Find $\int \ln(\tan(x))/(\sin(x) \cos(x))dx$

Let $u = \ln(\tan x)$. It's simpler than you think.

Put $\ln \tan x =t$

therefore $${\sec^2x \over \tan x} \, \mathrm dx=\mathrm dt$$

$$\mathrm dx={\mathrm dt \, \sin x \cos x}$$

Subsitute these values into the integral

$$I=\int t \, \mathrm dt$$

$$I={t^2 \over 2}+constant$$

$$I={(\ln (\tan x))^2 \over 2}+constant$$

\begin{eqnarray} \int\frac{\ln\tan x}{\sin x\cos x}dx&=&\int\frac{\ln\tan x}{\tan x}\sec^2xdx =\int\frac{\ln\tan x}{\tan x}d\tan x\\ &=&\int\ln\tan xd\ln\tan x=\frac{1}{2}(\ln\tan x)^2+C \end{eqnarray}