Use a change of variables to evaluate the following integral. \[ \int \sec 38 w \tan 38 w d w \]

To solve the integral \( \int \sec(38w) \tan(38w) \, dw \), we can use the change of variables method. Let's set: \[ u = 38w \quad \Rightarrow \quad du = 38 \, dw \quad \Rightarrow \quad dw = \frac{du}{38} \] Now, rewrite the integral in terms of \( u \): \[ \int \sec(38w) \tan(38w) \, dw = \int \sec(u) \tan(u) \cdot \frac{du}{38} = \frac{1}{38} \int \sec(u) \tan(u) \, du \] The integral \( \int \sec(u) \tan(u) \, du \) is a standard integral, and we know that: \[ \int \sec(u) \tan(u) \, du = \sec(u) + C \] Substituting this back into our expression gives: \[ \int \sec(38w) \tan(38w) \, dw = \frac{1}{38} \left( \sec(u) + C \right) = \frac{1}{38} \left( \sec(38w) + C \right) \] Thus, the final result is: \[ \int \sec(38w) \tan(38w) \, dw = \frac{1}{38} \sec(38w) + C \]