Use trigonometric substitution to evaluate \( I=\int \frac{2}{\sqrt{1-4 x^{2}}} d x \) A. \( \frac{1}{4} \sin ^{-1}(2 x)+C \) B. \( -\sin ^{-1}(2 x)+C x \) C. \( -\sin ^{-1}(2 x)-3 x+C \) D. None of the options E. \( \sin ^{-1}(2 x)+C \)

To evaluate the integral \[ I = \int \frac{2}{\sqrt{1 - 4x^2}} \, dx, \] we can use the trigonometric substitution \( x = \frac{1}{2} \sin(\theta) \). Then, we have: \[ dx = \frac{1}{2} \cos(\theta) \, d\theta. \] Substituting \( x \) into the integral, we get: \[ 1 - 4x^2 = 1 - 4\left(\frac{1}{2} \sin(\theta)\right)^2 = 1 - \sin^2(\theta) = \cos^2(\theta). \] Thus, the integral becomes: \[ I = \int \frac{2}{\sqrt{\cos^2(\theta)}} \cdot \frac{1}{2} \cos(\theta) \, d\theta = \int \frac{2}{|\cos(\theta)|} \cdot \frac{1}{2} \cos(\theta) \, d\theta. \] Since \( \cos(\theta) \) is positive in the range we are considering (where \( \theta \) is in the range of arcsin), we can drop the absolute value: \[ I = \int d\theta = \theta + C. \] Now, we need to express \( \theta \) in terms of \( x \). From our substitution \( x = \frac{1}{2} \sin(\theta) \), we have: \[ \sin(\theta) = 2x \implies \theta = \sin^{-1}(2x). \] Thus, we can write: \[ I = \sin^{-1}(2x) + C. \] Therefore, the final result is: \[ I = \sin^{-1}(2x) + C. \] The correct answer is: \[ \boxed{E. \sin^{-1}(2x) + C}. \]