The value of int_{0}^{2 pi} frac{x sin^{8} x} | vedclass

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(D) Let $I = \int_{0}^{2 \pi} \frac{x \sin^{8} x}{\sin^{8} x + \cos^{8} x} dx$. Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we

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$\pi^{2}$

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(D) Let $I = \int_{0}^{2 \pi} \frac{x \sin^{8} x}{\sin^{8} x + \cos^{8} x} dx$. Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we have: $I = \int_{0}^{2 \pi} \frac{(2 \pi - x) \sin^{8}(2 \pi - x)}{\sin^{8}(2 \pi - x) + \cos^{8}(2 \pi - x)} dx = \int_{0}^{2 \pi} \frac{(2 \pi - x) \sin^{8} x}{\sin^{8} x + \cos^{8} x} dx$. Adding the two expressions for $I$: $2I = \int_{0}^{2 \pi} \frac{2 \pi \sin^{8} x}{\sin^{8} x + \cos^{8} x} dx = 2 \pi \int_{0}^{2 \pi} \frac{\sin^{8} x}{\sin^{8} x + \cos^{8} x} dx$. $I = \pi \int_{0}^{2 \pi} \frac{\sin^{8} x}{\sin^{8} x + \cos^{8} x} dx$. Using the property $\int_{0}^{2a} f(x) dx = 2 \int_{0}^{a} f(x) dx$ if $f(2a-x) = f(x)$: $I = 2 \pi \int_{0}^{\pi} \frac{\sin^{8} x}{\sin^{8} x + \cos^{8} x} dx = 4 \pi \int_{0}^{\pi/2} \frac{\sin^{8} x}{\sin^{8} x + \cos^{8} x} dx$. Using $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$: $I = 4 \pi \int_{0}^{\pi/2} \frac{\cos^{8} x}{\cos^{8} x + \sin^{8} x} dx$. Adding these two: $2I = 4 \pi \int_{0}^{\pi/2} 1 dx = 4 \pi \cdot \frac{\pi}{2} = 2 \pi^{2}$. Thus,$I = \pi^{2}$.

The value of $\int_{0}^{2 \pi} \frac{x \sin^{8} x}{\sin^{8} x + \cos^{8} x} dx$ is equal to

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