1. **State the problem:** We want to transform the integral $$\int \frac{x^6}{\sqrt{1+x^2}} \, dx$$ using the substitution $$x = \sec(\theta)$$ where $$0 \leq \theta < \frac{\pi}{2}$$ or $$\pi \leq \theta < \frac{3\pi}{2}$$. 2. **Recall the substitution and derivatives:** Since $$x = \sec(\theta)$$, then $$dx = \sec(\theta)\tan(\theta) \, d\theta$$. 3. **Rewrite the integral in terms of $$\theta$$:** Substitute $$x = \sec(\theta)$$ and $$dx = \sec(\theta)\tan(\theta) \, d\theta$$ into the integral: $$\int \frac{(\sec(\theta))^6}{\sqrt{1 + (\sec(\theta))^2}} \cdot \sec(\theta)\tan(\theta) \, d\theta$$ 4. **Simplify the square root term:** Recall the identity: $$1 + \sec^2(\theta) = \tan^2(\theta) + 1 + 1 = \tan^2(\theta) + 2$$ But this is incorrect; the correct identity is: $$1 + \sec^2(\theta) = 1 + \frac{1}{\cos^2(\theta)} = \tan^2(\theta) + 1 + 1$$ Actually, the standard identity is: $$\tan^2(\theta) + 1 = \sec^2(\theta)$$ So: $$1 + \sec^2(\theta) = 1 + \sec^2(\theta)$$ (no simpler identity) But the problem is to simplify $$\sqrt{1 + x^2} = \sqrt{1 + \sec^2(\theta)}$$. Using the identity: $$\sec^2(\theta) - 1 = \tan^2(\theta)$$ So: $$1 + \sec^2(\theta) = 1 + \sec^2(\theta) = \tan^2(\theta) + 2$$ Therefore: $$\sqrt{1 + \sec^2(\theta)} = \sqrt{\tan^2(\theta) + 2}$$ 5. **Rewrite the integral fully:** $$\int \frac{\sec^6(\theta)}{\sqrt{\tan^2(\theta) + 2}} \cdot \sec(\theta)\tan(\theta) \, d\theta = \int \frac{\sec^7(\theta) \tan(\theta)}{\sqrt{\tan^2(\theta) + 2}} \, d\theta$$ 6. **Final transformed integral:** $$\int \frac{\sec^7(\theta) \tan(\theta)}{\sqrt{\tan^2(\theta) + 2}} \, d\theta$$ This is the integral after substitution. **Answer:** $$\int \frac{x^6}{\sqrt{1+x^2}} \, dx = \int \frac{\sec^7(\theta) \tan(\theta)}{\sqrt{\tan^2(\theta) + 2}} \, d\theta$$