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

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(C) माना $I = \int_0^\pi \frac{x \sin x}{\sin^2 x + 2 \cos^2 x} dx$. गुणधर्म $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ का उपयोग करने पर: $I = \int_0^\pi

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

Vedclass · Answer

(C) माना $I = \int_0^\pi \frac{x \sin x}{\sin^2 x + 2 \cos^2 x} dx$. गुणधर्म $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ का उपयोग करने पर: $I = \int_0^\pi \frac{(\pi - x) \sin(\pi - x)}{\sin^2(\pi - x) + 2 \cos^2(\pi - x)} dx = \int_0^\pi \frac{(\pi - x) \sin x}{\sin^2 x + 2 \cos^2 x} dx$. $I$ के दोनों व्यंजकों को जोड़ने पर: $2I = \int_0^\pi \frac{x \sin x + (\pi - x) \sin x}{\sin^2 x + 2 \cos^2 x} dx = \pi \int_0^\pi \frac{\sin x}{\sin^2 x + 2 \cos^2 x} dx$. चूंकि $\sin^2 x = 1 - \cos^2 x$,हर $1 - \cos^2 x + 2 \cos^2 x = 1 + \cos^2 x$ हो जाता है। $2I = \pi \int_0^\pi \frac{\sin x}{1 + \cos^2 x} dx$. माना $u = \cos x$,तब $du = -\sin x dx$. जब $x=0, u=1$; जब $x=\pi, u=-1$. $2I = \pi \int_1^{-1} \frac{-du}{1 + u^2} = \pi \int_{-1}^1 \frac{du}{1 + u^2} = \pi [	an^{-1} u]_{-1}^1$. $2I = \pi (	an^{-1}(1) - 	an^{-1}(-1)) = \pi (\frac{\pi}{4} - (-\frac{\pi}{4})) = \pi (\frac{\pi}{2}) = \frac{\pi^2}{2}$. अतः,$I = \frac{\pi^2}{4}$.

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

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