int frac{dx}{3 cos 2x + 5} equals | vedclass

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(C) Let $I = \int \frac{dx}{3 \cos 2x + 5}$.Using the identity $\cos 2x = \frac{1 - 	an^2 x}{1 + 	an^2 x}$,we get:$I = \int \frac{dx}{3 \left(\frac{

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$\frac{1}{4} 	an^{-1}\left(\frac{1}{2} 	an xight) + c$,where $c$ is the constant of integration.

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(C) Let $I = \int \frac{dx}{3 \cos 2x + 5}$.Using the identity $\cos 2x = \frac{1 - 	an^2 x}{1 + 	an^2 x}$,we get:$I = \int \frac{dx}{3 \left(\frac{1 - 	an^2 x}{1 + 	an^2 x}ight) + 5}$$I = \int \frac{1 + 	an^2 x}{3(1 - 	an^2 x) + 5(1 + 	an^2 x)} dx$$I = \int \frac{\sec^2 x}{3 - 3 	an^2 x + 5 + 5 	an^2 x} dx$$I = \int \frac{\sec^2 x}{8 + 2 	an^2 x} dx$Let $u = 	an x$,then $du = \sec^2 x dx$.$I = \int \frac{du}{8 + 2u^2} = \frac{1}{2} \int \frac{du}{4 + u^2}$Using the formula $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} 	an^{-1}\left(\frac{x}{a}ight) + c$:$I = \frac{1}{2} \cdot \frac{1}{2} 	an^{-1}\left(\frac{u}{2}ight) + c$$I = \frac{1}{4} 	an^{-1}\left(\frac{	an x}{2}ight) + c$.

$\int \frac{dx}{3 \cos 2x + 5}$ equals

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