If $I_n = \int_{-\pi}^{\pi} \frac{\sin nx}{(1+\pi^x)\sin x}\,dx$, $n = 0, 1, 2, \ldots$, then

If Iₙ = ∫_(-π)^π (sin nx)/((1+πˣ)sin x) dx, n = 0, 1, 2, …, then $I_n=\displaystyle\int_{-\pi}^{\pi}\dfrac{\sin nx}{(1+\pi^{x})\sin x}\,dx,\quad n=0,1,2,\ldots$ $(1)\ I_n=I_{n+2} \qquad (2)\ \sum_{m=1}^{10} I_{2m+1}=10\pi$ $(3)\ \sum_{m=1}^{10} I_{2m}=0 \qquad (4)\ I_n=I_{n+1}$ $\int_{-\pi}^{\pi} f(x)\,dx=\int_0^{\pi}\big(f(x)+f(-x)\big)\,dx$ $f(x)=\dfrac{\sin nx}{(1+\pi^{x})\sin x}$ $f(-x)=\dfrac{\sin nx}{(1+\pi^{-x})\sin x}$ $\dfrac{1}{1+\pi^{-x}}=\dfrac{\pi^{x}}{1+\pi^{x}}$ $f(x)+f(-x)=\dfrac{\sin nx}{\sin x}\left[\dfrac{1}{1+\pi^{x}}+\dfrac{\pi^{x}}{1+\pi^{x}}\right]=\dfrac{\sin nx}{\sin x}$ $I_n=\displaystyle\int_0^{\pi}\dfrac{\sin nx}{\sin x}\,dx$ $n \text{ odd}:\ \dfrac{\sin nx}{\sin x}=1+2\big(\cos 2x+\cos 4x+\cdots+\cos(n{-}1)x\big)$ $n \text{ even}:\ \dfrac{\sin nx}{\sin x}=2\big(\cos x+\cos 3x+\cdots+\cos(n{-}1)x\big)$ $\int_0^{\pi}\cos(kx)\,dx=0\quad(k\ge1)$ $I_n=\pi\ (n \text{ odd});\qquad I_n=0\ (n \text{ even})$ $I_1=\int_0^\pi 1\,dx=\pi,\qquad I_2=\int_0^\pi 2\cos x\,dx=0$ $(1)\ n,\,n+2 \text{ same parity} \Rightarrow I_n=I_{n+2}\ \textbf{TRUE}$ $(2)\ I_{2m+1}=\pi \Rightarrow \sum_{m=1}^{10} I_{2m+1}=10\pi\ \textbf{TRUE}$ $(3)\ I_{2m}=0 \Rightarrow \sum_{m=1}^{10} I_{2m}=0\ \textbf{TRUE}$ $(4)\ n,\,n+1 \text{ opposite parity} \Rightarrow I_n\neq I_{n+1}\ \textbf{FALSE}$ $\text{TRUE: }(1),\ (2),\ (3)$