# What is the arclength of (tant-sect*csct) on t in [pi/8,pi/3]?

Mar 18, 2018

$L = \cot \left(\frac{\pi}{8}\right) - \cot \left(\frac{\pi}{3}\right) + 4 {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{1}{16} ^ n \left\{\left(\begin{matrix}4 n - 2 \\ 2 n - 1\end{matrix}\right) \frac{5 \pi}{24} + {\sum}_{k = 0}^{2 n - 2} \left(\begin{matrix}4 n - 2 \\ k\end{matrix}\right) {\left(- 1\right)}^{2 n - 1 - k} / \left(2 n - 1 - k\right) \left(\sin \left(\left(4 n - 2 - 2 k\right) \frac{\pi}{3}\right) - \sin \left(\left(4 n - 2 - 2 k\right) \frac{\pi}{3}\right)\right)\right\}$ units.

#### Explanation:

$f \left(t\right) = \tan t - \sec t \cdot \csc t = - \cot t$

$f ' \left(t\right) = {\csc}^{2} t$

Arclength is given by:

$L = {\int}_{\frac{\pi}{8}}^{\frac{\pi}{3}} \sqrt{1 + {\csc}^{4} t} \mathrm{dt}$

Rearrange:

$L = {\int}_{\frac{\pi}{8}}^{\frac{\pi}{3}} {\csc}^{2} t \sqrt{1 + {\sin}^{4} t} \mathrm{dt}$

Since ${\sin}^{4} t < 1$, take the series expansion of the square root:

$L = {\int}_{\frac{\pi}{8}}^{\frac{\pi}{3}} {\csc}^{2} t \left\{{\sum}_{n = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\sin}^{4 n} t\right\} \mathrm{dt}$

Isolate the $n = 0$ term:

$L = {\int}_{\frac{\pi}{8}}^{\frac{\pi}{3}} {\csc}^{2} t \mathrm{dt} + {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\int}_{\frac{\pi}{8}}^{\frac{\pi}{3}} {\sin}^{4 n - 2} t \mathrm{dt}$

Apply the trigonometric power-reduction formula:

$L = {\left[- \cot t\right]}_{\frac{\pi}{8}}^{\frac{\pi}{3}} + {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\int}_{\frac{\pi}{8}}^{\frac{\pi}{3}} \left\{\frac{1}{2} ^ \left(4 n - 2\right) \left(\begin{matrix}4 n - 2 \\ 2 n - 1\end{matrix}\right) + \frac{2}{2} ^ \left(4 n - 2\right) {\sum}_{k = 0}^{2 n - 2} \left(\begin{matrix}4 n - 2 \\ k\end{matrix}\right) {\left(- 1\right)}^{2 n - 1 - k} \cos \left(\left(4 n - 2 - 2 k\right) t\right)\right\} \mathrm{dt}$

Simplify:

$L = \cot \left(\frac{\pi}{8}\right) - \cot \left(\frac{\pi}{3}\right) + 4 {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{1}{16} ^ n {\int}_{\frac{\pi}{8}}^{\frac{\pi}{3}} \left\{\left(\begin{matrix}4 n - 2 \\ 2 n - 1\end{matrix}\right) + 2 {\sum}_{k = 0}^{2 n - 2} \left(\begin{matrix}4 n - 2 \\ k\end{matrix}\right) {\left(- 1\right)}^{2 n - 1 - k} \cos \left(\left(4 n - 2 - 2 k\right) t\right)\right\} \mathrm{dt}$

Integrate directly:

$L = \cot \left(\frac{\pi}{8}\right) - \cot \left(\frac{\pi}{3}\right) + 4 {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{1}{16} ^ n {\left[\left(\begin{matrix}4 n - 2 \\ 2 n - 1\end{matrix}\right) t + {\sum}_{k = 0}^{2 n - 2} \left(\begin{matrix}4 n - 2 \\ k\end{matrix}\right) {\left(- 1\right)}^{2 n - 1 - k} / \left(2 n - 1 - k\right) \sin \left(\left(4 n - 2 - 2 k\right) t\right)\right]}_{\frac{\pi}{8}}^{\frac{\pi}{3}}$

Simplify:

$L = \cot \left(\frac{\pi}{8}\right) - \cot \left(\frac{\pi}{3}\right) + 4 {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{1}{16} ^ n \left\{\left(\begin{matrix}4 n - 2 \\ 2 n - 1\end{matrix}\right) \frac{5 \pi}{24} + {\sum}_{k = 0}^{2 n - 2} \left(\begin{matrix}4 n - 2 \\ k\end{matrix}\right) {\left(- 1\right)}^{2 n - 1 - k} / \left(2 n - 1 - k\right) \left(\sin \left(\left(4 n - 2 - 2 k\right) \frac{\pi}{3}\right) - \sin \left(\left(4 n - 2 - 2 k\right) \frac{\pi}{3}\right)\right)\right\}$