Question:

Step 1: Rewrite the integrand using trigonometric identities. The given integral is $I(x)=\displaystyle\int\frac{6}{\sin^2 x(1-\cot x)^2}\,dx$. We know that $\frac{1}{\sin^2 x} = \csc^2 x$. So, we can rewrite the integrand as: $$ I(x) = \int 6 \cdot \frac{1}{(1-\cot x)^2} \cdot \csc^2 x \, dx $$ This form suggests a substitution, as the derivative of $(1-\cot x)$ is $\csc^2 x$. Step 2: Perform a substitution to simplify the integral. Let $u = 1-\cot x$. Differentiating $u$ with respect to $x$, we get: $$ \frac{du}{dx} = \frac{d}{dx}(1-\cot x) = 0 - (-\csc^2 x) = \csc^2 x $$ So, $du = \csc^2 x \, dx$. Substituting $u$ and $du$ into the integral: $$ I(u) = \int \frac{6}{u^2} \, du $$ Step 3: Integrate the simplified expression. Now, we integrate $6u^{-2}$ with respect to $u$: $$ I(u) = 6 \int u^{-2} \, du = 6 \left( \frac{u^{-2+1}}{-2+1} \right) + C = 6 \left( \frac{u^{-1}}{-1} \right) + C = -\frac{6}{u} + C $$ Substitute back $u = 1-\cot x$ to express $I(x)$ in terms of $x$: $$ I(x) = -\frac{6}{1-\cot x} + C $$ Step 4: Use the initial condition to find the constant of integration $C$. We are given that $I(0)=3$. While $\cot x$ is undefined at $x=0$, this condition is used to determine the specific antiderivative. Substitute $x=0$ into the expression for $I(x)$ and set it equal to 3. In the context of such problems, $I(0)=3$ generally means that as $x \to 0$, $I(x) \to 3$. As $x \to 0^+$, $\cot x \to \infty$. Thus, $1-\cot x \to -\infty$. So, $\lim_{x \to 0^+} \left(-\frac{6}{1-\cot x}\right) = 0$. Therefore, $I(0) = 0 + C = 3$, which implies $C=3$. So, the specific integral function is: $$ I(x) = 3 - \frac{6}{1-\cot x} $$ Step 5: Evaluate $I\!\left(\dfrac{\pi}{12}\right)$. Now we substitute $x = \frac{\pi}{12}$ into the expression for $I(x)$. First, we need the value of $\cot\left(\frac{\pi}{12}\right)$, which is $\cot(15^\circ)$. We know that $\tan(15^\circ) = \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} = \frac{1 - 1/\sqrt{3}}{1 + 1/\sqrt{3}} = \frac{\sqrt{3}-1}{\sqrt{3}+1}$. Rationalizing this, we get $\tan(15^\circ) = \frac{(\sqrt{3}-1)^2}{(\sqrt{3}+1)(\sqrt{3}-1)} = \frac{3+1-2\sqrt{3}}{3-1} = \frac{4-2\sqrt{3}}{2} = 2-\sqrt{3}$. Therefore, $\cot(15^\circ) = \frac{1}{\tan(15^\circ)} = \frac{1}{2-\sqrt{3}} = \frac{1(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})} = \frac{2+\sqrt{3}}{4-3} = 2+\sqrt{3}$. Now, substitute this value into $I(x)$: $$ I\left(\frac{\pi}{12}\right) = 3 - \frac{6}{1-(2+\sqrt{3})} $$ $$ I\left(\frac{\pi}{12}\right) = 3 - \frac{6}{1-2-\sqrt{3}} $$ $$ I\left(\frac{\pi}{12}\right) = 3 - \frac{6}{-1-\sqrt{3}} $$ $$ I\left(\frac{\pi}{12}\right) = 3 + \frac{6}{1+\sqrt{3}} $$ To simplify, rationalize the denominator: $$ I\left(\frac{\pi}{12}\right) = 3 + \frac{6( \sqrt{3}-1)}{(1+\sqrt{3})(\sqrt{3}-1)} $$ $$ I\left(\frac{\pi}{12}\right) = 3 + \frac{6(\sqrt{3}-1)}{3-1} $$ $$ I\left(\frac{\pi}{12}\right) = 3 + \frac{6(\sqrt{3}-1)}{2} $$ $$ I\left(\frac{\pi}{12}\right) = 3 + 3(\sqrt{3}-1) $$ $$ I\left(\frac{\pi}{12}\right) = 3 + 3\sqrt{3} - 3 $$ $$ I\left(\frac{\pi}{12}\right) = 3\sqrt{3} $$ The final answer is $\boxed{3\sqrt{3}}$.