$\int \frac{{cosec}^{2} x - 2019}{\cos^{2019} x} d x =$

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$\frac{\cot x}{{(\cos x)}^{2019}} + c$

$\frac{\tan x}{{(\cos x)}^{2019}} + c$

$\frac{- \tan x}{{(\cos x)}^{2019}} + c$

$\frac{- \cot x}{{(\cos x)}^{2019}} + c$

answer is C.

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Detailed Solution

$\begin{matrix} \int \frac{{cosec}^{2} x - 2019}{\cos^{2019} x} d x = \int \cos^{- 2019} x \cdot {cosec}^{2} x d x - 2019 \int \frac{d x}{\cos^{2019} x} \\ = \cos^{- 2019} x (- \cot x) - \int (- 2019) \cos^{- 2020} x (- \sin x) (- \cot x) - 2019 \int \frac{d x}{\cos^{2019} x} \\ = - \cos^{- 2019} x \cdot \cot x + 2019 \int \cos^{- 2020} x \sin x \times \frac{\cos x}{\sin x} d x - 2019 \int \frac{d x}{\cos^{2019} x} \\ = \frac{- \cot x}{\cos 2019 x} + 2019 \int \cos^{- 2019} x d x - 2019 \int \cos^{- 2019} x d x \\ = \frac{- \cot x}{\cos^{2019} x} + c \end{matrix}$