If $$\int_{\pi/6}^{\pi/4} \left(\cot\left(x - \frac{\pi}{3}\right)\cot\left(x + \frac{\pi}{3}\right) + 1\right) dx = \alpha \log_e(\sqrt{3} - 1)$$, then $$9\alpha^2$$ is equal to __________.

This problem looks quite complex because of the $$\cot \cdot \cot$$ product, but we can simplify the integrand significantly using a standard trigonometric identity.

1. Simplify the Integrand

Recall the identity for $$\cot(A - B)$$:

$$\cot(A - B) = \frac{\cot A \cot B + 1}{\cot B - \cot A}$$

Rearranging this, we get:

$$\cot A \cot B + 1 = \cot(A - B)(\cot B - \cot A)$$

Let $$A = x + \frac{\pi}{3}$$ and $$B = x - \frac{\pi}{3}$$.

Then $$A - B = (x + \frac{\pi}{3}) - (x - \frac{\pi}{3}) = \frac{2\pi}{3}$$.

Substitute these into our expression:

$$\cot(x + \frac{\pi}{3}) \cot(x - \frac{\pi}{3}) + 1 = \cot(\frac{2\pi}{3}) \left[ \cot(x - \frac{\pi}{3}) - \cot(x + \frac{\pi}{3}) \right]$$

Since $$\cot(\frac{2\pi}{3}) = -\frac{1}{\sqrt{3}}$$, the integral $$I$$ becomes:

$$I = -\frac{1}{\sqrt{3}} \int_{\pi/6}^{\pi/4} \left[ \cot(x - \frac{\pi}{3}) - \cot(x + \frac{\pi}{3}) \right] dx$$

2. Integrate

The integral of $$\cot(u)$$ is $$\ln|\sin u|$$.

$$I = -\frac{1}{\sqrt{3}} \left[ \ln|\sin(x - \frac{\pi}{3})| - \ln|\sin(x + \frac{\pi}{3})| \right]_{\pi/6}^{\pi/4}$$

Using log properties:

$$I = -\frac{1}{\sqrt{3}} \left[ \ln \left| \frac{\sin(x - \frac{\pi}{3})}{\sin(x + \frac{\pi}{3})} \right| \right]_{\pi/6}^{\pi/4}$$

3. Evaluate the Limits

• At $$x = \frac{\pi}{4}$$:
• At $$x = \frac{\pi}{6}$$:

$$\frac{\sin(\frac{\pi}{4} - \frac{\pi}{3})}{\sin(\frac{\pi}{4} + \frac{\pi}{3})} = \frac{\sin(-15^\circ)}{\sin(105^\circ)} = \frac{-\sin(15^\circ)}{\cos(15^\circ)} = -\tan(15^\circ) = -(2 - \sqrt{3}) = \sqrt{3} - 2$$

$$\frac{\sin(\frac{\pi}{6} - \frac{\pi}{3})}{\sin(\frac{\pi}{6} + \frac{\pi}{3})} = \frac{\sin(-\pi/6)}{\sin(\pi/2)} = \frac{-1/2}{1} = -1/2$$

Substitute back (taking absolute values):

$$I = -\frac{1}{\sqrt{3}} \left[ \ln(2 - \sqrt{3}) - \ln(1/2) \right] = -\frac{1}{\sqrt{3}} \ln\left( \frac{2 - \sqrt{3}}{1/2} \right)$$

$$I = -\frac{1}{\sqrt{3}} \ln(4 - 2\sqrt{3})$$

Notice that $$4 - 2\sqrt{3} = (\sqrt{3} - 1)^2$$.

$$I = -\frac{1}{\sqrt{3}} \ln(\sqrt{3} - 1)^2 = -\frac{2}{\sqrt{3}} \ln(\sqrt{3} - 1)$$

4. Final Calculation

The given form is $$\alpha \ln(\sqrt{3} - 1)$$.

By comparison:

$$\alpha = -\frac{2}{\sqrt{3}}$$

We need to find $$9\alpha^2$$:

$$9\alpha^2 = 9 \left( -\frac{2}{\sqrt{3}} \right)^2 = 9 \left( \frac{4}{3} \right)$$

$$9\alpha^2 = 3 \times 4 = \mathbf{12}$$