1. **Problem Statement:** Solve the integral $$\int \frac{\sqrt{x^2 - 36}}{x} \, dx$$ using the trigonometric substitution $$x = 6 \sec \theta$$. 2. **Formula and Substitution:** We use the substitution $$x = 6 \sec \theta$$ because it simplifies the expression under the square root. Recall that $$\sec^2 \theta - 1 = \tan^2 \theta$$. 3. **Substitute and Simplify:** - Compute $$dx = 6 \sec \theta \tan \theta \, d\theta$$. - Substitute into the integral: $$\sqrt{x^2 - 36} = \sqrt{36 \sec^2 \theta - 36} = \sqrt{36(\sec^2 \theta - 1)} = 6 \tan \theta$$. - The integral becomes: $$\int \frac{6 \tan \theta}{6 \sec \theta} \cdot 6 \sec \theta \tan \theta \, d\theta = \int \tan \theta \cdot 6 \tan \theta \, d\theta = 6 \int \tan^2 \theta \, d\theta$$. 4. **Integral of $$\tan^2 \theta$$:** Recall that $$\tan^2 \theta = \sec^2 \theta - 1$$, so $$6 \int \tan^2 \theta \, d\theta = 6 \int (\sec^2 \theta - 1) \, d\theta = 6 \left( \int \sec^2 \theta \, d\theta - \int 1 \, d\theta \right)$$. 5. **Evaluate the integrals:** - $$\int \sec^2 \theta \, d\theta = \tan \theta + C$$ - $$\int 1 \, d\theta = \theta + C$$ So, $$6 (\tan \theta - \theta) + C$$. 6. **Back-substitute $$\theta$$:** Since $$x = 6 \sec \theta$$, then $$\sec \theta = \frac{x}{6}$$ and $$\theta = \sec^{-1} \left( \frac{x}{6} \right)$$. Also, $$\tan \theta = \sqrt{\sec^2 \theta - 1} = \sqrt{\left( \frac{x}{6} \right)^2 - 1} = \frac{\sqrt{x^2 - 36}}{6}$$. 7. **Final answer:** $$6 \left( \frac{\sqrt{x^2 - 36}}{6} - \sec^{-1} \left( \frac{x}{6} \right) \right) + C = \sqrt{x^2 - 36} - 6 \sec^{-1} \left( \frac{x}{6} \right) + C$$. This completes the solution.