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

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(B) माना $I = \int_{-\pi}^\pi \frac{2x(1+\sin x)}{1+\cos^2 x} dx$.हम समाकलन को दो भागों में विभाजित कर सकते हैं:$I = \int_{-\pi}^\pi \frac{2x}{1+\cos^

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$\pi^2$

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(B) माना $I = \int_{-\pi}^\pi \frac{2x(1+\sin x)}{1+\cos^2 x} dx$.हम समाकलन को दो भागों में विभाजित कर सकते हैं:$I = \int_{-\pi}^\pi \frac{2x}{1+\cos^2 x} dx + \int_{-\pi}^\pi \frac{2x\sin x}{1+\cos^2 x} dx = I_1 + I_2$.$I_1 = \int_{-\pi}^\pi \frac{2x}{1+\cos^2 x} dx$ के लिए,माना $f(x) = \frac{2x}{1+\cos^2 x}$.चूंकि $f(-x) = \frac{2(-x)}{1+\cos^2(-x)} = -\frac{2x}{1+\cos^2 x} = -f(x)$,इसलिए $f(x)$ एक विषम फलन है।अतः,$I_1 = 0$.$I_2 = \int_{-\pi}^\pi \frac{2x\sin x}{1+\cos^2 x} dx$ के लिए,माना $g(x) = \frac{2x\sin x}{1+\cos^2 x}$.चूंकि $g(-x) = \frac{2(-x)\sin(-x)}{1+\cos^2(-x)} = \frac{2x\sin x}{1+\cos^2 x} = g(x)$,इसलिए $g(x)$ एक सम फलन है।अतः,$I_2 = 2 \int_0^\pi \frac{2x\sin x}{1+\cos^2 x} dx = 4 \int_0^\pi \frac{x\sin x}{1+\cos^2 x} dx$.गुणधर्म $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ का उपयोग करने पर:$I_2 = 4 \int_0^\pi \frac{(\pi-x)\sin(\pi-x)}{1+\cos^2(\pi-x)} dx = 4 \int_0^\pi \frac{(\pi-x)\sin x}{1+\cos^2 x} dx$.$I_2$ के दोनों व्यंजकों को जोड़ने पर:$2I_2 = 4 \int_0^\pi \frac{\pi\sin x}{1+\cos^2 x} dx = 4\pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx$.माना $u = \cos x$,तो $du = -\sin x dx$.जब $x=0, u=1$; जब $x=\pi, u=-1$.$2I_2 = 4\pi \int_1^{-1} \frac{-du}{1+u^2} = 4\pi \int_{-1}^1 \frac{du}{1+u^2} = 4\pi [	an^{-1} u]_{-1}^1$.$2I_2 = 4\pi (\frac{\pi}{4} - (-\frac{\pi}{4})) = 4\pi (\frac{\pi}{2}) = 2\pi^2$.इस प्रकार,$I_2 = \pi^2$.अंत में,$I = I_1 + I_2 = 0 + \pi^2 = \pi^2$.

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

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