Evaluate the integral: int frac{dx}{2+cos x} | vedclass

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(A) To evaluate 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 i

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

Vedclass · Answer

(A) To evaluate 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:$I = \int \frac{dx}{2 + \frac{1-	an^2(x/2)}{1+	an^2(x/2)}} = \int \frac{(1+	an^2(x/2)) dx}{2(1+	an^2(x/2)) + 1 - 	an^2(x/2)} = \int \frac{\sec^2(x/2) dx}{3 + 	an^2(x/2)}$.Let $t = 	an(x/2)$,then $dt = \frac{1}{2} \sec^2(x/2) dx$,which implies $\sec^2(x/2) dx = 2 dt$.Substituting these into the integral:$I = \int \frac{2 dt}{3 + t^2} = 2 \int \frac{dt}{(\sqrt{3})^2 + t^2}$.Using the standard integral formula $\int \frac{dx}{a^2+x^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + c$,we get:$I = 2 \cdot \frac{1}{\sqrt{3}} 	an^{-1}\left(\frac{t}{\sqrt{3}}ight) + c = \frac{2}{\sqrt{3}} 	an^{-1}\left(\frac{1}{\sqrt{3}} 	an \frac{x}{2}ight) + c$.

Evaluate the integral: $\int \frac{dx}{2+\cos x}$

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