### Integration by Parts Example The problem is to evaluate the integral: \[ \int e^{2x} \cos x \, dx \] **Step 1: Choose \( u \) and \( dv \)** Let: \[ u = e^{2x} \] \[ dv = \cos x \, dx \] Then, the derivatives are: \[ du = 2e^{2x} \, dx \] \[ v = \sin x \] **Step 2: Apply Integration by Parts** Integration by parts formula: \[ \int u \, dv = uv - \int v \, du \] Applying the formula: \[ e^{2x} \sin x - \int \sin x \cdot 2e^{2x} \, dx \] **Step 3: Repeat Integration by Parts** Now we handle the integral: \[ \int \sin x \cdot 2e^{2x} \, dx \] Choose: \[ u = e^{2x} \] \[ dv = \sin x \, dx \] Then: \[ du = 2e^{2x} \, dx \] \[ v = -\cos x \] Apply integration by parts again: \[ e^{2x} \sin x - 2(e^{2x}(-\cos x) - \int -\cos x \cdot 2e^{2x} \, dx) \] Simplifies to: \[ e^{2x} \sin x - 2(-e^{2x} \cos x + 2 \int e^{2x} \cos x \, dx) \] \[ e^{2x} \sin x - 2(-e^{2x} \cos x - 2\int e^{2x} \cos x \, dx) \] **Final Result:** \[ \int e^{2x} \cos x \, dx = e^{2x} \sin x - 2(-e^{2x} \cos x - 2 \int e^{2x} \cos x \, dx) \] This integral requires further simplification or solving through methods such as algebraic manipulation or finding a pattern/semi-circular queue.