Nếu có x = c o t t thì:

Ta có:  \[x = \cot t \Rightarrow dx = {\left( {\cot t} \right)^\prime }dt = - \frac{1}{{{{\sin }^2}t}}dt = - \left( {1 + {{\cot }^2}t} \right)dt\]

Do

\[\begin{array}{*{20}{l}}{\frac{1}{{{{\sin }^2}t}} = \frac{{{{\sin }^2}x + {{\cos }^2}x}}{{{{\sin }^2}x}}}\\{ = 1 + {{\left( {\frac{{\cos x}}{{\sin x}}} \right)}^2} = 1 + {{\cot }^2}x}\end{array}\]