1. Stating the problem: Evaluate the definite integral $$\int_{\frac{\pi}{2}}^{0} \frac{1+\cos 2t}{2} \, dt$$. 2. Formula and rules: Recall the integral of cosine and the property of definite integrals that reversing limits changes the sign: $$\int_a^b f(t) \, dt = -\int_b^a f(t) \, dt$$. 3. Rewrite the integral by switching limits: $$\int_{\frac{\pi}{2}}^{0} \frac{1+\cos 2t}{2} \, dt = -\int_0^{\frac{\pi}{2}} \frac{1+\cos 2t}{2} \, dt$$. 4. Split the integral: $$-\int_0^{\frac{\pi}{2}} \frac{1}{2} \, dt - \int_0^{\frac{\pi}{2}} \frac{\cos 2t}{2} \, dt$$. 5. Integrate each term: - For the first term: $$\int_0^{\frac{\pi}{2}} \frac{1}{2} \, dt = \frac{1}{2} t \Big|_0^{\frac{\pi}{2}} = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}$$. - For the second term: $$\int_0^{\frac{\pi}{2}} \frac{\cos 2t}{2} \, dt = \frac{1}{2} \int_0^{\frac{\pi}{2}} \cos 2t \, dt$$. 6. Integrate $$\cos 2t$$: $$\int \cos 2t \, dt = \frac{\sin 2t}{2}$$. 7. Evaluate the second term: $$\frac{1}{2} \cdot \frac{\sin 2t}{2} \Big|_0^{\frac{\pi}{2}} = \frac{1}{4} (\sin \pi - \sin 0) = \frac{1}{4} (0 - 0) = 0$$. 8. Combine results: $$-\left( \frac{\pi}{4} + 0 \right) = -\frac{\pi}{4}$$. Final answer: $$\boxed{-\frac{\pi}{4}}$$