Evaluate the following integral using integration by parts. \[ \int e^{18 x} \cos x d x \]

To evaluate the integral \(\int e^{18x} \cos x \, dx\) using integration by parts, we will apply the technique repeatedly. Let: - \( u = \cos x \) \(\Rightarrow du = -\sin x \, dx\) - \( dv = e^{18x} \, dx \) \(\Rightarrow v = \frac{1}{18} e^{18x}\) Now, we apply integration by parts: \[ \int e^{18x} \cos x \, dx = uv - \int v \, du \] Substituting in our expressions, we have: \[ = \cos x \cdot \frac{1}{18} e^{18x} - \int \frac{1}{18} e^{18x} (-\sin x) \, dx \] \[ = \frac{1}{18} e^{18x} \cos x + \frac{1}{18} \int e^{18x} \sin x \, dx \] Next, we need to evaluate \(\int e^{18x} \sin x \, dx\). We will again use integration by parts: Let: - \( u = \sin x \) \(\Rightarrow du = \cos x \, dx\) - \( dv = e^{18x} \, dx \) \(\Rightarrow v = \frac{1}{18} e^{18x}\) Using integration by parts again: \[ \int e^{18x} \sin x \, dx = \sin x \cdot \frac{1}{18} e^{18x} - \int \frac{1}{18} e^{18x} \cos x \, dx \] \[ = \frac{1}{18} e^{18x} \sin x - \frac{1}{18} \int e^{18x} \cos x \, dx \] Now, substitute this back into our first integral: Let \( I = \int e^{18x} \cos x \, dx \). Then we have: \[ I = \frac{1}{18} e^{18x} \cos x + \frac{1}{18} \left( \frac{1}{18} e^{18x} \sin x - \frac{1}{18} I \right) \] \[ = \frac{1}{18} e^{18x} \cos x + \frac{1}{18^2} e^{18x} \sin x - \frac{1}{324} I \] Now, combine the terms: \[ I + \frac{1}{324} I = \frac{1}{18} e^{18x} \cos x + \frac{1}{18^2} e^{18x} \sin x \] \[ \left(1 + \frac{1}{324}\right) I = \frac{1}{18} e^{18x} \cos x + \frac{1}{324} e^{18x} \sin x \] \[ \frac{325}{324} I = \frac{1}{18} e^{18x} \cos x + \frac{1}{324} e^{18x} \sin x \] Now, multiply both sides by \(\frac{324}{325}\): \[ I = \frac{324}{325} \left( \frac{1}{18} e^{18x} \cos x + \frac{1}{324} e^{18x} \sin x \right) \] \[ = \frac{18}{325} e^{18x} \cos x + \frac{1}{325} e^{18x} \sin x \] So we have: \[ \int e^{18x} \cos x \, dx = \frac{18}{325} e^{18x} \cos x + \frac{1}{325} e^{18x} \sin x + C \] where \(C\) is the constant of integration.