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Question

(C) Let $I = \int \frac{dx}{1-2a \cos x + a^2}$.Using the identity $\cos x = \frac{1-	an^2(x/2)}{1+	an^2(x/2)}$,we have:$I = \int \frac{dx}{1-2a \le

Vedclass · Accepted Answer

$\frac{2}{1-a^2} 	an^{-1}\left[\frac{1+a}{1-a} 	an \frac{x}{2}ight] + c$

Vedclass · Answer

(C) Let $I = \int \frac{dx}{1-2a \cos x + a^2}$.Using the identity $\cos x = \frac{1-	an^2(x/2)}{1+	an^2(x/2)}$,we have:$I = \int \frac{dx}{1-2a \left(\frac{1-	an^2(x/2)}{1+	an^2(x/2)}ight) + a^2} = \int \frac{\sec^2(x/2) dx}{(1+a^2)(1+	an^2(x/2)) - 2a(1-	an^2(x/2))}$.Let $t = 	an(x/2)$,then $dt = \frac{1}{2} \sec^2(x/2) dx$,so $\sec^2(x/2) dx = 2 dt$.$I = \int \frac{2 dt}{(1+a^2)(1+t^2) - 2a(1-t^2)} = \int \frac{2 dt}{1+a^2+t^2+a^2t^2-2a+2at^2} = \int \frac{2 dt}{(1-a)^2 + t^2(1+a)^2}$.$I = \frac{2}{(1+a)^2} \int \frac{dt}{\left(\frac{1-a}{1+a}ight)^2 + t^2}$.Using $\int \frac{dx}{k^2+x^2} = \frac{1}{k} 	an^{-1}(\frac{x}{k})$,we get:$I = \frac{2}{(1+a)^2} \cdot \frac{1+a}{1-a} 	an^{-1}\left(t \cdot \frac{1+a}{1-a}ight) + c = \frac{2}{1-a^2} 	an^{-1}\left[\frac{1+a}{1-a} 	an \frac{x}{2}ight] + c$.

If $0 < a < 1$,then $\int \frac{dx}{1-2a \cos x + a^2} =$

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