Answer
$$
\int \tan x \ln (\cos x) d x=-\frac{1}{2}[\ln (\cos x)]^{2}+C
$$
where $C$ is an arbitrary constant.
Work Step by Step
$$
\int \tan x \ln (\cos x) d x
$$
$$\text { Let } u=\ln (\cos x). \text { Then } d u=\frac{-\sin x}{\cos x} d x=-\tan x d x
$$
substituting in the given integral we have :
$$
\begin{aligned}
\int \tan x \ln (\cos x) d x &=-\int u d u \\
&=-\frac{1}{2} u^{2}+C \\
&=-\frac{1}{2}[\ln (\cos x)]^{2}+C
\end{aligned}
$$
where $C$ is an arbitrary constant.