Let I(x)=int frac{6}{sin ^2 x(1-cot x)^2} d x | vedclass

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(B) Given $I(x)=\int \frac{6}{\sin ^2 x(1-\cot x)^2} d x$. We can rewrite the integral as $I(x)=\int \frac{6 \operatorname{cosec}^2 x}{(1-\cot x)^2} d

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$3 \sqrt{3}$

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(B) Given $I(x)=\int \frac{6}{\sin ^2 x(1-\cot x)^2} d x$. We can rewrite the integral as $I(x)=\int \frac{6 \operatorname{cosec}^2 x}{(1-\cot x)^2} d x$. Let $t = 1-\cot x$. Then $dt = \operatorname{cosec}^2 x d x$. Substituting these into the integral,we get $I(x) = \int \frac{6}{t^2} dt = -\frac{6}{t} + C = -\frac{6}{1-\cot x} + C$. Given $I(0) = 3$. However,$\cot(0)$ is undefined. Assuming the limit as $x 	o 0$,$I(x) = \frac{-6}{1-\cot x} + C$. Actually,evaluating the constant $C$ from the expression $I(x) = \frac{-6}{1-\cot x} + C$,we have $I(x) = \frac{6}{\cot x - 1} + C$. Given $I(0)=3$ is problematic,but typically such problems imply the constant $C$ is determined by the form. Let's re-evaluate: $I(x) = \frac{6}{\cot x - 1} + C$. At $x = \frac{\pi}{12}$,$\cot(\frac{\pi}{12}) = \cot(15^\circ) = 2+\sqrt{3}$. $I(\frac{\pi}{12}) = \frac{6}{2+\sqrt{3}-1} + C = \frac{6}{1+\sqrt{3}} + C = 3(\sqrt{3}-1) + C$. Assuming $C=3$,$I(\frac{\pi}{12}) = 3\sqrt{3}-3+3 = 3\sqrt{3}$.

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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