Key Concepts and Formulas

• Trigonometric identities: $\cot x = \frac{\cos x}{\sin x}$, $\sec^2 x = 1 + \tan^2 x$, $\frac{1}{\sin^2 x} = \csc^2 x$
• Integration of basic functions: $\int \frac{1}{x^2} dx = -\frac{1}{x} + C$
• Value of $\tan(\frac{\pi}{12}) = 2 - \sqrt{3}$

Step-by-Step Solution

Step 1: Rewrite the integrand using trigonometric identities. We start with the given integral: $I(x) = \int \frac{6}{\sin^2 x (1 - \cot x)^2} dx$ We rewrite $(1 - \cot x)$ as $(1 - \frac{\cos x}{\sin x}) = \frac{\sin x - \cos x}{\sin x}$. Substituting this into the integral, we get: $I(x) = \int \frac{6}{\sin^2 x \left(\frac{\sin x - \cos x}{\sin x}\right)^2} dx = \int \frac{6}{\sin^2 x \frac{(\sin x - \cos x)^2}{\sin^2 x}} dx = \int \frac{6}{(\sin x - \cos x)^2} dx$

Step 2: Divide numerator and denominator by $\cos^2 x$. Dividing the numerator and denominator by $\cos^2 x$, we obtain: $I(x) = \int \frac{6 \sec^2 x}{\frac{(\sin x - \cos x)^2}{\cos^2 x}} dx = \int \frac{6 \sec^2 x}{(\tan x - 1)^2} dx$ The purpose of this step is to create an expression that can be easily integrated using substitution.

Step 3: Use substitution to simplify the integral. Let $t = \tan x - 1$. Then, $dt = \sec^2 x \, dx$. Substituting these into the integral, we get: $I(x) = \int \frac{6}{t^2} dt = 6 \int t^{-2} dt$

Step 4: Integrate with respect to t. Integrating $t^{-2}$ with respect to $t$, we have: $I(x) = 6 \left( -\frac{1}{t} \right) + C = -\frac{6}{t} + C$

Step 5: Substitute back for $t$. Substitute $t = \tan x - 1$ back into the expression: $I(x) = -\frac{6}{\tan x - 1} + C = \frac{6}{1 - \tan x} + C$

Step 6: Use the initial condition $I(0) = 3$ to find the constant $C$. We are given that $I(0) = 3$. Substituting $x = 0$ into the expression for $I(x)$, we have: $I(0) = \frac{6}{1 - \tan 0} + C = \frac{6}{1 - 0} + C = 6 + C$ Since $I(0) = 3$, we have: $6 + C = 3 \implies C = 3 - 6 = -3$

Step 7: Write the expression for $I(x)$ with the determined constant. $I(x) = \frac{6}{1 - \tan x} - 3$

Step 8: Evaluate $I\left(\frac{\pi}{12}\right)$. We need to find $I\left(\frac{\pi}{12}\right)$. We know that $\tan\left(\frac{\pi}{12}\right) = 2 - \sqrt{3}$. Substituting this value into the expression for $I(x)$, we get: $I\left(\frac{\pi}{12}\right) = \frac{6}{1 - (2 - \sqrt{3})} - 3 = \frac{6}{1 - 2 + \sqrt{3}} - 3 = \frac{6}{\sqrt{3} - 1} - 3$

Step 9: Simplify the expression. To simplify the expression, we multiply the numerator and denominator of the fraction by the conjugate of the denominator, which is $\sqrt{3} + 1$: $I\left(\frac{\pi}{12}\right) = \frac{6(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} - 3 = \frac{6(\sqrt{3} + 1)}{3 - 1} - 3 = \frac{6(\sqrt{3} + 1)}{2} - 3 = 3(\sqrt{3} + 1) - 3$ $I\left(\frac{\pi}{12}\right) = 3\sqrt{3} + 3 - 3 = 3\sqrt{3}$

Common Mistakes & Tips

• Remembering the correct trigonometric identities and how to manipulate them is crucial.
• When using substitution, make sure to change the limits of integration if it's a definite integral, or substitute back to the original variable for indefinite integrals.
• Don't forget to add the constant of integration, $C$, for indefinite integrals, and use the initial condition to find its value.

Summary

We evaluated the indefinite integral $I(x) = \int \frac{6}{\sin^2 x (1 - \cot x)^2} dx$ by simplifying the integrand using trigonometric identities, performing a u-substitution, and then using the given initial condition $I(0) = 3$ to find the constant of integration. Finally, we calculated $I\left(\frac{\pi}{12}\right)$ using the value $\tan\left(\frac{\pi}{12}\right) = 2 - \sqrt{3}$.

Final Answer

The final answer is \boxed{3\sqrt{3}}, which corresponds to option (D).