A question regarding Czarnowski's cat function

Question

The Czarnowski cat function $c(x, n) = \sin^n(x) + \cos(x)$ gives the so-called Czarnowski's cats. There are infinitely many of these cats, and legend has it that Czarnowski created a clever mathematical naming scheme to give them all unique names.

The problem: for Czarnowski's principal cat (named Mruk, the one centered at the origin) find the area between the cats corresponding to $n = 18$ and $n = 26$.

Some background. In 1896, Polish mathematician Żurzysław Czarnowski (born 1875) discovered the cat function while doodling on his note pad during a particularly boring talk by his teacher, Belgian mathematician Per Hansel. It is rumored that one day Czarnowski presented the cat function to Hansel, who studied it and noticed that for very large $n$ the cats turn into devils. It is rumored that Hansel was so terrified and distraught by this discovery that he committed suicide. This prompted Czarnowski to find an optimal cat, i.e. a value $n$ for which the cats are "most cat-like and least devil-like". It is rumored that this $n$ lies somewhere between 18 and 26, and that if we can find the area between the curves corresponding to "cat 18" and "cat 26" we will have half of the solution.

Answer 1

We seek a reduction formula for integrands of the form $$\sin^{2n} x.$$ Integration by parts with the choice $$u = \sin^{2n-1} x, \quad du = (2n-1) \sin^{2n-2} x \cos x \, dx, \\ dv = \sin x \, dx, \quad v = -\cos x, $$ gives $$\begin{align*} I_n (x) &= \int \sin^{2n} x \, dx = -\sin^{2n-1} x \cos x + (2n-1) \int \sin^{2n-2} x \cos^2 x \, dx \\ &= -\sin^{2n-1} x \cos x + (2n-1) \int \sin^{2n-2} (1 - \sin^2 x) \, dx \\ &= -\sin^{2n-1} x \cos x + (2n-1) I_{n-1}(x) - (2n-1)I_n(x). \end{align*}$$ Consequently, $$I_n(x) = \frac{1}{2n} \left( -\sin^{2n-1} x \cos x + (2n-1) I_{n-1}(x) \right).$$ On the interval $[0,\pi]$, we then have $$I_n = \int_{x=0}^\pi I_n(x) \, dx = \frac{2n-1}{2n} I_{n-1}.$$ And for $n = 0$, we trivially have $$I_0(x) = 1, \quad I_0 = \pi.$$ It follows that $$I_n = \pi \prod_{k=1}^n \frac{2k-1}{2k} = \pi \prod_{k=1}^n \frac{(2k-1)(2k)}{(2k)^2} = \frac{(2n)!}{4^n (n!)^2} \pi = \binom{2n}{n}\frac{\pi}{4^n}.$$ Therefore, since $c(x,n+2) \le c(x,n)$, $$\begin{align*} \int_{x=-\pi}^\pi c(x,18) - c(x,26) \, dx &= 2 \int_{x=0}^\pi \sin^{18} x - \sin^{26} x \, dx \\ &= 2(I_{9} - I_{13}) \\ &= 2\pi \left(\binom{18}{9}\frac{1}{4^{9}} - \binom{26}{13} \frac{1}{4^{13}}\right) \\ &= \frac{\pi}{2^{25}}\left(\binom{18}{9} 4^4 - \binom{26}{13}\right) \\ &= \frac{255765 \pi }{4194304}. \end{align*}$$