Answer
$- \dfrac{1}{4} \ln |\sin (-4 x)| +C $
Work Step by Step
Our aim is to solve the integral $ \int \cot (-4 x) \ dx$
Use formula:$ \int \cot (ax+b) \ dx =\dfrac{1}{a} \ln |\sin (ax+b)|+C$
Let us consider that $a =-4$ and $b=0$
Now, $ \int \cot (-4 x) \ dx =\dfrac{1}{-4} \ln |\sin (-4 x)| +C $
or, $=- \dfrac{1}{4} \ln |\sin (-4 x)| +C $
