int frac{1}{3-2 cos 2 x} ,d x= (where C is co | vedclass

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(B) We have the integral $I = \int \frac{d x}{3-2 \cos 2 x}$.Using the identity $\cos 2x = \frac{1-	an^2 x}{1+	an^2 x}$,we get:$I = \int \frac{d x}{

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$\frac{1}{\sqrt{5}} 	an ^{-1}(\sqrt{5} 	an x)+C$

Vedclass · Answer

(B) We have the integral $I = \int \frac{d x}{3-2 \cos 2 x}$.Using the identity $\cos 2x = \frac{1-	an^2 x}{1+	an^2 x}$,we get:$I = \int \frac{d x}{3-2 \left(\frac{1-	an^2 x}{1+	an^2 x}ight)} = \int \frac{1+	an^2 x}{3(1+	an^2 x) - 2(1-	an^2 x)} \,d x$.Since $1+	an^2 x = \sec^2 x$,the integral becomes:$I = \int \frac{\sec^2 x}{3+3	an^2 x - 2 + 2	an^2 x} \,d x = \int \frac{\sec^2 x}{5	an^2 x + 1} \,d x$.Let $t = 	an x$,then $dt = \sec^2 x \,d x$.Substituting these into the integral:$I = \int \frac{dt}{5t^2 + 1} = \int \frac{dt}{(\sqrt{5}t)^2 + 1^2}$.Using the standard formula $\int \frac{dx}{x^2+a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$,we get:$I = \frac{1}{\sqrt{5}} 	an^{-1}(\sqrt{5}t) + C$.Substituting $t = 	an x$ back,we get:$I = \frac{1}{\sqrt{5}} 	an^{-1}(\sqrt{5} 	an x) + C$.

$\int \frac{1}{3-2 \cos 2 x} \,d x=$ (where $C$ is constant of integration.)

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