Answer
$$\left( {\sqrt 2 - \frac{{2\sqrt 3 }}{3}} \right)\ln \sqrt 3 - \ln \left( {\frac{{\sqrt 2 }}{4}} \right)$$
Work Step by Step
$$\eqalign{
& \int_{\pi /4}^{\pi /3} {\frac{{\ln \left( {\tan x} \right)}}{{\sin x\cos x}}} dx \cr
& = \int_{\pi /4}^{\pi /3} {\ln \left( {\tan x} \right)\left( {\frac{1}{{\sin x\cos x}}} \right)} dx \cr
& = \int_{\pi /4}^{\pi /3} {\ln \left( {\tan x} \right)\left( {\csc x\cot x} \right)} dx \cr
& {\text{Integrate by parts, }} \cr
& {\text{Let }}u = \ln \left( {\tan x} \right),{\text{ }}du = \frac{{{{\sec }^2}x}}{{\tan x}}dx,{\text{ }}du = \frac{1}{{\csc x\cot x}}dx,{\text{ }} \cr
& dv = \csc x\cot xdx,{\text{ }}v = - \csc x \cr
& {\text{Using }}\int {udv} = uv - \int {vdu} \cr
& \int_{\pi /4}^{\pi /3} {\frac{{\ln \left( {\tan x} \right)}}{{\sin x\cos x}}} dx = - \left[ {\csc x\ln \left( {\tan x} \right)} \right]_{\pi /4}^{\pi /3} + \int_{\pi /4}^{\pi /3} {\tan x} dx \cr
& \int_{\pi /4}^{\pi /3} {\frac{{\ln \left( {\tan x} \right)}}{{\sin x\cos x}}} dx = - \left[ {\csc x\ln \left( {\tan x} \right)} \right]_{\pi /4}^{\pi /3} - \left[ {\ln \left| {\cos x} \right|} \right]_{\pi /4}^{\pi /3} \cr
& {\text{Evaluating}} \cr
& = - \left[ {\csc \left( {\frac{\pi }{3}} \right)\ln \left( {\tan \frac{\pi }{3}} \right) - \csc \left( {\frac{\pi }{4}} \right)\ln \left( {\tan \frac{\pi }{4}} \right)} \right] \cr
& - \left[ {\ln \left| {\cos \frac{\pi }{3}} \right| - \ln \left| {\cos \frac{\pi }{4}} \right|} \right] \cr
& {\text{Simplifying}} \cr
& = - \left[ {\left( {\frac{{2\sqrt 3 }}{3}} \right)\ln \left( {\sqrt 3 } \right) - \sqrt 2 \ln \left( {\sqrt 3 } \right)} \right] - \left[ {\ln \left| {\frac{1}{2}} \right| - \ln \left| {\frac{{\sqrt 2 }}{2}} \right|} \right] \cr
& = - \frac{{2\sqrt 3 }}{3}\ln \sqrt 3 + \sqrt 2 \ln \sqrt 3 - \ln \left( {\frac{1}{2}} \right) - \ln \left( {\frac{{\sqrt 2 }}{2}} \right) \cr
& = \left( {\sqrt 2 - \frac{{2\sqrt 3 }}{3}} \right)\ln \sqrt 3 - \ln \left( {\frac{{\sqrt 2 }}{4}} \right) \cr} $$
