Chapter 7.7, Problem 42E

The figure shows a pendulum with length L that makes a maximum angle ${\theta }_0$ with the vertical. Using Newton’s Second Law, it can be shown that the period T (the time for one complete swing) is given by

$T=4\sqrt{\frac{L}{g}}{\displaystyle {\int }_0^{\pi /2}\frac{dx}{\sqrt{1-k^2{\sin }^{2}x}}}$

where $k=\sin \left(\frac{1}{2}{\theta }_0\right)$ and g is the acceleration due to gravity. If $L=1\text{ m}$ and ${\theta }_0={42}^0$, use Simpson’s Rule with $n=10$ to find the period.