Step 1: Simplify the Integral

The given integral is of the form $\int_{-a}^{a} f(x) dx$. We can use the property $\int_{-a}^{a} f(x) dx = \int_{0}^{a} [f(x) + f(-x)] dx$.

Let the integrand be $f(x) = \frac{\sin(nx)}{(1 + \pi^x)\sin x}$. Now, we find $f(-x)$: $f(-x) = \frac{\sin(n(-x))}{(1 + \pi^{-x})\sin(-x)} = \frac{-\sin(nx)}{(1 + 1/\pi^x)(-\sin x)} = \frac{\sin(nx)}{(\frac{\pi^x+1}{\pi^x})\sin x} = \frac{\pi^x \sin(nx)}{(1 + \pi^x)\sin x}$

Next, we compute the sum $f(x) + f(-x)$: $f(x) + f(-x) = \frac{\sin(nx)}{(1 + \pi^x)\sin x} + \frac{\pi^x \sin(nx)}{(1 + \pi^x)\sin x} = \frac{(1 + \pi^x)\sin(nx)}{(1 + \pi^x)\sin x} = \frac{\sin(nx)}{\sin x}$

So, the integral $I_n$ simplifies to: $I_n = \int_{0}^{\pi} \frac{\sin(nx)}{\sin x} dx$

Step 2: Establish a Recurrence Relation for $I_n$

Let's consider the difference $I_n - I_{n-2}$ for $n \ge 2$: $I_n - I_{n-2} = \int_{0}^{\pi} \frac{\sin(nx)}{\sin x} dx - \int_{0}^{\pi} \frac{\sin((n-2)x)}{\sin x} dx = \int_{0}^{\pi} \frac{\sin(nx) - \sin((n-2)x)}{\sin x} dx$

Using the trigonometric identity $\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$: $\sin(nx) - \sin((n-2)x) = 2 \cos\left(\frac{nx+(n-2)x}{2}\right) \sin\left(\frac{nx-(n-2)x}{2}\right) = 2 \cos((n-1)x) \sin x$

Substituting this back into the integral: $I_n - I_{n-2} = \int_{0}^{\pi} \frac{2 \cos((n-1)x) \sin x}{\sin x} dx = \int_{0}^{\pi} 2 \cos((n-1)x) dx$

For $n \ge 2$, $n-1 \ge 1$. Evaluating the integral: $I_n - I_{n-2} = 2 \left[ \frac{\sin((n-1)x)}{n-1} \right]_{0}^{\pi} = \frac{2}{n-1} [\sin((n-1)\pi) - \sin(0)] = \frac{2}{n-1}[0 - 0] = 0$

Thus, we have the recurrence relation $I_n = I_{n-2}$ for all $n \ge 2$. This can be rewritten as $I_{n+2} = I_n$ for $n \ge 0$.

Step 3: Calculate Base Cases and Find a General Formula for $I_n$

Let's calculate the first few values of $I_n$: For $n=0$: $I_0 = \int_{0}^{\pi} \frac{\sin(0x)}{\sin x} dx = \int_{0}^{\pi} 0 dx = 0$ For $n=1$: $I_1 = \int_{0}^{\pi} \frac{\sin(1x)}{\sin x} dx = \int_{0}^{\pi} 1 dx = [x]_{0}^{\pi} = \pi$

Using the recurrence relation $I_n = I_{n-2}$:

• If $n$ is an even number ($n=2k$ for $k \ge 0$), then $I_n = I_{2k} = I_{2(k-1)} = \dots = I_2 = I_0 = 0$.
• If $n$ is an odd number ($n=2k+1$ for $k \ge 0$), then $I_n = I_{2k+1} = I_{2(k-1)+1} = \dots = I_3 = I_1 = \pi$.

So, we have a general formula for $I_n$: $I_n = \begin{cases} 0 & \text{if } n \text{ is even} \\ \pi & \text{if } n \text{ is odd} \end{cases}$

Step 4: Evaluate the Given Options

A: ${I_n} = {I_{n + 2}}$ From our recurrence relation derived in Step 2, $I_{n+2} = I_n$ for $n \ge 0$. This is true for all non-negative integers $n$. Thus, option A is correct.

B: $\sum\limits_{m = 1}^{10} {{I_{2m + 1}}} = 10\pi$ The index $2m+1$ is always odd. For any odd integer $k$, $I_k = \pi$. Therefore, $I_{2m+1} = \pi$ for all $m=1, 2, \dots, 10$. The sum is: $\sum_{m=1}^{10} I_{2m+1} = \sum_{m=1}^{10} \pi = 10 \times \pi = 10\pi$ Thus, option B is correct.

C: $\sum\limits_{m = 1}^{10} {{I_{2m}}} = 0$ The index $2m$ is always even. For any even integer $k$, $I_k = 0$. Therefore, $I_{2m} = 0$ for all $m=1, 2, \dots, 10$. The sum is: $\sum_{m=1}^{10} I_{2m} = \sum_{m=1}^{10} 0 = 0$ Thus, option C is correct.

D: ${I_n} = {I_{n + 1}}$ This implies the sequence is constant. Let's check for $n=0$ and $n=1$. We have $I_0 = 0$ and $I_1 = \pi$. Since $I_0 \neq I_1$, this statement is not generally true. Thus, option D is incorrect.

Conclusion: Options A, B, and C are correct.