how to derive the fact that the integral of $1/\sin^2(x) = -\cot (x)$ I know how that the integral of  $\dfrac{1}{\sin^2(x)} = -\cot (x)$, but how does derive this fact? Can you use half-angle formula to do this integral?

$$
\frac{d}{dx} \cot x = \frac{d}{dx} \frac{\cos x}{\sin x} = \frac{[\text{etc. : Apply the quotient rule.}]}{\sin^2 x} = \frac{-1}{\sin^2 x}.
$$
