Step 1: Understanding the Concept:
This is a standard integral of the form $\int \frac{dx}{a + b\cos x}$.
The universal substitution for such integrals is to express $\cos x$ in terms of $\tan(x/2)$.
Step 2: Key Formula or Approach:
Use the half-angle substitution:
Let $t = \tan\left(\frac{x}{2}\right)$.
Then, $\cos x = \frac{1 - t^2}{1 + t^2}$ and $dx = \frac{2dt}{1 + t^2}$.
Also, use the standard integral $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + c$.
Step 3: Detailed Explanation:
Let $I = \int \frac{dx}{2+\cos x}$.
Substitute $t = \tan(x/2)$:
\[ I = \int \frac{\frac{2dt}{1+t^2}}{2 + \frac{1-t^2}{1+t^2}} \]
Multiply numerator and denominator by $(1+t^2)$:
\[ I = \int \frac{2dt}{2(1+t^2) + (1-t^2)} \]
\[ I = \int \frac{2dt}{2 + 2t^2 + 1 - t^2} \]
Combine like terms in the denominator:
\[ I = \int \frac{2dt}{t^2 + 3} \]
Factor out the 2 and rewrite the denominator in the form $x^2 + a^2$:
\[ I = 2 \int \frac{dt}{t^2 + (\sqrt{3})^2} \]
Now apply the standard formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right)$:
Here, $x$ is $t$ and $a$ is $\sqrt{3}$.
\[ I = 2 \left[ \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{t}{\sqrt{3}}\right) \right] + c \]
Substitute back $t = \tan(x/2)$:
\[ I = \frac{2}{\sqrt{3}} \tan^{-1}\left( \frac{\tan(x/2)}{\sqrt{3}} \right) + c \]
\[ I = \frac{2}{\sqrt{3}} \tan^{-1}\left( \frac{1}{\sqrt{3}} \tan\frac{x}{2} \right) + c \]
Step 4: Final Answer:
The evaluated integral is $\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{1}{\sqrt{3}}\tan\frac{x}{2}\right) + c$.