Use a change of variables or the table to evaluate the following indefinite integral. \( \int \sec (4 x-3) \tan (4 x-3) d x \) Click the icon to view the table of general integration formulas. \( \int \sec (4 x-3) \tan (4 x-3) d x=\square \)

To evaluate the integral \( \int \sec (4 x - 3) \tan (4 x - 3) \, dx \), we can use a change of variables. Let \( u = 4x - 3 \). Then, the differential \( du = 4 \, dx \), or equivalently \( dx = \frac{du}{4} \). Now, rewrite the integral: \[ \int \sec(u) \tan(u) \frac{du}{4} = \frac{1}{4} \int \sec(u) \tan(u) \, du \] The integral of \( \sec(u) \tan(u) \) is \( \sec(u) + C \). Thus: \[ \frac{1}{4} \int \sec(u) \tan(u) \, du = \frac{1}{4}(\sec(u) + C) = \frac{1}{4} \sec(4x - 3) + C \] So the final answer is: \[ \int \sec(4x - 3) \tan(4x - 3) \, dx = \frac{1}{4} \sec(4x - 3) + C \]