int_{ - pi }^{pi } {frac{{2x(1 + sin x)}}{{1 | 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 {\f

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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} $.For the first part,let $f(x) = \frac{2x}{1 + \cos^2 x}$. Since $f(-x) = \frac{-2x}{1 + \cos^2(-x)} = -f(x)$,the function is odd. Thus,$\int_{-\pi}^{\pi} f(x) dx = 0$.For the second part,let $g(x) = \frac{2x\sin x}{1 + \cos^2 x}$. Since $g(-x) = \frac{2(-x)\sin(-x)}{1 + \cos^2(-x)} = \frac{2x\sin x}{1 + \cos^2 x} = g(x)$,the function is even. Thus,$\int_{-\pi}^{\pi} g(x) dx = 2 \int_0^{\pi} g(x) dx$.So,$I = 2 \int_0^{\pi} \frac{2x\sin x}{1 + \cos^2 x} dx = 4 \int_0^{\pi} \frac{x\sin x}{1 + \cos^2 x} dx$ ... $(i)$.Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:$I = 4 \int_0^{\pi} \frac{(\pi - x)\sin(\pi - x)}{1 + \cos^2(\pi - x)} dx = 4 \int_0^{\pi} \frac{(\pi - x)\sin x}{1 + \cos^2 x} dx$ ... $(ii)$.Adding $(i)$ and $(ii)$:$2I = 4 \int_0^{\pi} \frac{x\sin x + (\pi - x)\sin x}{1 + \cos^2 x} dx = 4\pi \int_0^{\pi} \frac{\sin x}{1 + \cos^2 x} dx$.$I = 2\pi \int_0^{\pi} \frac{\sin x}{1 + \cos^2 x} dx$.Let $t = \cos x$,then $dt = -\sin x dx$. When $x=0, t=1$; when $x=\pi, t=-1$.$I = 2\pi \int_1^{-1} \frac{-dt}{1 + t^2} = 2\pi \int_{-1}^1 \frac{dt}{1 + t^2} = 2\pi [	an^{-1} t]_{-1}^1$.$I = 2\pi (	an^{-1}(1) - 	an^{-1}(-1)) = 2\pi (\frac{\pi}{4} - (-\frac{\pi}{4})) = 2\pi (\frac{\pi}{2}) = \pi^2$.

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

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