int_0^pi frac{x , dx}{4 cos^2 x + 9 sin^2 x} | vedclass

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(A) Let $I = \int_0^\pi \frac{x \, dx}{4 \cos^2 x + 9 \sin^2 x}$.Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get:$I = \int_0^\

Vedclass · Accepted Answer

$\frac{\pi^2}{12}$

Vedclass · Answer

(A) Let $I = \int_0^\pi \frac{x \, dx}{4 \cos^2 x + 9 \sin^2 x}$.Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get:$I = \int_0^\pi \frac{(\pi - x) \, dx}{4 \cos^2(\pi - x) + 9 \sin^2(\pi - x)} = \int_0^\pi \frac{(\pi - x) \, dx}{4 \cos^2 x + 9 \sin^2 x}$.Adding the two expressions for $I$:$2I = \int_0^\pi \frac{\pi \, dx}{4 \cos^2 x + 9 \sin^2 x} = \pi \int_0^\pi \frac{\sec^2 x \, dx}{4 + 9 	an^2 x}$.Using the property $\int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx$ if $f(2a-x) = f(x)$:$2I = 2\pi \int_0^{\pi/2} \frac{\sec^2 x \, dx}{4 + 9 	an^2 x} \Rightarrow I = \pi \int_0^{\pi/2} \frac{\sec^2 x \, dx}{4 + 9 	an^2 x}$.Let $t = 	an x$,then $dt = \sec^2 x \, dx$. Limits change from $[0, \pi/2]$ to $[0, \infty]$.$I = \pi \int_0^\infty \frac{dt}{4 + 9t^2} = \frac{\pi}{9} \int_0^\infty \frac{dt}{(2/3)^2 + t^2}$.Using $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a})$:$I = \frac{\pi}{9} \cdot \frac{3}{2} \left[ 	an^{-1}(\frac{3t}{2}) ight]_0^\infty = \frac{\pi}{6} \cdot \frac{\pi}{2} = \frac{\pi^2}{12}$.

$\int_0^\pi \frac{x \, dx}{4 \cos^2 x + 9 \sin^2 x} = $

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