1. **State the problem:** Evaluate the integral $$\int \frac{\sec \theta}{\cot \theta} \, d\theta$$. 2. **Rewrite the integrand using trigonometric identities:** Recall that $$\cot \theta = \frac{\cos \theta}{\sin \theta}$$ and $$\sec \theta = \frac{1}{\cos \theta}$$. 3. Substitute these into the integral: $$\int \frac{\sec \theta}{\cot \theta} \, d\theta = \int \frac{\frac{1}{\cos \theta}}{\frac{\cos \theta}{\sin \theta}} \, d\theta = \int \frac{1}{\cos \theta} \cdot \frac{\sin \theta}{\cos \theta} \, d\theta = \int \frac{\sin \theta}{\cos^2 \theta} \, d\theta$$. 4. **Use substitution:** Let $$u = \cos \theta$$, then $$du = -\sin \theta \, d\theta$$ or $$-du = \sin \theta \, d\theta$$. 5. Rewrite the integral in terms of $$u$$: $$\int \frac{\sin \theta}{\cos^2 \theta} \, d\theta = \int \frac{\sin \theta}{u^2} \, d\theta = \int \frac{-du}{u^2} = -\int u^{-2} \, du$$. 6. **Integrate:** $$-\int u^{-2} \, du = -\left(-u^{-1}\right) + C = \frac{1}{u} + C$$. 7. **Back-substitute:** $$\frac{1}{u} + C = \frac{1}{\cos \theta} + C = \sec \theta + C$$. **Final answer:** $$\int \frac{\sec \theta}{\cot \theta} \, d\theta = \sec \theta + C$$.