The average wave

Another wave is approaching the island, but no one knows which direction it’s coming from—for the moment, all directions are equally likely. On average, what is the length of the stretch of land directly under the wave halfway between when the wave first and last makes contact with the island?

Using our function $$\ell (\theta) = \min \left\{ \sqrt{(3 - \cos \theta) (1 + \cos \theta)}, \frac{ \tan \theta (1- \cos \theta) + \sqrt{(3-\cos \theta) (1 + \cos \theta)}}{2} \right\}$$ from the Classic answer along with the symmetry arguments that we made there to come to the conclusion that the average length of the stretch of land is \begin{align*}\bar{\ell} &= \int_0^{\pi/2} \ell(\theta) \frac{ 2d\theta }{ \pi } = \frac{2}{\pi} \int_0^{\cos^{-1} 1/3} \frac{ \tan \theta ( 1- \cos \theta) + \sqrt{(3 - \cos \theta) (1 + \cos \theta) }}{2} \,d\theta \\ &\quad\quad\quad\quad\quad\quad\quad\quad\quad + \frac{2}{\pi} \int_{\cos^{-1} 1/3}^{\pi/2} \sqrt{ (3 - \cos \theta) (1 + \cos \theta) } \,d\theta \end{align*}

Well that is certainly a mouthful, but we can break it into some pieces and come up with an analytical solution. First we see that if we rewrite the portion under the radical as $$\frac{\sqrt{(3 - \cos \theta)( 1 + \cos \theta) }}{2} = \sqrt{ 1- \left(\frac{1-\cos \theta}{2} \right)^2}$$ that we can do some trigonometrical simplifications and substitutions. So in particular we see that since $\sin \frac{\theta}{2} = \sqrt{ \frac{1-cos \theta}{2} }$ when $\theta \in (0, \frac{\pi}{2}),$ then we get \begin{align*}\int \sqrt{ 1- \left( \frac{ 1 - \cos \theta }{2} \right)^2 } \,d\theta &= \int \sqrt{ 1 - \sin^4 \frac{\theta}{2} } \,d\theta \\&= \int \sqrt{ \left( 1 - \sin^2 \frac{\theta}{2} \right) \left( 1 + \sin^2 \frac{\theta}{2} \right) } \,d\theta \\ &= \int \cos \frac{\theta}{2} \sqrt{ 1 + \sin^2 \frac{\theta}{2} } \,d\theta \\ &= 2\int \sqrt{1 + u^2} \,du,\end{align*} where the last equation involves the substitution of $u = \sin \frac{\theta}{2}.$ Using integration by parts, we get $$\int \sqrt{1+u^2} \,du = u \sqrt{1 + u^2} - \int \frac{u^2 \,du}{\sqrt{1+u^2}} = u \sqrt{1+u^2} + \int \frac{du}{\sqrt{1+u^2}} - \int \sqrt{ 1 + u^2 } \,du,$$ so that we get $$\int \sqrt{1+u^2} \,du = \frac{1}{2} \left( u \sqrt{1 + u^2} + \int \frac{du}{\sqrt{1+u^2}} \right) = \frac{1}{2} \left( u \sqrt{1+ u^2} + \sinh^{-1} u\right) + C.$$ Therefore we see that \begin{align*}\int \frac{ \sqrt{ (3-\cos \theta) (1 + \cos \theta) } }{2} \,d\theta &= \int \sqrt{ 1 - \left( \frac{1 - \cos \theta}{2} \right)^2 } \,d\theta \\ &= \sin \frac{\theta}{2} \sqrt{ 1 + \sin^2 \frac{\theta}{2} } + \sinh^{-1} \left(\sin \frac{\theta}{2}\right) + C.\end{align*} Similarly, but with a lot less fuss, we can get $$\int \frac{\tan \theta (1 - \cos \theta)}{2} \,d\theta = -\frac{1}{2} \ln | \cos \theta | + \frac{\cos \theta}{2} + C.$$

