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(C) Let $I = \int_0^\pi \frac{dx}{1-2a \cos x + a^2}$.Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we note that $\cos(\pi-x) = -\cos x$.

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $I = \int_0^\pi \frac{dx}{1-2a \cos x + a^2}$.Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we note that $\cos(\pi-x) = -\cos x$.Thus,$I = \int_0^\pi \frac{dx}{1+2a \cos x + a^2}$.Adding the two expressions for $I$:$2I = \int_0^\pi \left( \frac{1}{1-2a \cos x + a^2} + \frac{1}{1+2a \cos x + a^2} ight) dx$$2I = \int_0^\pi \frac{2(1+a^2)}{(1+a^2)^2 - 4a^2 \cos^2 x} dx$$I = (1+a^2) \int_0^\pi \frac{dx}{(1+a^2)^2 - 4a^2 \cos^2 x}$Since the integrand is symmetric about $\pi/2$,$I = 2(1+a^2) \int_0^{\pi/2} \frac{dx}{(1+a^2)^2 - 4a^2 \cos^2 x}$.Dividing numerator and denominator by $\cos^2 x$:$I = 2(1+a^2) \int_0^{\pi/2} \frac{\sec^2 x dx}{(1+a^2)^2 \sec^2 x - 4a^2}$Using $\sec^2 x = 1 + 	an^2 x$:$I = 2(1+a^2) \int_0^{\pi/2} \frac{\sec^2 x dx}{(1+a^2)^2 (1+	an^2 x) - 4a^2}$$I = 2(1+a^2) \int_0^{\pi/2} \frac{\sec^2 x dx}{(1+a^2)^2 	an^2 x + (1-a^2)^2}$Let $u = 	an x$,then $du = \sec^2 x dx$. As $x 	o 0, u 	o 0$ and as $x 	o \pi/2, u 	o \infty$.$I = 2(1+a^2) \int_0^{\infty} \frac{du}{(1+a^2)^2 u^2 + (1-a^2)^2}$$I = \frac{2(1+a^2)}{(1+a^2)^2} \int_0^{\infty} \frac{du}{u^2 + (\frac{1-a^2}{1+a^2})^2} = \frac{2}{1+a^2} \cdot \frac{1+a^2}{1-a^2} [\arctan(\frac{u(1+a^2)}{1-a^2})]_0^{\infty}$$I = \frac{2}{1-a^2} \cdot \frac{\pi}{2} = \frac{\pi}{1-a^2}$.

For $0 < a < 1$,the value of the integral $\int_0^\pi \frac{d x}{1-2 a \cos x+a^2}$ is:

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