If $I_n \int\limits_{-\pi}^\pi \frac{\sin n x}{\left(1+\pi^x\right) \sin x} d x, n=0,1,2,...$ then which one of the following is not true?

$I_n=I_{n+1}$

Using $\int\limits_{-a}^a f(x) d x=\int\limits_0^a\{f(x)+f(-x)\} d x$, we have

$I_n =\int\limits_{-\pi}^\pi \frac{\sin n x}{\left(1+\pi^x\right) \sin x} d x$

$\Rightarrow I_n =\int\limits_0^\pi\left\{\frac{\sin n x}{\left(1+\pi^x\right) \sin x}+\frac{\sin n x}{\left(1+\pi^{-x}\right) \sin x}\right\} d x$

$\Rightarrow I_n=\int\limits_0^\pi \frac{\sin n x}{\sin x}\left\{\frac{1}{1+\pi^x}+\frac{\pi^x}{1+\pi^x} d x\right\}$

$\Rightarrow I_n=\int\limits_0^\pi \frac{\sin n x}{\sin x} d x$

∴  $I_{n+2}-I_n=\int\limits_0^\pi \frac{\sin (n+2) x-\sin n x}{\sin x} d x$

$\Rightarrow I_{n+2}-I_n=2 \int\limits_0^\pi \cos (n+1) x d x$

$\Rightarrow I_{n+2}-I_n=2\left[\frac{\sin (n+1) x}{n+1}\right]_0^\pi$

$\Rightarrow I_{n+2}-I_n=0$

$\Rightarrow I_{n+2}=I_n$ for n = 0, 1, 2, 3, ...

$\Rightarrow I_n=I_{n-2}=I_{n-4}=...=I_2=I_0$, if n is even

and $I_n=I_{n-2}=I_{n-4}=...=I_3=I_1$, if n is odd

$\Rightarrow I_n=\left\{\begin{array}{ll}0, \text { if } n \text { is even } \\ \pi, \text { if } n \text { is odd }\end{array}\left[∵ I_0=0, I_1=\int\limits_0^\pi \frac{\sin x}{\sin x} d x=\pi\right]\right.$

$\Rightarrow \sum\limits_{m=1}^{10} I_{2 m}=0$ and $\sum\limits_{m=1}^{10} I_{2 m+1}=10 \pi$