int (tan x - cot x)^2 , dx = | vedclass

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(D) We need to evaluate the integral $I = \int (	an x - \cot x)^2 \, dx$.Expanding the square,we get:$I = \int (	an^2 x + \cot^2 x - 2 	an x \cot x

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(D) We need to evaluate the integral $I = \int (	an x - \cot x)^2 \, dx$.Expanding the square,we get:$I = \int (	an^2 x + \cot^2 x - 2 	an x \cot x) \, dx$Since $	an x \cot x = 1$,the expression becomes:$I = \int (	an^2 x + \cot^2 x - 2) \, dx$Using the trigonometric identities $	an^2 x = \sec^2 x - 1$ and $\cot^2 x = \csc^2 x - 1$,we substitute:$I = \int ((\sec^2 x - 1) + (\csc^2 x - 1) - 2) \, dx$$I = \int (\sec^2 x + \csc^2 x - 4) \, dx$Integrating term by term:$I = \int \sec^2 x \, dx + \int \csc^2 x \, dx - \int 4 \, dx$$I = 	an x - \cot x - 4x + c$.

$\int (\tan x - \cot x)^2 \, dx = $

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