Answer
$-\dfrac{[\ln (\cos x)]^2}{2}+c$
Work Step by Step
Solve $\int \tan x \ln (\cos x) dx$
Let us $p=\ln (\cos x)$ and $dp=- \tan x dx$
This implies $\int \tan x \ln (\cos x) dx=-\int p dp $
$= -\dfrac{p^2}{2}+c$
$= -\dfrac{[\ln (\cos x)]^2}{2}+c$
Hence, $\int \tan x \ln (\cos x) dx=-\dfrac{[\ln (\cos x)]^2}{2}+c$
