Is there a closed form for
$$f(2n-1) = \int_0^{2 \pi} \sin((2n-1)x) \ln(\sin(x)+5/4) dx $$
Of course this comes from searching the fourier series of $ \ln(\sin(x)+5/4) dx $.
In fact, it is equivalent to searching for that Fourier series since
$\int_0^{2 \pi} \sin((2n)x) \ln(\sin(x)+5/4) dx = \int_0^{2 \pi} \cos(nx) \ln(\sin(x)+5/4) dx = 0$
I know $f(4n + 1)$ has the opposite sign as $f(4n + 3)$.
Maybe using the fourier series of $\sin(x)^n$ and the taylor series of $\ln(\sin(x)+5/4) $ helps.
Or maybe we can use formulas such as $\cos(nx) = 2 \cos((n-1)x) \cos(x) - \cos((n-2) x) $ and the sine analogue.
Many thanks.
Without Fourier series.
Just a bit of patience with integrals to see obvious patterns. $$I_n=\int_0^{2\pi}\sin [(2 n-1) x]\, \log \left(\sin (x)+\frac{5}{4}\right)\,dx=\color{blue}{(-1)^{n+1}\frac{ 4^{1-n}}{2 n-1}\pi}$$