int_{0}^{pi} frac{x , dx}{a^2 cos^2 x + b^2 s | vedclass

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(D) Let $I = \int_0^\pi \frac{x \, dx}{a^2 \cos^2 x + b^2 \sin^2 x} \dots (i)$ Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we hav

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

Vedclass · Answer

(D) Let $I = \int_0^\pi \frac{x \, dx}{a^2 \cos^2 x + b^2 \sin^2 x} \dots (i)$ Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we have: $I = \int_0^\pi \frac{(\pi - x) \, dx}{a^2 \cos^2(\pi - x) + b^2 \sin^2(\pi - x)} = \int_0^\pi \frac{(\pi - x) \, dx}{a^2 \cos^2 x + b^2 \sin^2 x} \dots (ii)$ Adding $(i)$ and $(ii)$: $2I = \int_0^\pi \frac{\pi \, dx}{a^2 \cos^2 x + b^2 \sin^2 x} = \pi \int_0^\pi \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x}$ Since the integrand is symmetric about $\frac{\pi}{2}$,$I = \pi \int_0^{\pi/2} \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x}$ Dividing numerator and denominator by $\cos^2 x$: $I = \pi \int_0^{\pi/2} \frac{\sec^2 x \, dx}{a^2 + b^2 	an^2 x}$ Let $b 	an x = t$,then $b \sec^2 x \, dx = dt \implies \sec^2 x \, dx = \frac{dt}{b}$. As $x 	o 0, t 	o 0$ and as $x 	o \frac{\pi}{2}, t 	o \infty$. $I = \frac{\pi}{b} \int_0^\infty \frac{dt}{a^2 + t^2} = \frac{\pi}{b} \left[ \frac{1}{a} 	an^{-1} \left( \frac{t}{a} ight) ight]_0^\infty = \frac{\pi}{ab} \left( \frac{\pi}{2} - 0 ight) = \frac{\pi^2}{2ab}$.

$\int_{0}^{\pi} \frac{x \, dx}{a^2 \cos^2 x + b^2 \sin^2 x} = $

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