Question

How do you find the length of the curve for `y= ln(1-x²)` for (0, 1/2)?

Answer 1

The formula for the arc length of the curve `y` on the interval `[a,b]` is given by:

`s=int_a^bsqrt(1+(dy/dx)^2)dx`

Here, where `y=ln(1-x^2)`, then `dy/dx=(-2x)/(1-x^2)=(2x)/(x^2-1)`.

Thus, the arc length in question is:

`s=int_0^(1//2)sqrt(1+((2x)/(x^2-1))^2)dx`

`s=int_0^(1//2)sqrt(((x^2-1)^2+(2x)^2)/(x^2-1)^2)dx`

`s=int_0^(1//2)sqrt(x^4-2x^2+1+4x^2)/(x^2-1)dx`

`s=int_0^(1//2)sqrt(x^4+2x^2+1)/(x^2-1)dx`

Note this is factorable:

`s=int_0^(1//2)sqrt((x^2+1)^2)/(x^2-1)dx`

`s=int_0^(1//2)(x^2+1)/(x^2-1)dx`

Rewriting:

`s=int_0^(1//2)(x^2-1+2)/(x^2-1)dx`

`s=int_0^(1//2)(1+2/(x^2-1))dx`

Perform partial fractions on the second piece:

`2/(x^2-1)=A/(x-1)+B/(x+1)`

`2=A(x+1)+B(x-1)`

Letting `x=1` reveals that

`2=2A" "=>" "A=1`

And letting `x=-1` shows that

`2=-2B" "=>" "B=-1`

So

`2/(x^2-1)=1/(x-1)-1/(x+1)`

Then

`s=int_0^(1//2)(1+1/(x-1)-1/(x+1))dx`

Which are all easily integrateable:

`s=[x+lnabs(x-1)-lnabs(x+1)]_0^(1//2)`

`s=[x+lnabs((x-1)/(x+1))]_0^(1//2)`

`s=(1/2+lnabs((1/2-1)/(1/2+1)))-(0+lnabs((0-1)/(0+1)))`

`s=1/2+lnabs((-1/2)/(3/2))+lnabs(-1)`

`s=1/2+ln(1/3)`

Or

`s=1/2-ln3`