Evaluate the following integral using trigonometric substitution. \( \int_{0}^{1} \frac{d x}{\sqrt{4-x^{2}}} \) Rewrite the given integral using this substitution. \( \frac{1}{1} \frac{d x}{\sqrt{4-x^{2}}}=\int_{0}^{0} \) (Type exact answers.) ( \( \theta \)

To solve the integral \( \int_{0}^{1} \frac{dx}{\sqrt{4-x^{2}}} \) using trigonometric substitution, we start with the substitution \( x = 2\sin(\theta) \). This gives us \( dx = 2\cos(\theta)d\theta \) and changes the limits: when \( x = 0 \), \( \theta = 0 \); when \( x = 1 \), \( \theta = \arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6} \). Now, substituting into the integral, we get: \[ \int_{0}^{1} \frac{dx}{\sqrt{4 - x^{2}}} = \int_{0}^{\frac{\pi}{6}} \frac{2\cos(\theta)}{\sqrt{4 - 4\sin^{2}(\theta)}} d\theta = \int_{0}^{\frac{\pi}{6}} \frac{2\cos(\theta)}{2\cos(\theta)} d\theta = \int_{0}^{\frac{\pi}{6}} d\theta \] This simplifies to: \[ \int_{0}^{\frac{\pi}{6}} d\theta = \left[ \theta \right]_{0}^{\frac{\pi}{6}} = \frac{\pi}{6} - 0 = \frac{\pi}{6} \] Thus, the integral evaluates to \( \frac{\pi}{6} \). Now let's rewrite the integral in terms of \( \theta \): The relationship is as follows: \[ \frac{1}{1} \frac{dx}{\sqrt{4 - x^{2}}} = \int_{0}^{\frac{\pi}{6}} d\theta \]