## Calculus: Early Transcendentals 8th Edition

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