int_0^pi frac{dx}{1 - 2acos x + a^2} = | vedclass

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(C) Let $I = \int_0^\pi \frac{dx}{1 - 2a\cos x + a^2}$.Using the identity $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$ and $1 = \cos^2 \frac{x}{

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$\frac{\pi}{1 - a^2}$

Vedclass · Answer

(C) Let $I = \int_0^\pi \frac{dx}{1 - 2a\cos x + a^2}$.Using the identity $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$ and $1 = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2}$,we get:$I = \int_0^\pi \frac{dx}{(1+a^2)(\cos^2 \frac{x}{2} + \sin^2 \frac{x}{2}) - 2a(\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2})}$$I = \int_0^\pi \frac{dx}{(1-a)^2 \cos^2 \frac{x}{2} + (1+a)^2 \sin^2 \frac{x}{2}}$Divide numerator and denominator by $\cos^2 \frac{x}{2}$:$I = \int_0^\pi \frac{\sec^2 \frac{x}{2} dx}{(1-a)^2 + (1+a)^2 	an^2 \frac{x}{2}}$Let $t = 	an \frac{x}{2}$,then $dt = \frac{1}{2} \sec^2 \frac{x}{2} dx$,so $\sec^2 \frac{x}{2} dx = 2 dt$.As $x 	o 0, t 	o 0$ and as $x 	o \pi, t 	o \infty$.$I = \int_0^\infty \frac{2 dt}{(1-a)^2 + (1+a)^2 t^2} = \frac{2}{(1+a)^2} \int_0^\infty \frac{dt}{(\frac{1-a}{1+a})^2 + t^2}$Using $\int \frac{dx}{k^2 + x^2} = \frac{1}{k} 	an^{-1}(\frac{x}{k})$:$I = \frac{2}{(1+a)^2} \cdot \frac{1+a}{1-a} [	an^{-1}(\frac{1+a}{1-a} t)]_0^\infty = \frac{2}{1-a^2} [\frac{\pi}{2} - 0] = \frac{\pi}{1-a^2}$.

$\int_0^\pi \frac{dx}{1 - 2a\cos x + a^2} = $

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