Answer
$\int tan~x~ln(cos~x)~dx = -\frac{1}{2}[ln(cos~x)]^2+C$
Work Step by Step
$\int tan~x~ln(cos~x)~dx$
Let $u = ln(cos~x)$
$\frac{du}{dx} = -\frac{sin~x}{cos~x}$
$\frac{du}{dx} = -tan~x$
$dx = -\frac{du}{tan~x}$
$\int (tan~x)~(u)~(-\frac{du}{tan~x})$
$=\int -u~du$
$=-\frac{1}{2}u^2+C$
$=-\frac{1}{2}[ln(cos~x)]^2+C$