If int_0^pi frac{x sin x}{4 cos^2 x + 3 sin^2 | vedclass

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(A) $I = \int_0^\pi \frac{x \sin x}{4 \cos^2 x + 3 \sin^2 x} dx$ Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$: $I = \int_0^\pi \frac{(\p

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$\frac{\pi^2}{6 \sqrt{3}}$

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(A) $I = \int_0^\pi \frac{x \sin x}{4 \cos^2 x + 3 \sin^2 x} dx$ Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$: $I = \int_0^\pi \frac{(\pi - x) \sin x}{4 \cos^2(\pi - x) + 3 \sin^2(\pi - x)} dx = \int_0^\pi \frac{(\pi - x) \sin x}{4 \cos^2 x + 3 \sin^2 x} dx$ Adding the two expressions for $I$: $2I = \int_0^\pi \frac{\pi \sin x}{4 \cos^2 x + 3 \sin^2 x} dx = \pi \int_0^\pi \frac{\sin x}{4 \cos^2 x + 3(1 - \cos^2 x)} dx$ $2I = \pi \int_0^\pi \frac{\sin x}{\cos^2 x + 3} dx$ Let $t = \cos x$,then $dt = -\sin x dx$. When $x=0, t=1$; when $x=\pi, t=-1$. $2I = -\pi \int_1^{-1} \frac{dt}{t^2 + 3} = \pi \int_{-1}^1 \frac{dt}{t^2 + 3}$ $2I = \pi \left[ \frac{1}{\sqrt{3}} 	an^{-1} \left( \frac{t}{\sqrt{3}} ight) ight]_{-1}^1 = \frac{\pi}{\sqrt{3}} \left( 	an^{-1} \frac{1}{\sqrt{3}} - 	an^{-1} \left( -\frac{1}{\sqrt{3}} ight) ight)$ $2I = \frac{\pi}{\sqrt{3}} \left( \frac{\pi}{6} - (-\frac{\pi}{6}) ight) = \frac{\pi}{\sqrt{3}} \left( \frac{\pi}{3} ight) = \frac{\pi^2}{3 \sqrt{3}}$ $I = \frac{\pi^2}{6 \sqrt{3}}$

If $\int_0^\pi \frac{x \sin x}{4 \cos^2 x + 3 \sin^2 x} dx = $

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