1. **State the problem:** Evaluate the definite integral $$\int_{\sqrt{2}}^{2} \frac{1}{t^3 \sqrt{t^2 - 1}} \, dt.$$\n\n2. **Identify the integral type and substitution:** The integrand involves $t^3$ and $\sqrt{t^2 - 1}$, suggesting a trigonometric substitution using $t = \sec \theta$ because $\sec^2 \theta - 1 = \tan^2 \theta$.\n\n3. **Substitution:** Let $$t = \sec \theta,$$ then $$dt = \sec \theta \tan \theta \, d\theta.$$ Also, $$\sqrt{t^2 - 1} = \sqrt{\sec^2 \theta - 1} = \tan \theta.$$\n\n4. **Change limits:** When $t = \sqrt{2}$, $$\sec \theta = \sqrt{2} \implies \theta = \frac{\pi}{4}.$$ When $t = 2$, $$\sec \theta = 2 \implies \theta = \sec^{-1}(2).$$\n\n5. **Rewrite the integral:**\n$$\int_{\pi/4}^{\sec^{-1}(2)} \frac{1}{(\sec \theta)^3 \cdot \tan \theta} \cdot \sec \theta \tan \theta \, d\theta = \int_{\pi/4}^{\sec^{-1}(2)} \frac{\cancel{\sec \theta} \cancel{\tan \theta}}{\sec^3 \theta \cancel{\tan \theta}} \, d\theta = \int_{\pi/4}^{\sec^{-1}(2)} \frac{1}{\sec^2 \theta} \, d\theta.$$\n\n6. **Simplify:** $$\frac{1}{\sec^2 \theta} = \cos^2 \theta.$$ So the integral becomes $$\int_{\pi/4}^{\sec^{-1}(2)} \cos^2 \theta \, d\theta.$$\n\n7. **Use power-reduction formula:** $$\cos^2 \theta = \frac{1 + \cos 2\theta}{2}.$$\n\n8. **Integral becomes:**\n$$\int_{\pi/4}^{\sec^{-1}(2)} \frac{1 + \cos 2\theta}{2} \, d\theta = \frac{1}{2} \int_{\pi/4}^{\sec^{-1}(2)} (1 + \cos 2\theta) \, d\theta = \frac{1}{2} \left[ \theta + \frac{\sin 2\theta}{2} \right]_{\pi/4}^{\sec^{-1}(2)}.$$\n\n9. **Evaluate at bounds:**\nCalculate $$\theta + \frac{\sin 2\theta}{2}$$ at $\theta = \sec^{-1}(2)$ and $\theta = \pi/4$.\n\n- At $\theta = \sec^{-1}(2)$, $\sec \theta = 2$, so $\cos \theta = \frac{1}{2}$, thus $\theta = \frac{\pi}{3}$. Then $$\sin 2\theta = \sin \left(2 \cdot \frac{\pi}{3}\right) = \sin \frac{2\pi}{3} = \frac{\sqrt{3}}{2}.$$\n\n- At $\theta = \pi/4$, $$\sin 2\theta = \sin \frac{\pi}{2} = 1.$$\n\n10. **Substitute values:**\n$$\left( \frac{\pi}{3} + \frac{\sqrt{3}/2}{2} \right) - \left( \frac{\pi}{4} + \frac{1}{2} \right) = \left( \frac{\pi}{3} + \frac{\sqrt{3}}{4} \right) - \left( \frac{\pi}{4} + \frac{1}{2} \right) = \frac{\pi}{3} - \frac{\pi}{4} + \frac{\sqrt{3}}{4} - \frac{1}{2}.$$\n\n11. **Simplify differences:**\n$$\frac{\pi}{3} - \frac{\pi}{4} = \frac{4\pi - 3\pi}{12} = \frac{\pi}{12}.$$\n\n12. **Final value:**\n$$\frac{1}{2} \left( \frac{\pi}{12} + \frac{\sqrt{3}}{4} - \frac{1}{2} \right) = \frac{\pi}{24} + \frac{\sqrt{3}}{8} - \frac{1}{4}.$$\n\n**Answer:** $$\boxed{\frac{\pi}{24} + \frac{\sqrt{3}}{8} - \frac{1}{4}}.$$