Therefore the the first integral \begin{align*}I_1 &= \frac{2}{\pi} \int_0^{\cos^{-1} 1/3} \frac{ \tan \theta (1 - \cos \theta) }{2} + \sqrt{ 1 - \left( \frac{1-\cos \theta}{2} \right)^2 } \,d\theta \\ &= \frac{2}{\pi}\left[ -\frac{1}{2} \ln | \cos \theta | - \frac{\cos \theta}{2} + \sin \frac{\theta}{2} \sqrt{ 1 + \sin^2 \frac{\theta}{2} } + \sinh^{-1} \left( \sin \frac{\theta}{2} \right) \right]^{\theta = \cos^{-1} 1/3}_{\theta = 0} \\ &= \frac{2}{\pi}\left( -\frac{1}{2} \ln \frac{1}{3} + \frac{1}{6} + \sqrt{ \frac{ 1 - \frac{1}{3}}{2}} \sqrt{ 1 + \left(\sqrt{ \frac{ 1 - \frac{1}{3} }{2} }\right)^2 } + \sinh^{-1} \sqrt{ \frac{1 - \frac{1}{3}}{2} } \right) \\ &\quad\quad\quad\quad\quad - \frac{2}{\pi}(0 + \frac{1}{2} + 0 + 0) \\ &= \frac{2}{\pi} \left(\frac{1}{2} \ln 3 + \frac{1}{3} + \sinh^{-1} \sqrt{\frac{1}{3}} \right) \\ &= \frac{2}{\pi}\ln 3 + \frac{2}{3\pi}, \end{align*} since $\sinh^{-1} t = \ln ( t + \sqrt{ 1 + t^2} ),$ so $$\sinh^{-1} \sqrt{\frac{1}{3}} = \ln \left( \sqrt{\frac{1}{3}} + \sqrt{1 + \frac{1}{3}} \right) = \ln \left( \sqrt{\frac{1}{3}} + \sqrt{ \frac{4}{3}} \right) = \ln \left( 3 \sqrt{\frac{1}{3}} \right) = \ln \sqrt{3} = \frac{1}{2} \ln 3.$$ The second integral is \begin{align*}I_2 &= \frac{4}{\pi} \int_{\cos^{-1} 1/3}^{\pi/2} \sqrt{ 1 - \left( \frac{1 - \cos \theta}{2} \right)^2 } \,d\theta \\ &= \frac{4}{\pi} \left[ \sin \frac{\theta}{2} \sqrt{ 1 + \sin^2 \frac{\theta}{2} } + \sinh^{-1} ( \sin \frac{\theta}{2} ) \right]_{\theta=\cos^{-1} 1/3}^{\theta=\pi/2} \\ &= \frac{4}{\pi} \left( \frac{1}{\sqrt{2}} \cdot \sqrt{1+\frac{1}{2}} + \sinh^{-1}\frac{1}{ \sqrt{2}} \right) \\ &\quad\quad\quad - \frac{4}{\pi} \left( \sqrt{ \frac{ 1 - \frac{1}{3}}{2}} \sqrt{ 1 + \left(\sqrt{ \frac{ 1 - \frac{1}{3} }{2} }\right)^2 } + \sinh^{-1} \sqrt{ \frac{1 - \frac{1}{3}}{2} } \right) \\ &= \frac{4}{\pi} \left( \frac{\sqrt{3}}{2} + \sinh^{-1} \frac{1}{\sqrt{2}} - \frac{2}{3} - \frac{1}{2} \ln 3 \right)\end{align*} Putting this all together and taking advantage of some beneficial cancelling, we get that the average length of the stretch of land covered by the wave at 10:05 a.m. if all directions for it to approach Semicircle Island are equally likely is $$\bar{\ell} = I_1 + I_2 = \frac{4}{\pi} \left( \frac{ \sqrt{3}}{2} + \sinh^{-1} \frac{1}{\sqrt{2}} \right) - \frac{2}{\pi} \approx 1.30443945503\dots$$ miles.