int_{-pi}^pi frac{x sin x}{1+cos^2 x} dx = | vedclass

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(D) Let $I = \int_{-\pi}^\pi \frac{x \sin x}{1+\cos^2 x} dx$. Since $f(x) = \frac{x \sin x}{1+\cos^2 x}$ is an even function because $f(-x) = \frac{(-

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

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(D) Let $I = \int_{-\pi}^\pi \frac{x \sin x}{1+\cos^2 x} dx$. Since $f(x) = \frac{x \sin x}{1+\cos^2 x}$ is an even function because $f(-x) = \frac{(-x) \sin(-x)}{1+\cos^2(-x)} = \frac{(-x)(-\sin x)}{1+\cos^2 x} = f(x)$,we can write: $I = 2 \int_0^\pi \frac{x \sin x}{1+\cos^2 x} dx$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$: $I = 2 \int_0^\pi \frac{(\pi-x) \sin(\pi-x)}{1+\cos^2(\pi-x)} dx = 2 \int_0^\pi \frac{(\pi-x) \sin x}{1+\cos^2 x} dx$. $I = 2\pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx - 2 \int_0^\pi \frac{x \sin x}{1+\cos^2 x} dx$. $I = 2\pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx - I$. $2I = 2\pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx \Rightarrow I = \pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx$. Let $\cos x = t$,then $-\sin x dx = dt$. When $x=0, t=1$; when $x=\pi, t=-1$. $I = \pi \int_1^{-1} \frac{-dt}{1+t^2} = \pi \int_{-1}^1 \frac{dt}{1+t^2} = \pi [	an^{-1} t]_{-1}^1$. $I = \pi (	an^{-1}(1) - 	an^{-1}(-1)) = \pi (\frac{\pi}{4} - (-\frac{\pi}{4})) = \pi (\frac{\pi}{2}) = \frac{\pi^2}{2}$.

$\int_{-\pi}^\pi \frac{x \sin x}{1+\cos^2 x} dx =$

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