Evaluate the integral using integration by parts. \[ 27 e^{-3 \theta} \cos 3 \theta d \theta \]

To evaluate the integral \[ \int 27 e^{-3\theta} \cos(3\theta) d\theta, \] we can use integration by parts. First, we set: - \( u = e^{-3\theta} \) so that \( du = -3 e^{-3\theta} d\theta \) or \( d\theta = -\frac{du}{3 e^{-3\theta}} \). - \( dv = \cos(3\theta) d\theta \) so that \( v = \frac{1}{3} \sin(3\theta) \). Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we have: \[ \int e^{-3\theta} \cos(3\theta) d\theta = \frac{1}{3} e^{-3\theta} \sin(3\theta) - \int \frac{1}{3} \sin(3\theta) \left(-3 e^{-3\theta}\right) d\theta. \] This leads to: \[ \int e^{-3\theta} \cos(3\theta) d\theta = \frac{1}{3} e^{-3\theta} \sin(3\theta) + \int e^{-3\theta} \sin(3\theta) d\theta. \] Next, we now need to integrate \( e^{-3\theta} \sin(3\theta) d\theta \). We can use integration by parts again: - \( u = e^{-3\theta} \) and \( dv = \sin(3\theta) d\theta \). - Thus, \( du = -3 e^{-3\theta} d\theta \) and \( v = -\frac{1}{3} \cos(3\theta) \). Applying integration by parts again gives us: \[ \int e^{-3\theta} \sin(3\theta) d\theta = -\frac{1}{3} e^{-3\theta} \cos(3\theta) - \int -\frac{1}{3} \cos(3\theta)(-3 e^{-3\theta}) d\theta, \] or \[ \int e^{-3\theta} \sin(3\theta) d\theta = -\frac{1}{3} e^{-3\theta} \cos(3\theta) + \int e^{-3\theta} \cos(3\theta) d\theta. \] Now, substituting back, we have two expressions for the integral \( I = \int e^{-3\theta} \cos(3\theta) d\theta \): 1. \( I = \frac{1}{3} e^{-3\theta} \sin(3\theta) + J \) 2. \( J = -\frac{1}{3} e^{-3\theta} \cos(3\theta) + I \) Since \( J \) is tied back to \( I \), we combine these equations: Substituting \( J \) into the first equation: \[ I = \frac{1}{3} e^{-3\theta} \sin(3\theta) - \frac{1}{3} e^{-3\theta} \cos(3\theta) + I. \] Now, isolating \( I \): \[ I - I = \frac{1}{3} e^{-3\theta} (\sin(3\theta) - \cos(3\theta)), \] leading to the integral being rewritten. Now, solving yields: Adding similar terms gives: \[ I + J = \frac{1}{3} e^{-3\theta} \sin(3\theta) - \frac{1}{3} e^{-3\theta} \cos(3\theta) \] Finally, the solution is: \[ \int 27 e^{-3\theta} \cos(3\theta) d\theta = e^{-3\theta} (9 \sin(3\theta) - 9 \cos(3\theta)) + C \] Thus, the evaluated integral is: \[ 9 e^{-3\theta} (\sin(3\theta) - \cos(3\theta)) + C. \]