Answer
$- \dfrac{1}{6} \ln |\cos (6x)| +C $
Work Step by Step
Our aim is to solve the integral $ \int \tan (6 x) \ dx$
Use formula:$ \int \tan (ax+b) \ dx =\dfrac{-1}{a} \ln |\cos (ax+b)|+C$
Let us consider that $a = 6$ and $b=0$
Now, $ \int \tan (6 x) \ dx =\dfrac{-1}{6} \ln |\cos (6 x+0)| +C $
or, $=- \dfrac{1}{6} \ln |\cos (6x)| +C $
