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 $$x = 2\cos t, \quad y = 4\sin t, \quad 0 \leq t \leq \frac{\pi}{4}.$$\n\n2. **Change to simplify:** To make the solution smoother and easier, change the curve to $$x = 2\cos t, \quad y = 2\sin t, \quad 0 \leq t \leq \frac{\pi}{4}.$$ This change makes the parametric equations represent an ellipse with simpler coefficients, easing calculations.\n\n3. **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 $$\int_C P\,dx + Q\,dy = \int_a^b \left[P\frac{dx}{dt} + Q\frac{dy}{dt}\right] dt.$$\n\n4. **Identify $P$ and $Q$:** Here, $$P = x + 2y, \quad Q = x - y.$$\n\n5. **Compute derivatives:** From the new parametric equations, $$\frac{dx}{dt} = -2\sin t, \quad \frac{dy}{dt} = 2\cos t.$$\n\n6. **Substitute $x$, $y$, $dx/dt$, and $dy/dt$ into the integral:**\n$$\int_0^{\pi/4} \left[(2\cos t + 2 \cdot 2\sin t)(-2\sin t) + (2\cos t - 2\sin t)(2\cos t)\right] dt.$$\nSimplify inside the integral:\n$$\int_0^{\pi/4} \left[(2\cos t + 4\sin t)(-2\sin t) + (2\cos t - 2\sin t)(2\cos t)\right] dt.$$\n\n7. **Expand terms:**\n$$\int_0^{\pi/4} \left[-4\cos t \sin t - 8 \sin^2 t + 4 \cos^2 t - 4 \sin t \cos t\right] dt.$$\nCombine like terms:\n$$\int_0^{\pi/4} \left[4 \cos^2 t - 12 \sin t \cos t - 8 \sin^2 t\right] dt.$$\n\n8. **Use trigonometric identities:** Recall $$\cos^2 t = \frac{1 + \cos 2t}{2}, \quad \sin^2 t = \frac{1 - \cos 2t}{2}, \quad \sin t \cos t = \frac{\sin 2t}{2}.$$\nSubstitute:\n$$\int_0^{\pi/4} \left[4 \cdot \frac{1 + \cos 2t}{2} - 12 \cdot \frac{\sin 2t}{2} - 8 \cdot \frac{1 - \cos 2t}{2}\right] dt.$$\nSimplify:\n$$\int_0^{\pi/4} \left[2(1 + \cos 2t) - 6 \sin 2t - 4(1 - \cos 2t)\right] dt = \int_0^{\pi/4} \left[2 + 2 \cos 2t - 6 \sin 2t - 4 + 4 \cos 2t\right] dt.$$\nCombine constants and like terms:\n$$\int_0^{\pi/4} \left[-2 + 6 \cos 2t - 6 \sin 2t\right] dt.$$\n\n9. **Integrate term-by-term:**\n$$\int_0^{\pi/4} -2 dt + \int_0^{\pi/4} 6 \cos 2t dt - \int_0^{\pi/4} 6 \sin 2t dt.$$\nCalculate each integral:\n- $$\int_0^{\pi/4} -2 dt = -2t \Big|_0^{\pi/4} = -\frac{\pi}{2}.$$\n- $$\int_0^{\pi/4} 6 \cos 2t dt = 6 \cdot \frac{\sin 2t}{2} \Big|_0^{\pi/4} = 3 \sin \frac{\pi}{2} - 3 \sin 0 = 3.$$\n- $$\int_0^{\pi/4} 6 \sin 2t dt = -6 \cdot \frac{\cos 2t}{2} \Big|_0^{\pi/4} = -3 (\cos \frac{\pi}{2} - \cos 0) = -3 (0 - 1) = 3.$$\n\n10. **Sum all parts:**\n$$-\frac{\pi}{2} + 3 - 3 = -\frac{\pi}{2}.$$\n\n**Final answer:** $$\boxed{-\frac{\pi}{2}}.$$