int_0^pi frac{x sin x}{sin ^2 x+2 cos ^2 x} d | vedclass

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(C) Let $I = \int_0^\pi \frac{x \sin x}{\sin^2 x + 2 \cos^2 x} dx$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have: $I = \int_0^\p

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$\frac{\pi^2}{4}$

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(C) Let $I = \int_0^\pi \frac{x \sin x}{\sin^2 x + 2 \cos^2 x} dx$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have: $I = \int_0^\pi \frac{(\pi - x) \sin(\pi - x)}{\sin^2(\pi - x) + 2 \cos^2(\pi - x)} dx = \int_0^\pi \frac{(\pi - x) \sin x}{\sin^2 x + 2 \cos^2 x} dx$. Adding the two expressions for $I$: $2I = \int_0^\pi \frac{x \sin x + (\pi - x) \sin x}{\sin^2 x + 2 \cos^2 x} dx = \pi \int_0^\pi \frac{\sin x}{\sin^2 x + 2 \cos^2 x} dx$. Since $\sin^2 x = 1 - \cos^2 x$,the denominator becomes $1 - \cos^2 x + 2 \cos^2 x = 1 + \cos^2 x$. $2I = \pi \int_0^\pi \frac{\sin x}{1 + \cos^2 x} dx$. Let $u = \cos x$,then $du = -\sin x dx$. When $x=0, u=1$; when $x=\pi, u=-1$. $2I = \pi \int_1^{-1} \frac{-du}{1 + u^2} = \pi \int_{-1}^1 \frac{du}{1 + u^2} = \pi [	an^{-1} u]_{-1}^1$. $2I = \pi (	an^{-1}(1) - 	an^{-1}(-1)) = \pi (\frac{\pi}{4} - (-\frac{\pi}{4})) = \pi (\frac{\pi}{2}) = \frac{\pi^2}{2}$. Therefore,$I = \frac{\pi^2}{4}$.

$\int_0^\pi \frac{x \sin x}{\sin ^2 x+2 \cos ^2 x} d x=$

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