If int frac{d x}{3-2 cos 2 x}=frac{tan ^{-1}( | vedclass

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

Vedclass · Accepted Answer

$\sqrt{5}$

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(B) Let $I = \int \frac{dx}{3-2 \cos 2x}$.Using the identity $\cos 2x = \frac{1-	an^2 x}{1+	an^2 x}$,we have:$I = \int \frac{dx}{3-2(\frac{1-	an^2 x}{1+	an^2 x})} = \int \frac{(1+	an^2 x) dx}{3(1+	an^2 x) - 2(1-	an^2 x)}$.$I = \int \frac{\sec^2 x dx}{3+3	an^2 x - 2 + 2	an^2 x} = \int \frac{\sec^2 x dx}{1+5	an^2 x}$.Let $t = 	an x$,then $dt = \sec^2 x dx$.$I = \int \frac{dt}{1+(\sqrt{5}t)^2} = \frac{1}{\sqrt{5}} 	an^{-1}(\sqrt{5}t) + c$.Substituting $t = 	an x$,we get $I = \frac{1}{\sqrt{5}} 	an^{-1}(\sqrt{5} 	an x) + c$.Comparing this with $\frac{	an^{-1}(f(x))}{\sqrt{5}} + c$,we find $f(x) = \sqrt{5} 	an x$.Therefore,$f(\pi/4) = \sqrt{5} 	an(\pi/4) = \sqrt{5} 	imes 1 = \sqrt{5}$.

If $\int \frac{d x}{3-2 \cos 2 x}=\frac{\tan ^{-1}(f(x))}{\sqrt{5}}+c$,(where $c$ is the constant of integration),then $f(\pi / 4)$ has the value:

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