Step-by-step Solution:

We are given the integral: \[ \int \sec^2 x \, \csc^4 x \, dx \] This can be written as: \[ \int \frac{1}{\cos^2 x \, \sin^4 x} \, dx = \int \frac{1}{\cot^2 x \, \sin^6 x} \, dx = \int \frac{\csc^6 x}{\cot^2 x} \, dx \] Next, we express the integrand in a different form: \[ = \int \frac{(1 + \cot^2 x)^2 \csc^2 x}{\cot^2 x} \, dx. \] Now, let \( \cot x = t \), so that: \[ - \csc^2 x \, dx = dt \] Substitute this into the integral: \[ - \int \frac{(1 + t^2)^2}{t^2} \, dt. \] This simplifies to: \[ - \int \frac{(1 + t^2)^2}{t^2} \, dt = - \frac{t^3}{3} - 2t + \frac{1}{t}. \] Substituting back \( t = \cot x \), we get: \[ - \frac{1}{3} \cot^3 x - 2 \cot x + \tan x. \] Thus, the value of \( k \) is: \[ k = 1. \]