1. The problem appears to involve the expression \(\sin 7 \cos 3 x x dx\), which is ambiguous. Assuming you want to integrate the function \(f(x) = \sin(7) \cos(3x) x\) with respect to \(x\). 2. First, note that \(\sin(7)\) is a constant since 7 is a constant angle in radians. So the integral becomes \(\sin(7) \int x \cos(3x) dx\). 3. To integrate \(\int x \cos(3x) dx\), use integration by parts: let \(u = x\) and \(dv = \cos(3x) dx\). Then \(du = dx\) and \(v = \frac{\sin(3x)}{3}\). 4. Applying integration by parts formula \(\int u dv = uv - \int v du\), we get: $$\int x \cos(3x) dx = x \cdot \frac{\sin(3x)}{3} - \int \frac{\sin(3x)}{3} dx$$ 5. Now integrate \(\int \sin(3x) dx = -\frac{\cos(3x)}{3} + C\), so $$\int \frac{\sin(3x)}{3} dx = \frac{1}{3} \left(-\frac{\cos(3x)}{3}\right) = -\frac{\cos(3x)}{9} + C$$ 6. Substitute back: $$\int x \cos(3x) dx = \frac{x \sin(3x)}{3} + \frac{\cos(3x)}{9} + C$$ 7. Multiply by \(\sin(7)\): $$\int \sin(7) x \cos(3x) dx = \sin(7) \left( \frac{x \sin(3x)}{3} + \frac{\cos(3x)}{9} \right) + C$$ This is the integral of the given expression assuming the interpretation above.