int frac{dx}{2 + cos x} = | vedclass

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(B) We know that $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$ and $1 = \sin^2 \frac{x}{2} + \cos^2 \frac{x}{2}$.Substituting these into the inte

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$\frac{2}{\sqrt{3}} 	an^{-1} \left( \frac{1}{\sqrt{3}} 	an \frac{x}{2} ight) + c$

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(B) We know that $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$ and $1 = \sin^2 \frac{x}{2} + \cos^2 \frac{x}{2}$.Substituting these into the integral,we get:$\int \frac{dx}{2 + \cos x} = \int \frac{dx}{2(\sin^2 \frac{x}{2} + \cos^2 \frac{x}{2}) + (\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2})}$$= \int \frac{dx}{\sin^2 \frac{x}{2} + 3\cos^2 \frac{x}{2}}$Divide numerator and denominator by $\cos^2 \frac{x}{2}$:$= \int \frac{\sec^2 \frac{x}{2}}{	an^2 \frac{x}{2} + 3} dx$Let $	an \frac{x}{2} = t$,then $\frac{1}{2} \sec^2 \frac{x}{2} dx = dt$,which implies $\sec^2 \frac{x}{2} dx = 2dt$.Substituting this into the integral:$= \int \frac{2 dt}{t^2 + 3} = 2 \int \frac{dt}{t^2 + (\sqrt{3})^2}$Using the formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} 	an^{-1} \frac{x}{a} + c$:$= 2 \cdot \frac{1}{\sqrt{3}} 	an^{-1} \left( \frac{t}{\sqrt{3}} ight) + c = \frac{2}{\sqrt{3}} 	an^{-1} \left( \frac{	an \frac{x}{2}}{\sqrt{3}} ight) + c$.

$\int \frac{dx}{2 + \cos x} = $

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