$$\begin{aligned} & I(x)=\int \frac{6}{\sin ^2 x(1-\cot x)^2} d x \\ & I(x)=\int \frac{6}{(\sin x-\cos x)^2} d x \\ & =\int \frac{6 \sec ^2 x}{(\tan x-1)^2} d x \end{aligned}$$

$$\begin{aligned} & \text { Let } \tan x=t \Rightarrow \sec ^2 x d x=d t \\ & =\int \frac{6 d t}{(t-1)^2} \\ & =-\frac{6}{(t-1)}+c \\ & =\frac{-6}{(\tan x-1)}+c \\ & I(x)=\frac{6}{1-\tan x}+c \\ & I(0)=3 \\ & \frac{6}{1-\tan 0}+c=3 \\ & c=-3 \end{aligned}$$

$$\begin{aligned} I(x) & =\frac{6}{1-\tan x}-3 \\ I\left(\frac{\pi}{12}\right) & =\frac{6}{1-\tan \left(\frac{\pi}{12}\right)}-3 \\ & =\frac{6}{1-(2-\sqrt{3})}-3 \\ & =\frac{6}{\sqrt{3}-1}-3 \\ & =\frac{6-3 \sqrt{3}+3}{\sqrt{3}-1} \\ & =\frac{9-3 \sqrt{3}}{\sqrt{3}-1} \\ & =\frac{3 \sqrt{3}(\sqrt{3}-1)}{\sqrt{3}-1} \\ & =3 \sqrt{3} \end{aligned}$$