroban]

Re: arc length

Hello, ezrajoelmicah!

Your algebra is bit off . . .

\(\displaystyle y \:=\:\ln\left(\frac{e^x\,+\,1}{e^x\,-\,1}\right)\)

I take the derivative and then do the integral of: \(\displaystyle \,\sqrt{1\,+\,\left(\frac{dy}{dx}\right)^2}\;\) . . . Right!

We have: \(\displaystyle \,y \:=\:\ln\left(\frac{e^x\,+\,1}{e^x\,-\,1}\right) \:=\:\ln(e^x\,+\,1)\,-\,\ln(e^x\,-\,1}\)

Then: \(\displaystyle \L\,\frac{dy}{dx} \;=\;\frac{e^x}{e^x\,+\,1}\,-\,\frac{e^x}{e^x\,-\,1} \;=\;\frac{-2e^x}{e^{2x}\,-\,1}\)

And: \(\displaystyle \L\,1\,+\,\left(\frac{dy}{dx}\right)^2\;= \;1\,+\,\frac{4e^{2x}}{(e^{2x}\,-\,1)^2} \;= \;\frac{(e^{2x}\,-\,1)^2\,+\,4e^{2x}}{(e^{2x}\,-\,1)^2}\)

. . \(\displaystyle \L=\;\frac{e^{4x}\,-\,2e^{2x}\,+\,1\,+\,4e^{2x}}{(e^{2x}\,-\,1)^2} \;= \;\frac{e^{4x}\,+\,2e^{2x}\,+\,1}{(e^{2x}\,-\,1)^2} \;=\;\frac{(e^{2x}\,+\,1)^2}{(e^{2x}\,-\,1)^2}\)

Then: \(\displaystyle \L\:\sqrt{1\,+\,\left(\frac{dy}{dx}\right)^2} \;= \;\sqrt{\frac{(e^{2x}\,+\,1)^2}{(e^{2x}\,-\,1)^2}} \;= \;\frac{e^{2x}\,+\,1}{e^{2x}\,-\,1}\)

Therefore: \(\displaystyle \L\:L\;=\;\int^{\;\;\;b}_a\frac{e^{2x}\,+\,1}{e^{2x}\,-\,1}\,dx\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

For the second one: \(\displaystyle \:\frac{dy}{dx}\:=\:\frac{-\sin x}{\cos x}\:=\:-\tan x\)

Then: \(\displaystyle \:1\,+\,\left(\frac{dy}{dx}\right)^2\:=\:1\,+\,(-\tan x)^2\:=\:1\,+\,\tan^2x\:=\:\sec^2x\)

Hence: \(\displaystyle \,\sqrt{1\,+\,\left(\frac{dy}{dx}\right)^2} \:=\:\sqrt{\sec^2x} \:=\:\sec x\)

Got it?