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

Jun 27, 2018

$L = \frac{\pi}{24} + {\sum}_{n = 1}^{\infty} {\sum}_{m = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \left(\begin{matrix}- 1 \\ m\end{matrix}\right) \frac{1}{1 + 4 n + 2 m} \left({\tan}^{1 + 4 n + 2 m} \left(\frac{\pi}{8}\right) - {\tan}^{1 + 4 n + 2 m} \left(\frac{\pi}{12}\right)\right)$ units.

#### Explanation:

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

$f ' \left(x\right) = - {\tan}^{2} x$

Arc length is given by:

$L = {\int}_{\frac{\pi}{12}}^{\frac{\pi}{8}} \sqrt{1 + {\tan}^{4} x} \mathrm{dx}$

For $x \in \left[\frac{\pi}{12} , \frac{\pi}{8}\right]$, ${\tan}^{4} x < 1$. Take the series expansion of the square root:

$L = {\int}_{\frac{\pi}{12}}^{\frac{\pi}{8}} {\sum}_{n = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\tan}^{4 n} x \mathrm{dx}$

Isolate the $n = 0$ term and simplify:

$L = {\int}_{\frac{\pi}{12}}^{\frac{\pi}{8}} \mathrm{dx} + {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\int}_{\frac{\pi}{12}}^{\frac{\pi}{8}} {\tan}^{4 n} x \mathrm{dx}$

Apply the substitution $\tan x = u$:

$L = \frac{\pi}{8} - \frac{\pi}{12} + {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \int {u}^{4 n} / \left({u}^{2} + 1\right) \mathrm{du}$

Take the series expansion of $\frac{1}{{u}^{2} + 1}$:

L=pi/24+sum_(n=1)^oo((1/2),(n))intu^(4n){sum_(m=0)^oo((-1),(m))u^(2m))}du

Simplify:

$L = \frac{\pi}{24} + {\sum}_{n = 1}^{\infty} {\sum}_{m = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \left(\begin{matrix}- 1 \\ m\end{matrix}\right) \int {u}^{4 n + 2 m} \mathrm{du}$

Integrate directly:

$L = \frac{\pi}{24} + {\sum}_{n = 1}^{\infty} {\sum}_{m = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \left(\begin{matrix}- 1 \\ m\end{matrix}\right) \frac{\left[{u}^{1 + 4 n + 2 m}\right]}{1 + 4 n + 2 m}$

Reverse the last substitution:

$L = \frac{\pi}{24} + {\sum}_{n = 1}^{\infty} {\sum}_{m = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \left(\begin{matrix}- 1 \\ m\end{matrix}\right) \frac{1}{1 + 4 n + 2 m} {\left[{\tan}^{1 + 4 n + 2 m} x\right]}_{\frac{\pi}{12}}^{\frac{\pi}{8}}$

Insert the limits of integration:

$L = \frac{\pi}{24} + {\sum}_{n = 1}^{\infty} {\sum}_{m = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \left(\begin{matrix}- 1 \\ m\end{matrix}\right) \frac{1}{1 + 4 n + 2 m} \left({\tan}^{1 + 4 n + 2 m} \left(\frac{\pi}{8}\right) - {\tan}^{1 + 4 n + 2 m} \left(\frac{\pi}{12}\right)\right)$