int_{-pi}^pi frac{2 x(1+sin x)}{1+cos ^2 x} d | vedclass

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(B) Let $I = \int_{-\pi}^\pi \frac{2x(1+\sin x)}{1+\cos^2 x} dx$.We can split the integral into two parts:$I = \int_{-\pi}^\pi \frac{2x}{1+\cos^2 x} d

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$\pi^2$

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(B) Let $I = \int_{-\pi}^\pi \frac{2x(1+\sin x)}{1+\cos^2 x} dx$.We can split the integral into two parts:$I = \int_{-\pi}^\pi \frac{2x}{1+\cos^2 x} dx + \int_{-\pi}^\pi \frac{2x\sin x}{1+\cos^2 x} dx = I_1 + I_2$.For $I_1 = \int_{-\pi}^\pi \frac{2x}{1+\cos^2 x} dx$,let $f(x) = \frac{2x}{1+\cos^2 x}$.Since $f(-x) = \frac{2(-x)}{1+\cos^2(-x)} = -\frac{2x}{1+\cos^2 x} = -f(x)$,$f(x)$ is an odd function.Therefore,$I_1 = 0$.For $I_2 = \int_{-\pi}^\pi \frac{2x\sin x}{1+\cos^2 x} dx$,let $g(x) = \frac{2x\sin x}{1+\cos^2 x}$.Since $g(-x) = \frac{2(-x)\sin(-x)}{1+\cos^2(-x)} = \frac{2x\sin x}{1+\cos^2 x} = g(x)$,$g(x)$ is an even function.Therefore,$I_2 = 2 \int_0^\pi \frac{2x\sin x}{1+\cos^2 x} dx = 4 \int_0^\pi \frac{x\sin x}{1+\cos^2 x} dx$.Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:$I_2 = 4 \int_0^\pi \frac{(\pi-x)\sin(\pi-x)}{1+\cos^2(\pi-x)} dx = 4 \int_0^\pi \frac{(\pi-x)\sin x}{1+\cos^2 x} dx$.Adding the two expressions for $I_2$:$2I_2 = 4 \int_0^\pi \frac{\pi\sin x}{1+\cos^2 x} dx = 4\pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx$.Let $u = \cos x$,then $du = -\sin x dx$.When $x=0, u=1$; when $x=\pi, u=-1$.$2I_2 = 4\pi \int_1^{-1} \frac{-du}{1+u^2} = 4\pi \int_{-1}^1 \frac{du}{1+u^2} = 4\pi [	an^{-1} u]_{-1}^1$.$2I_2 = 4\pi (\frac{\pi}{4} - (-\frac{\pi}{4})) = 4\pi (\frac{\pi}{2}) = 2\pi^2$.Thus,$I_2 = \pi^2$.Finally,$I = I_1 + I_2 = 0 + \pi^2 = \pi^2$.

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

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