Let I(x)=\int \frac{6}{\sin ^2 x(1-\cot x)^2} d x. If I(0)=3, then I\left(\frac{\pi}{12}\right) is equal to

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Answer: D. 3\sqrt3 \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}{(	an x-1)^2} d x \end{aligned} \begin{aligned} & 	ext { Let } 	an 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}{(	an x-1)}+c \ & I(x)=\frac{6}{1-	an x}+c \ & I(0)=3 \ & \frac{6}{1-	an 0}+c=3 \ & c=-3 \end{aligned} \begin{aligned} I(x) & =\frac{6}{1-	an x}-3 \ I\left(\frac{\pi}{12}ight) & =\frac{6}{1-	an \left(\frac{\pi}{12}ight)}-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}

Let $$I(x)=\int \frac{6}{\sin ^2 x(1-\cot x)^2} d x$$. If $$I(0)=3$$, then $$I\left(\frac{\pi}{12}\right)$$ is equal to

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