Evaluate the following integral using integration by parts. \[ \int 14 x \sin x \cos x d x \]

To evaluate the integral \( \int 14 x \sin x \cos x \, dx \) using integration by parts, we first simplify the integrand. We can use the identity \( \sin x \cos x = \frac{1}{2} \sin(2x) \). Thus, we rewrite the integral: \[ \int 14 x \sin x \cos x \, dx = 7 \int x \sin(2x) \, dx. \] Now, we will use integration by parts on \( \int x \sin(2x) \, dx \). Let: - \( u = x \) which gives \( du = dx \). - \( dv = \sin(2x) \, dx \) which can be integrated to give \( v = -\frac{1}{2} \cos(2x) \). According to the integration by parts formula, \( \int u \, dv = uv - \int v \, du \), we proceed as follows: \[ \int x \sin(2x) \, dx = -\frac{1}{2} x \cos(2x) - \int -\frac{1}{2} \cos(2x) \, dx. \] Now let's compute the integral \( \int \cos(2x) \, dx \): \[ \int \cos(2x) \, dx = \frac{1}{2} \sin(2x). \] Now substituting this back, we have: \[ \int x \sin(2x) \, dx = -\frac{1}{2} x \cos(2x) + \frac{1}{2} \cdot \frac{1}{2} \sin(2x) + C = -\frac{1}{2} x \cos(2x) + \frac{1}{4} \sin(2x) + C. \] Now recall that we had \( 7 \int x \sin(2x) \, dx \): \[ 7 \int x \sin(2x) \, dx = 7 \left(-\frac{1}{2} x \cos(2x) + \frac{1}{4} \sin(2x) + C\right). \] Distributing the 7 gives: \[ -\frac{7}{2} x \cos(2x) + \frac{7}{4} \sin(2x) + C. \] Thus, the final answer to the integral \( \int 14 x \sin x \cos x \, dx \) is: \[ \int 14 x \sin x \cos x \, dx = -\frac{7}{2} x \cos(2x) + \frac{7}{4} \sin(2x) + C. \]