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

1 Answer
Jun 2, 2018

$L = \csc \left(\frac{\pi}{12}\right) - \csc \left(\frac{\pi}{8}\right) + {\sum}_{n = 1}^{\infty} {\sum}_{k = 0}^{2 n - 1} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \left(\begin{matrix}2 n - 1 \\ k\end{matrix}\right) {\left(- 1\right)}^{k} {\int}_{\frac{\pi}{12}}^{\frac{\pi}{8}} {\sec}^{2 \left(n - k\right) - 1} x \mathrm{dx}$ units.

Explanation:

$f \left(x\right) = - \csc x$

$f ' \left(x\right) = \csc x \cot x$

Arc length is given by:

$L = {\int}_{\frac{\pi}{12}}^{\frac{\pi}{8}} \sqrt{1 + {\csc}^{2} x {\cot}^{2} x} \mathrm{dx}$

Rearrange:

$L = {\int}_{\frac{\pi}{12}}^{\frac{\pi}{8}} \csc x \cot x \sqrt{1 + {\sin}^{2} x {\tan}^{2} x} \mathrm{dx}$

For $x \in \left[\frac{\pi}{12} , \frac{\pi}{8}\right]$, ${\sin}^{2} x {\tan}^{2} x < 1$. Take the series expansion of the square root:

$L = {\int}_{\frac{\pi}{12}}^{\frac{\pi}{8}} \csc x \cot x \left\{{\sum}_{n = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(\sin x \tan x\right)}^{2 n}\right\} \mathrm{dx}$

Isolate the $n = 0$ term:

$L = {\int}_{\frac{\pi}{12}}^{\frac{\pi}{8}} \csc x \cot x \mathrm{dx} + {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\int}_{\frac{\pi}{12}}^{\frac{\pi}{8}} {\left(\sin x \tan x\right)}^{2 n - 1} \mathrm{dx}$

Rearrange:

$L = {\left[- \csc x\right]}_{\frac{\pi}{12}}^{\frac{\pi}{8}} + {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\int}_{\frac{\pi}{12}}^{\frac{\pi}{8}} {\left(\sec x - \frac{1}{\sec} x\right)}^{2 n - 1} \mathrm{dx}$

Apply binominal expansion:

$L = \csc \left(\frac{\pi}{12}\right) - \csc \left(\frac{\pi}{8}\right) + {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\int}_{\frac{\pi}{12}}^{\frac{\pi}{8}} {\sum}_{k = 0}^{2 n - 1} \left(\begin{matrix}2 n - 1 \\ k\end{matrix}\right) {\left(\sec x\right)}^{2 n - 1 - k} {\left(- \frac{1}{\sec} x\right)}^{k} \mathrm{dx}$

Rearrange:

$L = \csc \left(\frac{\pi}{12}\right) - \csc \left(\frac{\pi}{8}\right) + {\sum}_{n = 1}^{\infty} {\sum}_{k = 0}^{2 n - 1} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \left(\begin{matrix}2 n - 1 \\ k\end{matrix}\right) {\left(- 1\right)}^{k} {\int}_{\frac{\pi}{12}}^{\frac{\pi}{8}} {\sec}^{2 \left(n - k\right) - 1} x \mathrm{dx}$