What is the arc length of f(x) = ln(x^2) on x in [1,3] ?

1 Answer
Apr 2, 2016

sqrt13-sqrt5+2*ln((sqrt65+2*sqrt13-2*sqrt5-4))~=3.006

Explanation:

To do this we need to apply the formula for the length of the curve
mentioned in:
How do you find the length of a curve using integration?

We start from
L=int_a^b sqrt(1+(f' (x))^2 )dx
f(x)=lnx^2=2lnx => f'(x)=2/x
Then
L=int_1^3 sqrt(1+4/x^2)dx=int_1^3 sqrt(x^2+4)/xdx=F(x=3)-F(x=1)

Making
x=2tany
dx=2sec^2y*dy
we get
F(x)=int sqrt(x^2+4)/xdx=int(2secy*cancel2sec^2y)/(cancel2tany)dy=2int (1/cos^3y)(cosy/siny)dy=2int dy/(cos^2y*siny)

But
1/(cos^2y*siny)=(siny)/(cos^2y)+1/siny

So
F(x)=2int siny/cos^2ydy+2int dy/siny

Making cosy=z => siny*dy=-dz
The first part becomes
-2int dz/z^2=2/z=2/cosy

Therefore
F(x)=2/cosy+2ln |csc y-coty| +const.
But
x=2tany => siny=x/2cosy
sin^2y+cos^2y=1 => (x^2/4+1)cos^2y=1 => cosy=2/sqrt(x^2+4)
-> siny=x/2.(2/sqrt(x^2+4)) => siny=x/sqrt(x^2+4)
So
F(x)=sqrt(x^2+4)+2ln |sqrt(x^2+4)/x-2/x|+const.
F(x)=sqrt(x^2+4)+2ln |sqrt(x^2+4)-2|-2ln|x|+const.

Finally
L=F(x=3)-F(x=1)=sqrt13+2ln (sqrt13-2)-2ln3-(sqrt5+2ln(sqrt5-2)-2ln1)
L=sqrt13-sqrt5+ln((sqrt13-2)/(3(sqrt5-2)))
But
(sqrt13-2)/(sqrt5-2)*(sqrt5+2)/(sqrt5+2)=sqrt65+2sqrt13-2sqrt5-4
So

L=sqrt13-sqrt5+ln((sqrt65+2sqrt13-2sqrt5-4)/3)~=3.006