Evaluate the indefinite integral. \( \int \frac{\sin \left(\frac{8}{x}\right)}{x^{2}} d x=\square+C \)

To evaluate the integral: \[ \int \frac{\sin\left(\frac{8}{x}\right)}{x^{2}} \, dx \] **Step 1: Choose a substitution** Let’s set: \[ u = \frac{8}{x} \] Then, the derivative of \( u \) with respect to \( x \) is: \[ du = -\frac{8}{x^{2}} \, dx \quad \Rightarrow \quad -\frac{du}{8} = \frac{dx}{x^{2}} \] **Step 2: Rewrite the integral in terms of \( u \)** Substituting into the original integral: \[ \int \frac{\sin\left(\frac{8}{x}\right)}{x^{2}} \, dx = \int \sin(u) \left(-\frac{du}{8}\right) = -\frac{1}{8} \int \sin(u) \, du \] **Step 3: Integrate** \[ -\frac{1}{8} \int \sin(u) \, du = -\frac{1}{8} (-\cos(u)) + C = \frac{\cos(u)}{8} + C \] **Step 4: Substitute back \( u = \frac{8}{x} \)** \[ \frac{\cos\left(\frac{8}{x}\right)}{8} + C \] **Final Answer:** \[ \int \frac{\sin\left(\frac{8}{x}\right)}{x^{2}} \, dx = \frac{\cos\left(\frac{8}{x}\right)}{8} + C \]