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

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

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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 $I = \int \frac{d x}{7+5 \cos x}$.Using the identities $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$ and $1 = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2}$, we get:$I = \int \frac{d x}{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{d x}{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 $\int \frac{du}{a^2 + u^2} = \frac{1}{a} 	an^{-1} \frac{u}{a} + c$:$I = \frac{1}{\sqrt{6}} 	an^{-1} \left(\frac{z}{\sqrt{6}}ight) + c = \frac{1}{\sqrt{6}} 	an^{-1} \left(\frac{1}{\sqrt{6}} 	an \frac{x}{2}ight) + c$.

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

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