Let's solve the integral step-by-step: ## Step 1: First Integration by Parts Let \( u x^2 \) ⟹ \( du = 2x\, dx \) \( dv = \cos(6x)\, \) ⟹ \( v = \frac{1}{6} \sin(6x) \) Apply integration by parts: \[ \int u\, dv = uv - \int v\, du \] So, \[ \int x^2 \cos(6x)\, dx = x^2 \cdot \frac{1}{6} \sin(6x) - \int \frac{1}{6} \sin(6x) \cdot 2x\, dx \] \[ = \frac{x^2}{6} \sin(6x) - \frac{2}{6} \int x \sin(6x)\, dx \] \[ = \frac{x^2}{6} \sin(6x) - \frac{1}{3} \int x \sin(6x)\, dx \] **Explanation:** We used integration by parts, treating \(x^2\) as \(u\) and \(\cos(6x)\) as \(dv\). This reduced the power of \(x\) by one. --- ## Step 2: Second Integration by Parts Now, integrate \(\int x \sin(6x)\, dx\) by parts: Let \( u = x \) ⟹ \( du = dx \) \( dv = \sin(6x)\, dx \) ⟹ \( v = -\frac{1}{6} \cos(6x) \) So, \[ \int x \sin(6x)\, dx = -\frac{x}{6} \cos(6x) + \int \frac{1}{6} \cos(6x)\, dx \] \[ = -\frac{x}{6} \cos(6x) + \frac{1}{36} \sin(6x) \] Plug this back into the result from Step 1: \[ \int x^2 \cos(6x)\, dx = \frac{x^2}{6} \sin(6x) - \frac{1}{3} \left( -\frac{x}{6} \cos(6x) + \frac{1}{36} \sin(6x) \right) \] \[ = \frac{x^2}{6} \sin(6x) + \frac{x}{18} \cos(6x) - \frac{1}{108} \sin(6x) \] **Explanation:** Another integration by parts on the remaining integral, now with \(x\) and \(\sin(6x)\), and then combine the results. --- ## Final Answer \[ \boxed{ \int x^2 \cos(6x)\, dx = \frac{x^2}{6} \sin(6x) + \frac{x}{18} \cos(6x) - \frac{1}{108} \sin(6x) + C } \]