1. **Problem Statement:** Evaluate the line integral $$\int_C (x+2y)\,dx + (x - y)\,dy$$ where the curve $$C$$ is given by the parametric equations $$x=2\cos t, y=2\sin t, 0 \leq t \leq \frac{\pi}{4}$$. 2. **Formula for line integral:** For a vector field $$\mathbf{F} = P\mathbf{i} + Q\mathbf{j}$$, the line integral along curve $$C$$ parameterized by $$t$$ is $$\int_C P\,dx + Q\,dy = \int_a^b \left[P \frac{dx}{dt} + Q \frac{dy}{dt}\right] dt.$$ 3. **Identify** $$P$$ and $$Q$$: Here, $$P = x + 2y$$ and $$Q = x - y$$. 4. **Compute derivatives:** From the parametric equations, $$\frac{dx}{dt} = -2\sin t, \quad \frac{dy}{dt} = 2\cos t.$$ 5. **Substitute** $$x, y, \frac{dx}{dt}, \frac{dy}{dt}$$ into the integral: $$\int_0^{\pi/4} \left[(2\cos t + 2 \cdot 2\sin t)(-2\sin t) + (2\cos t - 2\sin t)(2\cos t)\right] dt.$$ Simplify inside the integral: $$\int_0^{\pi/4} \left[(2\cos t + 4\sin t)(-2\sin t) + (2\cos t - 2\sin t)(2\cos t)\right] dt.$$ 6. **Expand terms:** $$\int_0^{\pi/4} \left[-4\cos t \sin t - 8\sin^2 t + 4\cos^2 t - 4\sin t \cos t\right] dt.$$ Combine like terms: $$\int_0^{\pi/4} \left[4\cos^2 t - 12\sin t \cos t - 8\sin^2 t\right] dt.$$ 7. **Use trigonometric identities:** $$\cos^2 t = \frac{1 + \cos 2t}{2}, \quad \sin^2 t = \frac{1 - \cos 2t}{2}, \quad \sin t \cos t = \frac{\sin 2t}{2}.$$ Substitute: $$\int_0^{\pi/4} \left[4 \cdot \frac{1 + \cos 2t}{2} - 12 \cdot \frac{\sin 2t}{2} - 8 \cdot \frac{1 - \cos 2t}{2}\right] dt.$$ Simplify: $$\int_0^{\pi/4} \left[2(1 + \cos 2t) - 6 \sin 2t - 4(1 - \cos 2t)\right] dt = \int_0^{\pi/4} \left[2 + 2\cos 2t - 6 \sin 2t - 4 + 4 \cos 2t\right] dt.$$ Combine constants and like terms: $$\int_0^{\pi/4} \left[-2 + 6 \cos 2t - 6 \sin 2t\right] dt.$$ 8. **Integrate term-by-term:** $$\int_0^{\pi/4} -2 dt + \int_0^{\pi/4} 6 \cos 2t dt - \int_0^{\pi/4} 6 \sin 2t dt.$$ Calculate each integral: - $$\int_0^{\pi/4} -2 dt = -2t \Big|_0^{\pi/4} = -\frac{\pi}{2}.$$ - $$\int_0^{\pi/4} 6 \cos 2t dt = 6 \cdot \frac{\sin 2t}{2} \Big|_0^{\pi/4} = 3 \sin \frac{\pi}{2} - 3 \sin 0 = 3.$$ - $$\int_0^{\pi/4} 6 \sin 2t dt = -6 \cdot \frac{\cos 2t}{2} \Big|_0^{\pi/4} = -3 (\cos \frac{\pi}{2} - \cos 0) = -3 (0 - 1) = 3.$$ 9. **Sum all parts:** $$-\frac{\pi}{2} + 3 - 3 = -\frac{\pi}{2}.$$ **Final answer:** $$-\frac{\pi}{2}.$$