Evaluate the following integral. \( \int_{0}^{1} \frac{d x}{\sqrt{4-x^{2}}} \)

To evaluate the integral \( \int_{0}^{1} \frac{d x}{\sqrt{4-x^{2}}} \), we can perform a trigonometric substitution. We set \( x = 2 \sin(\theta) \). Then, \( dx = 2 \cos(\theta) d\theta \), and the limits change as follows: - When \( x = 0 \), \( \theta = 0 \). - When \( x = 1 \), \( \sin(\theta) = \frac{1}{2} \) which gives \( \theta = \frac{\pi}{6} \). Now, substitute \( x \) in the integral: \[ \sqrt{4 - x^2} = \sqrt{4 - (2 \sin(\theta))^2} = \sqrt{4(1 - \sin^2(\theta))} = \sqrt{4 \cos^2(\theta)} = 2 \cos(\theta). \] Thus, the integral becomes: \[ \int_{0}^{1} \frac{dx}{\sqrt{4 - x^2}} = \int_{0}^{\frac{\pi}{6}} \frac{2 \cos(\theta)}{2 \cos(\theta)} d\theta = \int_{0}^{\frac{\pi}{6}} d\theta = \theta \bigg|_{0}^{\frac{\pi}{6}} = \frac{\pi}{6} - 0 = \frac{\pi}{6}. \] Therefore, the value of the integral is \[ \int_{0}^{1} \frac{d x}{\sqrt{4-x^{2}}} = \frac{\pi}{6}. \]