1. The problem is to find the integral of $\sec w$ with respect to $w$, i.e., $\int \sec w \, dw$. 2. The formula used for this integral is a standard result in calculus: $\int \sec w \, dw = \ln|\sec w + \tan w| + C$, where $C$ is the constant of integration. 3. To understand why, recall the identity $\frac{d}{dw}(\ln|\sec w + \tan w|) = \sec w$. 4. This is because the derivative of $\sec w$ is $\sec w \tan w$ and the derivative of $\tan w$ is $\sec^2 w$, which combine to give the integrand when applying the chain rule. 5. Therefore, the integral evaluates to: $$\int \sec w \, dw = \ln|\sec w + \tan w| + C$$ This is the final answer.