int frac{dx}{7+5 cos x} is equal to | vedclass

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(B) Let $I = \int \frac{dx}{7+5 \cos x}$.Using the identity $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$ and $1 = \cos^2 \frac{x}{2} + \sin^2 \f

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$\frac{1}{\sqrt{6}} 	an ^{-1}\left(\frac{1}{\sqrt{6}} 	an \frac{x}{2}ight)+c$

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(B) Let $I = \int \frac{dx}{7+5 \cos x}$.Using the identity $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$ and $1 = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2}$,we get:$I = \int \frac{dx}{7(\cos^2 \frac{x}{2} + \sin^2 \frac{x}{2}) + 5(\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2})}$$I = \int \frac{dx}{12 \cos^2 \frac{x}{2} + 2 \sin^2 \frac{x}{2}}$Divide numerator and denominator by $\cos^2 \frac{x}{2}$:$I = \int \frac{\sec^2 \frac{x}{2} dx}{12 + 2 	an^2 \frac{x}{2}} = \frac{1}{2} \int \frac{\sec^2 \frac{x}{2} dx}{6 + 	an^2 \frac{x}{2}}$Let $	an \frac{x}{2} = z$,then $\frac{1}{2} \sec^2 \frac{x}{2} dx = dz$.$I = \int \frac{dz}{6 + z^2} = \int \frac{dz}{(\sqrt{6})^2 + z^2}$Using the formula $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} 	an^{-1} \frac{x}{a} + c$:$I = \frac{1}{\sqrt{6}} 	an^{-1} \frac{z}{\sqrt{6}} + c = \frac{1}{\sqrt{6}} 	an^{-1} \left(\frac{1}{\sqrt{6}} 	an \frac{x}{2}ight) + c$.

$\int \frac{dx}{7+5 \cos x}$ is equal to

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