Chọn: Đáp án C

Đặt $u=\ln \left( \sin x \right)\Rightarrow du=\frac{\text{cos}x}{\sin x}dx$

$dv=\frac{dx}{{{\cos }^{2}}x}$ chọn $v=\tan x$

Vậy $I=\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}}{\frac{\ln \left( \sin x \right)}{{{\cos }^{2}}x}}dx=\left[ \tan x.\ln x\left( \sin x \right) \right]_{\frac{\pi }{6}}^{\frac{\pi }{3}}-\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}}{dx}$

            $=\sqrt{3}\ln \left( \frac{\sqrt{3}}{2} \right)-\frac{\sqrt{3}}{2}\ln \frac{1}{2}-\left( \frac{\pi }{3}-\frac{\pi }{2} \right)=\sqrt{3}\ln \left( \frac{\sqrt{3}}{\sqrt[3]{4}} \right)-\frac{\pi }{6}\Rightarrow a=\sqrt{3};b=\frac{1}{6}\Rightarrow A=-1$