Evaluate the integral using any appropriate algebraic method or trigonometric identity. \( \int(\csc x-\tan x)^{2} d x \)

To evaluate the integral \( \int(\csc x - \tan x)^{2} \, dx \), we can start by expanding the integrand: \[ (\csc x - \tan x)^{2} = \csc^{2} x - 2 \csc x \tan x + \tan^{2} x. \] Now we know from trigonometric identities that: \[ \csc^{2} x = 1 + \cot^{2} x \quad \text{and} \quad \tan^{2} x = \sec^{2} x - 1. \] Thus, the expression can be simplified as follows: \[ \int(\csc^{2} x - 2 \csc x \tan x + \tan^{2} x) \, dx = \int \csc^{2} x \, dx - 2 \int \csc x \tan x \, dx + \int \tan^{2} x \, dx. \] Now we integrate each term separately: 1. \( \int \csc^{2} x \, dx = -\cot x + C_1 \). 2. \( \int \csc x \tan x \, dx = -\csc x + C_2 \) so \( -2 \int \csc x \tan x \, dx = 2 \csc x + C_3 \). 3. For \( \int \tan^{2} x \, dx \), we use the identity \( \tan^{2} x = \sec^{2} x - 1 \): \[ \int \tan^{2} x \, dx = \int(\sec^{2} x - 1) \, dx = \int \sec^{2} x \, dx - \int 1 \, dx = \tan x - x + C_4. \] Combining all this, we have: \[ \int(\csc x - \tan x)^{2} \, dx = -\cot x + 2\csc x + (\tan x - x) + C. \] Thus, the final answer is: \[ -\cot x + 2\csc x + \tan x - x + C, \] where \( C \) is the constant of integration.