Question

What is the arc length of `f(x)= sqrt(x-1)` on `x in [1,2]`?

Answer 1

`-ln(sqrt5-2)/4+sqrt5/2~=1.4789`

Explanation:

To solve this problem we should use the formula
`L=int_a^b sqrt(1+(f^'(x)^2))*dx`

`f(x)=sqrt(x-1)`
`f^'(x)=(1/2)(x-1)^(-1/2)`
`f^'(x)=(1/2)(x-1)^(-1)`

So we have the indefinite integral that solves the problem:
`F(x)=int sqrt (1-1/4(1/(x-1)))*dx=1/2int sqrt(4-(1/(x-1)))*dx`

Lets begin by using a trigonometrical substitution:
`(1/(x-1))=4*tan^2y`
`-dx/(x-1)^2*dx=4*2tanysec^2y*dy`
`-> -16tan^4y*dx=8tanysec^2y*dy` => `dx=-(sec^2y)/(2tan^3y)dy`

Then the indefinite integral becomes
`1/2int cancel(2)secy(-(sec^2y)/(cancel(2)tan^3y))dy=-1/2int sec^3y/(tan^3y)dy`
`=-1/2int 1/cancel(cos^3y)(cancel(cos^3y)/sin^3y)*dy=-1/2 int csc^3y*dy`

We find this last integral in the more complete tables. Anyway, I'll solve it:
`-1/2int csc^3y*dy=-1/2int cscy*dy-1/2int cscy*cot^2y*dy`

I call this last result as expression 1, and I go on to solve its last term:
`int cscy*cot^2y*dy=int (1/(siny))(cos^2y/sin^2y)dy`
Making
`sin y=z`
`cosy*dy=dz`
So
`=int sqrt(1-z^2)/z^3dz`
Using the rule

`int udv=uv-int vdu`
`-> u=(1-z^2)^(1/2)` => `du=-2z(1/2)(1-z^2)^(-1/2)=-z(1-z^2)^(-1/2)`
`-> dv=dz/z^3` => `v=-1/(2z^2)`
we get
`=-(1-z^2)^(1/2)/(2z^2)-1/2int dz/(z*sqrt(1-z^2))dz`

Calling the result above as expression 2 and proceeding to solve its last term
`-1/2int dz/(z*sqrt(1-z^2))dz=`
`z=sin alpha`
`dz=cos alpha*dalpha`
Therefore
`=-1/2int cancel(cos alpha)/(sin alpha*cancel(cos alpha))dalpha=1/2 int csc alpha*dalpha`
`=-1/2*ln|csc alpha-cot alpha|=-1/2*ln|1/z-sqrt(1-z^2)/z|=-1/2*ln|(1-sqrt(1-z^2))/z|`

Returning to expression 2 :
`=-(1-z^2)^(1/2)/(2z^2)-1/2*ln|(1-sqrt(1-z^2))/z|`
But `z=siny`
`-> =-cosy/(2sin^2y)-1/2*ln|(1-cosy)/siny|=-1/2coty*cscy-1/2*ln|cscy-coty|`

Returning to expression 1 ::
`=-1/2ln|cscy-cot y|+1/4*coty*cscy+1/4*ln|cscy-coty|`
`=-1/4ln|cscy-cot y|+1/4*coty*cscy`
Since
`tan^2y=1/(4(x-1))`
we find that
`cscy=sqrt(4x-3); cot y=2sqrt(x-1);siny=1/sqrt(4x-3) and cosy=2sqrt((x-1)/(4x-3))`

So

`F(x)=-1/4*ln|sqrt(4x-3)-2sqrt(x-1)|+(1/4)2sqrt(x-1)*sqrt(4x-3)`
`F(x)= -1/4*ln|sqrt(4x-3)-2sqrt(x-1)|+1/2*sqrt((x-1)(4x-3))`

Finally
`L=F(x=3)-F(x=1)`
`L=-ln(sqrt 5-2)/4+sqrt5/2~=1.4789`