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))#