Question

What is the arc length of `f(x) =x -tanx` on `x in [pi/12,(pi)/8]`?

Answer 1

`L=pi/24+sum_(n=1)^oosum_(m=0)^oo((1/2),(n))((-1),(m))1/(1+4n+2m)(tan^(1+4n+2m)(pi/8)-tan^(1+4n+2m)(pi/12))` units.

Explanation:

`f(x)=x-tanx`

`f'(x)=-tan^2x`

Arc length is given by:

`L=int_(pi/12)^(pi/8)sqrt(1+tan^4x)dx`

For `x in [pi/12,pi/8]`, `tan^4x<1`. Take the series expansion of the square root:

`L=int_(pi/12)^(pi/8)sum_(n=0)^oo((1/2),(n))tan^(4n)xdx`

Isolate the `n=0` term and simplify:

`L=int_(pi/12)^(pi/8)dx+sum_(n=1)^oo((1/2),(n))int_(pi/12)^(pi/8)tan^(4n)xdx`

Apply the substitution `tanx=u`:

`L=pi/8-pi/12+sum_(n=1)^oo((1/2),(n))intu^(4n)/(u^2+1)du`

Take the series expansion of `1/(u^2+1)`:

`L=pi/24+sum_(n=1)^oo((1/2),(n))intu^(4n){sum_(m=0)^oo((-1),(m))u^(2m))}du`

Simplify:

`L=pi/24+sum_(n=1)^oosum_(m=0)^oo((1/2),(n))((-1),(m))intu^(4n+2m)du`

Integrate directly:

`L=pi/24+sum_(n=1)^oosum_(m=0)^oo((1/2),(n))((-1),(m))([u^(1+4n+2m)])/(1+4n+2m)`

Reverse the last substitution:

`L=pi/24+sum_(n=1)^oosum_(m=0)^oo((1/2),(n))((-1),(m))1/(1+4n+2m)[tan^(1+4n+2m)x]_(pi/12)^(pi/8)`

Insert the limits of integration:

`L=pi/24+sum_(n=1)^oosum_(m=0)^oo((1/2),(n))((-1),(m))1/(1+4n+2m)(tan^(1+4n+2m)(pi/8)-tan^(1+4n+2m)(pi/12))`