Integrate wave function squared - M. Chester text

[GreyNoise]

Homework Statement

given: A wire loop with a circumference of L has a bead that moves freely around it. The momentum state function for the bead is $\psi(x) = \sqrt{\frac{2}{L}} \sin \left (\frac{4\pi}{L}x \right )$
find: The probability of finding the bead between $\textstyle \frac{L}{24}$ and $\textstyle \frac{L}{8}$

Homework Equations

$\int_{a}^{b}|<x| \psi >|^2 dx = \int_{a}^{b} | \psi(x) |^2 dx$

$\psi(x) = \sqrt{\frac{2}{L}} \sin \left ( \frac{4\pi}{L}x \right ) \hspace{10mm}$ the state function

The Attempt at a Solution

$\displaystyle \int_{a}^{b}|\lt x|\psi\gt|^2 dx = \int_{a}^{b} | \psi(x) |^2 dx$

${\displaystyle \int_{\frac{L}{24}}^{\frac{L}{8}} \left [ \sqrt{\frac{2}{L}} \sin \left ( \frac{4\pi}{L}x \right ) \right ]^2 dx } \hspace{10mm}$ sub $\textstyle \psi(x) = \sqrt{\frac{2}{L}} \sin \left ( \frac{4\pi}{L}x \right )$ and integration limits

$\displaystyle \int_{\frac{L}{24}}^{\frac{L}{8}} \left [ \sqrt{\frac{2}{L}} \sin \left ( \frac{4\pi}{L}x \right ) \right ]^2 dx = \left. \frac{x}{L} - \frac{ \sin \left ( \frac{8\pi}{L}x \right )}{8\pi} \right |_{\frac{L}{24}}^{\frac{L}{8}}$

$\displaystyle = \frac{1}{L}\frac{L}{8} - \frac{ \sin \left ( \frac{8\pi}{L}\frac{L}{8} \right ) }{8\pi} - \left [ \frac{1}{L}\frac{L}{24} - \frac{ \sin \left ( \frac{8\pi}{L}\frac{L}{24} \right ) }{8\pi} \right ] = \frac{1}{8} - \frac{1}{24} + \frac{ \sin \left ( \frac{\pi}{3} \right ) }{8\pi}$

$= \displaystyle \frac{1}{12} + \frac{\sqrt{3}}{16\pi}$

The answer given in the text is $\frac{1}{12} + \frac{1}{16\pi}$. I cannot shake the $\sqrt{3}$ in the second term. I even checked my evaluation of the integral on the Wolfram site, and it returned the same integral solution as I got. The book is Primer of Quantum Mechanics by Marvin Chester (Dover Publ). It appears to be a first edition, so I guess it's plausible that the text's answer is a typo, but I felt the need to consult the community. Can anyone show me what I am doing wrong?

 

[Charles Link]

In a quick check, I think your calculations are correct, and that the book is in error.

 

[GreyNoise]

Thnx Charles. I am confident of my own answer then.