Consider the following integrals
$\int_0^{2 \pi} \ln(\sin(x)^{2n+1} + C_n) = 0$
All of the $C_n$ are algebraic numbers.
In fact all these $C_n$ can be given as zero's of some integer polynomial with degree $v$ where $0<v<2n+2$.
Conjecture 1 : All these $C_n$ are algebraic numbers of degree exactly $2n+1$.
I know how to compute these $C_n$ yet I do not immediately see how to prove conjecture 1.
Also I do not see the pattern in the polynomials that define $C_n$.
This obviously probably relates to all the ways we can solve those integrals.
$$C_0 = \frac{5}{4}$$
( see the related Why is $\sup f_- (n) \inf f_+ (m) = \frac{5}{4} $? )
$$C_1 = 1.1792756...$$
where $C_1$ is the solution to
$$1/3 * \sqrt{x^2 - 1}(x^2 + 2) - \sqrt{\frac{1}{2}} = 0 $$
or
$$ 17+11*x+3*x^2+5*x^3 = 0 $$
I assume a simple recursion for these polynomials of degree $2n+1$ exists.
They are not the chebyshev polynomials though.
I am aware there are some connections to the gamma function and hypergeo functions.
One way to solve the integral is kinda like this :
differentiation under the integral sign.
substitution y = sin(x)
And then you end up with something like
$$A_m = \int (x^m + y)(1 - x^2)^{-\frac{1}{2}} dx $$
for odd integer $m$.
For instance the case $C_1$ is related to
$$A_3 = (-(\sqrt{1 - x^2} (2 + x^2)))/3 + y \arcsin[x]$$
From where we get
$$1/3 * \sqrt{x^2 - 1}(x^2 + 2) - \sqrt{\frac{1}{2}} = 0 $$
or
$$ 17+11*x+3*x^2+5*x^3 = 0 $$
Example for $C_0$
$$I(u) = \int_0^{2 \pi} \ln (u + \sin x) dx = \pi \ln(u+\sqrt(u²-1)) + c$$
by using differentiation under the integral sign and then integrating.
Computing $c$ by letting $u$ go to infinity and using $u(\sin(x)/u + C_0/u)$
$I(u) = \int \ln (u + \sin x) dx$
$= \int \ln (u (1 + (\sin x)/u)) dx$
$= \int \ln u + \ln (1 + (\sin x)/u) dx$
$= \pi \ln u + \int \ln (1 + (\sin x)/u) dx)$
As $u-> \infty$,
$(\sin x)/u -> 0$
=> $1 + (\sin x)/u -> 1$
=> $\ln (1 + (\sin x)/u) -> 0$
convergence being uniform in all cases, thus the integral approaches 0. Thus, putting equations the above together, as $u-> \infty$
$\pi \ln u -> \pi \ln(u+\sqrt(u²-1)) + c$
$=> c = \lim_{u \to \infty} [\pi \ln u - \pi \ln(u+\sqrt(u²-1))]$ $= \pi \lim_{u \to\infty} [\ln u - \ln(u+\sqrt(u²-1))]$ $= \pi \lim_{u \to \infty} [\ln u - \ln(u+\sqrt(u²))]$ $= \pi \lim_{u \to \infty} [\ln u - \ln(2 u)]$ $= -\pi \ln 2.$
Therefore
$$I(u) = \pi \ln(u+\sqrt{u²-1}) - \pi \ln 2$$
giving $u = 5/4$.
And similar for the other $C_n$.
This is just one way of looking at it ofcourse.
