Let N be the set of natural numbers. For n in | vedclass

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(C) Given $I_n = \int_0^\pi \frac{x \sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} dx \quad \dots (i)$Using the property $\int_0^a f(x) dx = \int_0^a f(a-x)

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$I_m = I_n$ for all $m 
eq n$

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(C) Given $I_n = \int_0^\pi \frac{x \sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} dx \quad \dots (i)$Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:$I_n = \int_0^\pi \frac{(\pi - x) \sin^{2n}(\pi - x)}{\sin^{2n}(\pi - x) + \cos^{2n}(\pi - x)} dx = \int_0^\pi \frac{(\pi - x) \sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} dx \quad \dots (ii)$Adding $(i)$ and $(ii)$:$2I_n = \int_0^\pi \frac{\pi \sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} dx = \pi \int_0^\pi \frac{\sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} dx$Using the property $\int_0^{2a} f(x) dx = 2 \int_0^a f(x) dx$ if $f(2a-x) = f(x)$:$2I_n = 2\pi \int_0^{\pi/2} \frac{\sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} dx \implies I_n = \pi \int_0^{\pi/2} \frac{\sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} dx \quad \dots (iii)$Also,$I_n = \pi \int_0^{\pi/2} \frac{\cos^{2n} x}{\cos^{2n} x + \sin^{2n} x} dx \quad \dots (iv)$Adding $(iii)$ and $(iv)$:$2I_n = \pi \int_0^{\pi/2} \frac{\sin^{2n} x + \cos^{2n} x}{\sin^{2n} x + \cos^{2n} x} dx = \pi \int_0^{\pi/2} 1 dx = \pi \cdot \frac{\pi}{2} = \frac{\pi^2}{2}$Thus,$I_n = \frac{\pi^2}{4}$,which is independent of $n$. Therefore,$I_m = I_n$ for all $m, n \in N$.

Let $N$ be the set of natural numbers. For $n \in N$,define $I_n = \int_0^\pi \frac{x \sin^{2n}(x)}{\sin^{2n}(x) + \cos^{2n}(x)} dx$. Then,for $m, n \in N$,which of the following is true?

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