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

Question

1730 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-02T15:25:19

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

$f(x)=-cscx$

$f'(x)=cscxcotx$

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:

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

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