Evaluate the definite integral. \[ \int_{0}^{1 / 3} \frac{3}{\sqrt{1-9 x^{2}}} d x \]

To evaluate the definite integral \[ I = \int_{0}^{1/3} \frac{3}{\sqrt{1 - 9x^2}} \, dx, \] we can start by recognizing the form of the integrand, which resembles the derivative of the inverse sine function. Let's perform a substitution to simplify this integration process. Set: \[ u = 3x \quad \Rightarrow \quad du = 3 \, dx \quad \Rightarrow \quad dx = \frac{du}{3}. \] Next, we also need to change the limits of integration. When \( x = 0 \), \( u = 3 \cdot 0 = 0 \), and when \( x = \frac{1}{3} \), \( u = 3 \cdot \frac{1}{3} = 1 \). Thus, the integral becomes: \[ I = \int_{0}^{1} \frac{3}{\sqrt{1 - u^2}} \cdot \frac{du}{3} = \int_{0}^{1} \frac{1}{\sqrt{1 - u^2}} \, du. \] Now, we recognize that \[ \int \frac{1}{\sqrt{1 - u^2}} \, du = \arcsin(u) + C. \] Evaluating our definite integral, we get: \[ I = \left[ \arcsin(u) \right]_{0}^{1} = \arcsin(1) - \arcsin(0). \] We know that \[ \arcsin(1) = \frac{\pi}{2} \quad \text{and} \quad \arcsin(0) = 0. \] Therefore, \[ I = \frac{\pi}{2} - 0 = \frac{\pi}{2}. \] Thus, the value of the definite integral is \[ \boxed{\frac{\pi}{2}}. \]