Find the indefinite integral. (Remember the constant of integration.) \[ \int \frac{9}{\sqrt{16-(x+4)^{2}}} d x \]

To solve the integral \[ \int \frac{9}{\sqrt{16-(x+4)^{2}}} \, dx, \] we can use a trigonometric substitution. Notice that \(16 - (x + 4)^2\) can be rewritten in terms of sine. Let's set: \[ x + 4 = 4\sin(\theta). \] Then, the differential \(dx\) becomes \[ dx = 4\cos(\theta) \, d\theta. \] Substituting this into the integral, we also have: \[ \sqrt{16 - (x + 4)^2} = \sqrt{16 - 16\sin^2(\theta)} = \sqrt{16(1 - \sin^2(\theta))} = \sqrt{16\cos^2(\theta)} = 4\cos(\theta). \] Now, substituting these into the integral gives us: \[ \int \frac{9}{4\cos(\theta)} \cdot 4\cos(\theta) \, d\theta = \int 9 \, d\theta. \] This simplifies to: \[ 9\theta + C, \] where \(C\) is the constant of integration. Now, we need to convert back to \(x\) using our substitution \(x + 4 = 4\sin(\theta)\). Therefore, \[ \sin(\theta) = \frac{x + 4}{4}, \] and thus, \[ \theta = \arcsin\left(\frac{x + 4}{4}\right). \] Substituting back gives us: \[ 9\arcsin\left(\frac{x + 4}{4}\right) + C. \] Thus, the indefinite integral is: \[ \int \frac{9}{\sqrt{16-(x+4)^{2}}} \, dx = 9\arcsin\left(\frac{x + 4}{4}\right) + C. \]