Find \( \int x^{2} \sin 4 x d x \)

To solve the integral \( \int x^{2} \sin 4 x \, dx \), we can use the method of integration by parts multiple times. Let's start with the first integration by parts: Let: - \( u = x^2 \) (so \( du = 2x \, dx \)) - \( dv = \sin 4x \, dx \) (so \( v = -\frac{1}{4} \cos 4x \)) Now applying integration by parts: \[ \int x^2 \sin 4x \, dx = uv - \int v \, du \] Substituting for \( u \), \( v \), and \( du \): \[ = -\frac{1}{4} x^2 \cos 4x - \int -\frac{1}{4} \cos 4x \cdot 2x \, dx \] This simplifies to: \[ = -\frac{1}{4} x^2 \cos 4x + \frac{1}{2} \int x \cos 4x \, dx \] Next, we need to evaluate \( \int x \cos 4x \, dx \) using integration by parts again. Let: - \( u = x \) (then \( du = dx \)) - \( dv = \cos 4x \, dx \) (then \( v = \frac{1}{4} \sin 4x \)) Applying integration by parts again: \[ \int x \cos 4x \, dx = uv - \int v \, du \] So we have: \[ = x \cdot \frac{1}{4} \sin 4x - \int \frac{1}{4} \sin 4x \, dx \] Now, the integral \( \int \sin 4x \, dx = -\frac{1}{4} \cos 4x \): Therefore, \[ = \frac{1}{4} x \sin 4x + \frac{1}{16} \cos 4x \] Now substituting back into our previous result for \( \int x^2 \sin 4x \, dx \): \[ = -\frac{1}{4} x^2 \cos 4x + \frac{1}{2} \left( \frac{1}{4} x \sin 4x + \frac{1}{16} \cos 4x \right) \] This gives: \[ = -\frac{1}{4} x^2 \cos 4x + \frac{1}{8} x \sin 4x + \frac{1}{32} \cos 4x \] Finally, we combine everything to produce the final answer: \[ \int x^{2} \sin 4x \, dx = -\frac{1}{4} x^2 \cos 4x + \frac{1}{8} x \sin 4x + \frac{1}{32} \cos 4x + C \] where \( C \) is the constant of integration.