Evaluate using z-substitution \( I=\int \frac{1}{2-\cos x} d x \) A. \( \frac{2}{\sqrt{3}} \tan ^{-1}\left(\sqrt{3} \cos \frac{x}{2}\right)+C \) B. \( 3 \tan ^{-1}\left(\sqrt{3} \tan \frac{x}{2}\right)+C \) C. \( \frac{2}{\sqrt{3}} \tan ^{-1}\left(\sqrt{3} \tan \frac{x}{2}\right)+C \) D. None of the options E. \( \frac{2}{\sqrt{2}} \tan ^{-1}\left(\sqrt{3} \sin \frac{x}{2}\right)+C \)

To evaluate the integral \[ I = \int \frac{1}{2 - \cos x} \, dx, \] we can use the substitution \( z = \tan\left(\frac{x}{2}\right) \). This substitution is useful because it transforms trigonometric functions into rational functions. The relationships we will use are: \[ \cos x = \frac{1 - z^2}{1 + z^2}, \quad dx = \frac{2}{1 + z^2} \, dz. \] Now, substituting these into the integral, we have: \[ I = \int \frac{1}{2 - \frac{1 - z^2}{1 + z^2}} \cdot \frac{2}{1 + z^2} \, dz. \] First, simplify the expression in the integral: \[ 2 - \cos x = 2 - \frac{1 - z^2}{1 + z^2} = 2 - \frac{1 - z^2}{1 + z^2} = \frac{(2(1 + z^2) - (1 - z^2))}{1 + z^2} = \frac{2 + 2z^2 - 1 + z^2}{1 + z^2} = \frac{1 + 3z^2}{1 + z^2}. \] Thus, we can rewrite the integral as: \[ I = \int \frac{1 + z^2}{1 + 3z^2} \cdot \frac{2}{1 + z^2} \, dz = \int \frac{2}{1 + 3z^2} \, dz. \] Now, we can evaluate this integral: \[ I = 2 \int \frac{1}{1 + 3z^2} \, dz. \] The integral \( \int \frac{1}{1 + 3z^2} \, dz \) can be solved using the formula for the integral of the form \( \int \frac{1}{a^2 + u^2} \, du = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C \). Here, \( a^2 = 1/3 \) so \( a = \frac{1}{\sqrt{3}} \): \[ \int \frac{1}{1 + 3z^2} \, dz = \frac{1}{\sqrt{3}} \tan^{-1}(\sqrt{3} z) + C. \] Thus, we have: \[ I = 2 \cdot \frac{1}{\sqrt{3}} \tan^{-1}(\sqrt{3} z) + C = \frac{2}{\sqrt{3}} \tan^{-1}(\sqrt{3} z) + C. \] Now, substituting back \( z = \tan\left(\frac{x}{2}\right) \): \[ I = \frac{2}{\sqrt{3}} \tan^{-1}\left(\sqrt{3} \tan\left(\frac{x}{2}\right)\right) + C. \] Thus, the final result is: \[ I = \frac{2}{\sqrt{3}} \tan^{-1}\left(\sqrt{3} \tan\left(\frac{x}{2}\right)\right) + C. \] Comparing this with the options provided, we find that the correct answer is: **C.** \( \frac{2}{\sqrt{3}} \tan^{-1}\left(\sqrt{3} \tan \frac{x}{2}\right)+C \)