Evaluate $\displaystyle\int \dfrac{\cot x}{\text{cosec }x - \cot x} dx$
$\begin{aligned}
\text{Let } I & = \int \dfrac{\cot x}{\text{cosec }x - \cot x} dx \\\\
& = \int \dfrac{\cot x \left(\text{cosec }x + \cot x\right)}{\left(\text{cosec }x - \cot x\right) \left(\text{cosec }x + \cot x\right)} dx \\\\
& = \int \dfrac{\cot x \;\; \text{cosec }x + \cot^2 x}{\text{cosec}^2 x - \cot^2 x} dx \\\\
& = \int \cot x \;\; \text{cosec }x \;\; dx + \int \cot^2 x \;\; dx \;\;\;\; \left[\text{Note: }\text{cosec}^2 x - \cot^2 x = 1\right]
\end{aligned}$
$\begin{aligned}
\text{Now, } \int \cot^2 x \;\; dx & = \int \dfrac{\cos^2 x}{\sin^2 x} \;\; dx \\\\
& = \int \dfrac{1 - \sin^2 x}{\sin^2 x} \;\; dx \\\\
& = \int \dfrac{1}{\sin^2 x} \;\; dx - \int dx \\\\
& = \int \text{cosec}^2 x \;\; dx - \int dx
\end{aligned}$
$\begin{aligned}
\therefore I & = \int \cot x \;\; \text{cosec }x \;\; dx + \int \text{cosec}^2 x \;\; dx - \int dx \\\\
& = - \text{cosec }x - \cot x - x + c
\end{aligned}$