The integral int_0^pi frac{8 x , dx}{4 cos^2 | vedclass

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(A) Let $I = \int_0^\pi \frac{8 x \, dx}{4 \cos^2 x + \sin^2 x}$.Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get:$I = \int_0^\

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$2 \pi^2$

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(A) Let $I = \int_0^\pi \frac{8 x \, dx}{4 \cos^2 x + \sin^2 x}$.Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get:$I = \int_0^\pi \frac{8(\pi - x) \, dx}{4 \cos^2(\pi - x) + \sin^2(\pi - x)} = \int_0^\pi \frac{8(\pi - x) \, dx}{4 \cos^2 x + \sin^2 x}$.Adding the two expressions for $I$:$2I = \int_0^\pi \frac{8x + 8\pi - 8x}{4 \cos^2 x + \sin^2 x} \, dx = 8\pi \int_0^\pi \frac{dx}{4 \cos^2 x + \sin^2 x}$.Since the integrand is symmetric about $\pi/2$,$2I = 8\pi 	imes 2 \int_0^{\pi/2} \frac{dx}{4 \cos^2 x + \sin^2 x}$.Divide numerator and denominator by $\cos^2 x$:$I = 8\pi \int_0^{\pi/2} \frac{\sec^2 x \, dx}{4 + 	an^2 x}$.Let $t = 	an x$,then $dt = \sec^2 x \, dx$. As $x 	o 0, t 	o 0$ and as $x 	o \pi/2, t 	o \infty$.$I = 8\pi \int_0^\infty \frac{dt}{4 + t^2} = 8\pi \left[ \frac{1}{2} 	an^{-1}\left(\frac{t}{2}ight) ight]_0^\infty$.$I = 4\pi \left( \frac{\pi}{2} - 0 ight) = 2\pi^2$.

The integral $\int_0^\pi \frac{8 x \, dx}{4 \cos^2 x + \sin^2 x}$ is equal to

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