Question

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

Answer 1

`L=csc(pi/12)-csc(pi/8)+sum_(n=1)^oosum_(k=0)^(2n-1)((1/2),(n))((2n-1),(k))(-1)^kint_(pi/12)^(pi/8)sec^(2(n-k)-1)xdx` units.

Explanation:

`f(x)=-cscx`

`f'(x)=cscxcotx`

Arc length is given by:

`L=int_(pi/12)^(pi/8)sqrt(1+csc^2xcot^2x)dx`

Rearrange:

`L=int_(pi/12)^(pi/8)cscxcotxsqrt(1+sin^2xtan^2x)dx`

For `x in [pi/12,pi/8]`, `sin^2xtan^2x<1`. Take the series expansion of the square root:

`L=int_(pi/12)^(pi/8)cscxcotx{sum_(n=0)^oo((1/2),(n))(sinxtanx)^(2n)}dx`

Isolate the `n=0` term:

`L=int_(pi/12)^(pi/8)cscxcotxdx+sum_(n=1)^oo((1/2),(n))int_(pi/12)^(pi/8)(sinxtanx)^(2n-1)dx`

Rearrange:

`L=[-cscx]_ (pi/12)^(pi/8)+sum_(n=1)^oo((1/2),(n))int_(pi/12)^(pi/8)(secx-1/secx)^(2n-1)dx`

Apply binominal expansion:

`L=csc(pi/12)-csc(pi/8)+sum_(n=1)^oo((1/2),(n))int_(pi/12)^(pi/8)sum_(k=0)^(2n-1)((2n-1),(k))(secx)^(2n-1-k)(-1/secx)^kdx`

Rearrange:

`L=csc(pi/12)-csc(pi/8)+sum_(n=1)^oosum_(k=0)^(2n-1)((1/2),(n))((2n-1),(k))(-1)^kint_(pi/12)^(pi/8)sec^(2(n-k)-1)xdx`