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

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Answer 1 · 2018-06-27T18:46:19

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

$f(x)=x-tanx$

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

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:

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

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