Answer
$$\frac{(\ln (3))^2}{8}$$
Work Step by Step
Given
$$\int_{\pi / 4}^{\pi / 3} \frac{\ln (\tan x)}{\sin x \cos x} d x$$
Let
$$ u = \ln (\tan (x) ) \ \ \ \ \ \ du =\frac{\sec^2 x}{\tan x }dx=\frac{dx}{\sin x\cos x}$$
and at $x= \pi /4\to u =\ln(1) = 0,\ \ \ x= \pi/3\ \ \to \ u= \ln(\sqrt{3} )$
The
\begin{aligned}
\int_{\pi / 4}^{\pi / 3} \frac{\ln (\tan x)}{\sin x \cos x} d x&=\int_{0}^{\ln ( \sqrt{3})} udu \\
&=\frac{1}{2}u^2\bigg|_{0}^{\ln \sqrt{3}}
\\
&= \frac{1}{2}(\ln \sqrt{3})^2\\
&=\frac{(\ln (3))^2}{8}
\end{aligned}
