Question:

Step 1: Simplify the integral $I_n$. Let the integrand be $G(x) = \dfrac{1}{1+2^{\sin\left(\frac{x}{2}\right)}} \left(\dfrac{\sin\left(\frac{nx}{2}\right)}{\sin\left(\frac{x}{2}\right)}\right)^2$. Let $f(x) = \dfrac{1}{1+2^{\sin\left(\frac{x}{2}\right)}}$ and $h(x) = \left(\dfrac{\sin\left(\frac{nx}{2}\right)}{\sin\left(\frac{x}{2}\right)}\right)^2$. We observe that $h(x)$ is an even function, since $h(-x) = \left(\dfrac{\sin\left(\frac{-nx}{2}\right)}{\sin\left(\frac{-x}{2}\right)}\right)^2 = \left(\dfrac{-\sin\left(\frac{nx}{2}\right)}{-\sin\left(\frac{x}{2}\right)}\right)^2 = \left(\dfrac{\sin\left(\frac{nx}{2}\right)}{\sin\left(\frac{x}{2}\right)}\right)^2 = h(x)$. For $f(x)$, we have: $$f(-x) = \dfrac{1}{1+2^{\sin\left(\frac{-x}{2}\right)}} = \dfrac{1}{1+2^{-\sin\left(\frac{x}{2}\right)}} = \dfrac{1}{1+\frac{1}{2^{\sin\left(\frac{x}{2}\right)}}} = \dfrac{1}{\frac{2^{\sin\left(\frac{x}{2}\right)}+1}{2^{\sin\left(\frac{x}{2}\right)}}} = \dfrac{2^{\sin\left(\frac{x}{2}\right)}}{1+2^{\sin\left(\frac{x}{2}\right)}}$$ Thus, $f(x) + f(-x) = \dfrac{1}{1+2^{\sin\left(\frac{x}{2}\right)}} + \dfrac{2^{\sin\left(\frac{x}{2}\right)}}{1+2^{\sin\left(\frac{x}{2}\right)}} = \dfrac{1+2^{\sin\left(\frac{x}{2}\right)}}{1+2^{\sin\left(\frac{x}{2}\right)}} = 1$. Using the property $\int_{-a}^a F(x) dx = \int_0^a (F(x) + F(-x)) dx$ for an integral of the form $\int_{-a}^a f(x)h(x) dx$ where $h(x)$ is even: $$2I_n = \int_{-\pi}^{\pi} \left(f(x)h(x) + f(-x)h(-x)\right) dx = \int_{-\pi}^{\pi} \left(f(x)h(x) + f(-x)h(x)\right) dx$$ $$2I_n = \int_{-\pi}^{\pi} (f(x) + f(-x))h(x) dx = \int_{-\pi}^{\pi} (1)h(x) dx = \int_{-\pi}^{\pi} \left(\dfrac{\sin\left(\frac{nx}{2}\right)}{\sin\left(\frac{x}{2}\right)}\right)^2 dx$$ Since $h(x)$ is an even function, $\int_{-\pi}^{\pi} h(x) dx = 2\int_0^\pi h(x) dx$. Therefore, $2I_n = 2\int_0^\pi \left(\dfrac{\sin\left(\frac{nx}{2}\right)}{\sin\left(\frac{x}{2}\right)}\right)^2 dx$, which simplifies to: $$I_n = \int_0^\pi \left(\dfrac{\sin\left(\frac{nx}{2}\right)}{\sin\left(\frac{x}{2}\right)}\right)^2 dx$$ Step 2: Compute $I_{n+1} - I_n$. $$I_{n+1} - I_n = \int_0^\pi \left[\left(\dfrac{\sin\left(\frac{(n+1)x}{2}\right)}{\sin\left(\frac{x}{2}\right)}\right)^2 - \left(\dfrac{\sin\left(\frac{nx}{2}\right)}{\sin\left(\frac{x}{2}\right)}\right)^2\right] dx$$ $$I_{n+1} - I_n = \int_0^\pi \dfrac{\sin^2\left(\frac{(n+1)x}{2}\right) - \sin^2\left(\frac{nx}{2}\right)}{\sin^2\left(\frac{x}{2}\right)} dx$$ Using the trigonometric identity $\sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B)$, with $A = \frac{(n+1)x}{2}$ and $B = \frac{nx}{2}$: $A+B = \frac{(n+1)x}{2} + \frac{nx}{2} = \frac{(2n+1)x}{2}$ $A-B = \frac{(n+1)x}{2} - \frac{nx}{2} = \frac{x}{2}$ So the numerator becomes $\sin\left(\frac{(2n+1)x}{2}\right)\sin\left(\frac{x}{2}\right)$. $$I_{n+1} - I_n = \int_0^\pi \dfrac{\sin\left(\frac{(2n+1)x}{2}\right)\sin\left(\frac{x}{2}\right)}{\sin^2\left(\frac{x}{2}\right)} dx = \int_0^\pi \dfrac{\sin\left(\frac{(2n+1)x}{2}\right)}{\sin\left(\frac{x}{2}\right)} dx$$ Step 3: Evaluate the integral for $I_{n+1} - I_n$. Let $t = \frac{x}{2}$. Then $x = 2t$ and $dx = 2dt$. When $x=0$, $t=0$. When $x=\pi$, $t=\frac{\pi}{2}$. $$I_{n+1} - I_n = \int_0^{\pi/2} \dfrac{\sin((2n+1)t)}{\sin t} (2dt) = 2\int_0^{\pi/2} \dfrac{\sin((2n+1)t)}{\sin t} dt$$ It is a known result that for any integer $k \ge 0$, $\int_0^{\pi/2} \dfrac{\sin((2k+1)t)}{\sin t} dt = \dfrac{\pi}{2}$. Here, $k=n$. Therefore, $$I_{n+1} - I_n = 2 \cdot \dfrac{\pi}{2} = \pi$$ Step 4: Determine the properties of the sequence $I_n$. Since $I_{n+1} - I_n = \pi$ for all $n = 0, 1, 2, 3, \ldots$, the sequence $I_0, I_1, I_2, \ldots, I_n$ forms an arithmetic progression with a common difference of $\pi$. Step 5: Calculate $I_0$ and the general term $I_n$. For $n=0$, the integrand is $\left(\dfrac{\sin(0)}{\sin(x/2)}\right)^2$. For $x \in (0, \pi]$, $\sin(0)=0$, so the integrand is $0$. At $x=0$, we evaluate the limit: $\lim_{x \to 0} \dfrac{\sin(0 \cdot x/2)}{\sin(x/2)} = \lim_{x \to 0} \dfrac{0}{x/2} = 0$. Thus, $I_0 = \int_0^\pi 0^2 dx = 0$. Since $I_n$ is an arithmetic progression with first term $I_0=0$ and common difference $\pi$, the general term is $I_n = I_0 + n\pi = 0 + n\pi = n\pi$. Step 6: Calculate $\displaystyle\sum_{m=0}^{9} I_{2m}$. Using $I_n = n\pi$, we have $I_{2m} = (2m)\pi$. $$\sum_{m=0}^{9} I_{2m} = \sum_{m=0}^{9} (2m)\pi = 2\pi \sum_{m=0}^{9} m = 2\pi \left(\dfrac{9(9+1)}{2}\right) = 2\pi \left(\dfrac{9 \cdot 10}{2}\right) = 2\pi \cdot 45 = 90\pi$$ Step 7: Calculate $\displaystyle\sum_{m=0}^{10} I_m$. Using $I_n = n\pi$: $$\sum_{m=0}^{10} I_m = \sum_{m=0}^{10} m\pi = \pi \sum_{m=0}^{10} m = \pi \left(\dfrac{10(10+1)}{2}\right) = \pi \left(\dfrac{10 \cdot 11}{2}\right) = \pi \cdot 55 = 55\pi$$