int frac{dx}{2+cos x} = (Where C is a consta | vedclass

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(D) To solve the integral $I = \int \frac{dx}{2+\cos x}$,we use the substitution $\cos x = \frac{1-	an^2(x/2)}{1+	an^2(x/2)}$.Substituting this into

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

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(D) To solve the integral $I = \int \frac{dx}{2+\cos x}$,we use the substitution $\cos x = \frac{1-	an^2(x/2)}{1+	an^2(x/2)}$.Substituting this into the integral,we get:$I = \int \frac{dx}{2 + \frac{1-	an^2(x/2)}{1+	an^2(x/2)}} = \int \frac{1+	an^2(x/2)}{2(1+	an^2(x/2)) + 1-	an^2(x/2)} dx$$I = \int \frac{\sec^2(x/2)}{3+	an^2(x/2)} dx$.Let $u = 	an(x/2)$,then $du = \frac{1}{2} \sec^2(x/2) dx$,which implies $\sec^2(x/2) dx = 2 du$.Substituting these into the integral:$I = \int \frac{2 du}{3+u^2} = 2 \int \frac{du}{(\sqrt{3})^2 + u^2}$.Using the standard formula $\int \frac{dx}{a^2+x^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$,we get:$I = 2 \cdot \frac{1}{\sqrt{3}} 	an^{-1}\left(\frac{u}{\sqrt{3}}ight) + C = \frac{2}{\sqrt{3}} 	an^{-1}\left(\frac{	an(x/2)}{\sqrt{3}}ight) + C$.

$\int \frac{dx}{2+\cos x} = $ (Where $C$ is a constant of integration.)

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