Find the indefinite integral. (Remember the constant of integration.) \[ \int \frac{d x}{\sqrt{16-x^{2}}} \]

To solve the indefinite integral \[ \int \frac{dx}{\sqrt{16 - x^2}}, \] we can use a trigonometric substitution. Notice that \( \sqrt{16 - x^2} \) suggests a substitution of the form \( x = 4 \sin(\theta) \). Then, the differential \( dx \) becomes \( dx = 4 \cos(\theta) d\theta \). When substituting \( x = 4 \sin(\theta) \), we find \[ \sqrt{16 - x^2} = \sqrt{16 - (4 \sin(\theta))^2} = \sqrt{16(1 - \sin^2(\theta))} = \sqrt{16 \cos^2(\theta)} = 4 \cos(\theta). \] Now our integral transforms as follows: \[ \int \frac{4 \cos(\theta) d\theta}{4 \cos(\theta)} = \int d\theta. \] This simplifies to \[ \theta + C, \] where \( C \) is the constant of integration. Now we need to convert back to \( x \): Recall that \( x = 4 \sin(\theta) \), which implies \( \sin(\theta) = \frac{x}{4} \). Therefore, \( \theta = \arcsin\left(\frac{x}{4}\right) \). Finally, we get \[ \int \frac{dx}{\sqrt{16 - x^2}} = \arcsin\left(\frac{x}{4}\right) + C. \]