Define the integral $$I_n = \int_0^{\pi} ((\sin{x})^3+(\cos{x})^3)^n dx$$ for any natural number $n$. I am trying to show that
$$I_n\text{ is rational } \iff n \text{ is odd}$$
My first idea was to create a recursive relation between $I_n$ and $I_{n-2}$ and use the result that $I_1 = \frac{4}{3}$ is rational and $I_2 = \frac{5}{8}\pi$ is irrational. However, I was not able to make any progress. Running this integral in mathematica 14.0 gives a closed form output in terms of Hypergeom function, but I can't make head or tails of it. Any ideas?
Best Answer
Not an answer, but an idea of how to get it.
Consider:
$$ I_n = \int_0^\pi (\sin^3{x} + \cos^3{x})^n dx = \sum_{k=0}^n \int_0^\pi \binom{n}{k} \sin^{3k}x \cos^{3(n-k)}xdx $$
Now we have several facts:
$\displaystyle \int_0^\pi \sin^{3k}x \cos^{3(n-k)}xdx \propto B\left(\frac{3k + 1}{2}, \frac{3(n-k) + 1}{2}\right)$.
For $n = 2m$, we have either zero term, or $C \pi$ term (why?).
For $n = 2m + 1$, we have either zero term, or $q \in \mathbb{Q}$ term (why?).