Use \( z \)-substitution to evaluate \( I=\int \frac{2 \sin x}{1-\cos 2 x} d x \) A. \( -2 \ln \left|\tan \frac{x}{2}\right|+C \) B. \( \ln \left|\tan ^{-1} \frac{x}{2}\right|+C \) C. \( \ln \left|\cos \frac{x}{2}\right|+C \) D. None of the options E. \( \ln \left|\tan \frac{x}{2}\right|+C \)

To evaluate the integral \[ I = \int \frac{2 \sin x}{1 - \cos 2x} \, dx, \] we can start by simplifying the integrand. We know that \[ \cos 2x = 1 - 2\sin^2 x, \] which gives us \[ 1 - \cos 2x = 2\sin^2 x. \] Thus, we can rewrite the integral as: \[ I = \int \frac{2 \sin x}{2 \sin^2 x} \, dx = \int \frac{\sin x}{\sin^2 x} \, dx = \int \frac{1}{\sin x} \, dx = \int \csc x \, dx. \] The integral of \(\csc x\) is known to be: \[ \int \csc x \, dx = -\ln |\csc x + \cot x| + C. \] Next, we can express \(\csc x\) and \(\cot x\) in terms of \(\tan \frac{x}{2}\) using the half-angle identities: \[ \csc x = \frac{1}{\sin x} = \frac{2}{1 - \tan^2 \frac{x}{2}} \quad \text{and} \quad \cot x = \frac{\cos x}{\sin x} = \frac{1 - \tan^2 \frac{x}{2}}{2 \tan \frac{x}{2}}. \] Thus, we have: \[ \csc x + \cot x = \frac{2}{1 - \tan^2 \frac{x}{2}} + \frac{1 - \tan^2 \frac{x}{2}}{2 \tan \frac{x}{2}}. \] However, for our purposes, we can directly use the result of the integral: \[ I = -\ln |\csc x + \cot x| + C. \] Now, we can express \(\csc x + \cot x\) in terms of \(\tan \frac{x}{2}\): \[ \csc x + \cot x = \frac{1 + \tan^2 \frac{x}{2}}{2 \tan \frac{x}{2}} = \frac{1}{\sin x} + \frac{\cos x}{\sin x} = \frac{1 + \cos x}{\sin x}. \] This expression can be simplified further, but we can also check the options given in the problem. The integral simplifies to: \[ I = -\ln |\csc x + \cot x| + C = -\ln \left| \frac{1 + \cos x}{\sin x} \right| + C. \] Using the identity \( \tan \frac{x}{2} = \frac{1 - \cos x}{\sin x} \), we can relate this back to \(\tan \frac{x}{2}\). After evaluating, we find that the integral can be expressed in terms of \(\tan \frac{x}{2}\): \[ I = \ln \left| \tan \frac{x}{2} \right| + C. \] Thus, the correct answer is: \[ \boxed{E. \ln \left|\tan \frac{x}{2}\right| + C}. \]