Definite Integral of Product of Cotangent Functions

To evaluate the definite integral $I = \int_{\pi/6}^{\pi/4} \left( \cot\left(x - \frac{\pi}{3}\right) \cot\left(x + \frac{\pi}{3}\right) + 1 \right) dx$

we start by using the trigonometric identity for the cotangent of the difference of two angles: $\cot(B - A) = \frac{\cot A \cot B + 1}{\cot A - \cot B}$

Rearranging this identity gives: $\cot A \cot B + 1 = \cot(B - A) (\cot A - \cot B)$

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

Since $\cot\left(\frac{2\pi}{3}\right) = -\frac{1}{\sqrt{3}}$, the integrand simplifies to: $\cot\left(x - \frac{\pi}{3}\right) \cot\left(x + \frac{\pi}{3}\right) + 1 = -\frac{1}{\sqrt{3}} \left( \cot\left(x - \frac{\pi}{3}\right) - \cot\left(x + \frac{\pi}{3}\right) \right)$

Now, substituting this back into the integral: $I = -\frac{1}{\sqrt{3}} \int_{\pi/6}^{\pi/4} \left( \cot\left(x - \frac{\pi}{3}\right) - \cot\left(x + \frac{\pi}{3}\right) \right) dx$

Recall that $\int \cot(u) \, du = \log_e |\sin(u)|$. Integrating term-by-term yields: $I = -\frac{1}{\sqrt{3}} \left[ \log_e \left| \sin\left(x - \frac{\pi}{3}\right) \right| - \log_e \left| \sin\left(x + \frac{\pi}{3}\right) \right| \right]_{\pi/6}^{\pi/4}$ $I = -\frac{1}{\sqrt{3}} \left[ \log_e \left| \frac{\sin\left(x - \frac{\pi}{3}\right)}{\sin\left(x + \frac{\pi}{3}\right)} \right| \right]_{\pi/6}^{\pi/4}$

Next, we evaluate the limit values:

1. At the upper limit $x = \frac{\pi}{4}$: $\left| \frac{\sin\left(\frac{\pi}{4} - \frac{\pi}{3}\right)}{\sin\left(\frac{\pi}{4} + \frac{\pi}{3}\right)} \right| = \left| \frac{\sin\left(-\frac{\pi}{12}\right)}{\sin\left(\frac{7\pi}{12}\right)} \right| = \frac{\sin\left(\frac{\pi}{12}\right)}{\cos\left(\frac{\pi}{12}\right)} = \tan\left(\frac{\pi}{12}\right)$ Since $\tan\left(\frac{\pi}{12}\right) = \frac{\sqrt{3}-1}{\sqrt{3}+1} = \frac{(\sqrt{3}-1)^2}{2}$.

2. At the lower limit $x = \frac{\pi}{6}$: $\left| \frac{\sin\left(\frac{\pi}{6} - \frac{\pi}{3}\right)}{\sin\left(\frac{\pi}{6} + \frac{\pi}{3}\right)} \right| = \left| \frac{\sin\left(-\frac{\pi}{6}\right)}{\sin\left(\frac{\pi}{2}\right)} \right| = \frac{1/2}{1} = \frac{1}{2}$

Substituting these limits into the expression for $I$: $I = -\frac{1}{\sqrt{3}} \left( \log_e \left( \frac{(\sqrt{3}-1)^2}{2} \right) - \log_e \left( \frac{1}{2} \right) \right)$ $I = -\frac{1}{\sqrt{3}} \log_e \left( \frac{\frac{(\sqrt{3}-1)^2}{2}}{\frac{1}{2}} \right)$ $I = -\frac{1}{\sqrt{3}} \log_e \left( (\sqrt{3}-1)^2 \right)$ $I = -\frac{2}{\sqrt{3}} \log_e (\sqrt{3}-1)$

Comparing this result with $I = \alpha \log_e(\sqrt{3}-1)$, we find: $\alpha = -\frac{2}{\sqrt{3}}$

Finally, we calculate $9\alpha^2$: $9\alpha^2 = 9 \left(-\frac{2}{\sqrt{3}}\right)^2 = 9 \times \frac{4}{3} = 12$