If \(\int_{\pi / 6}^{\pi / 4}\left(\cot \left(x-\frac{\pi}{3}\right) \cot \left(x+\frac{\pi}{3}\right)+1\right) d x=\alpha \log _{e}(\sqrt{3}-1)\), then \(9 \alpha^{2}\) is equal to ____.

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Definite Integrals

\int_{π /6}^{π /4} (cot (x - \frac{π}{3}) cot (x + \frac{π}{3}) + 1) d x = α lo g_{e} (3 - 1)

,
then

9 α^{2}

is equal to ____.

Answer: 12

Set

A = x - \frac{π}{3}

and

B = x + \frac{π}{3}

, so

B - A = \frac{2 π}{3}

. From

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

and

cot \frac{2 π}{3} = - \frac{1}{3}

cot (x - \frac{π}{3}) cot (x + \frac{π}{3}) + 1 = - \frac{1}{3} (cot (x - \frac{π}{3}) - cot (x + \frac{π}{3})) .

Integrating term by term,

\int_{π /6}^{π /4} (cot (x - \frac{π}{3}) cot (x + \frac{π}{3}) + 1) d x = - \frac{1}{3} [ln \frac{sin ( x - \frac{π}{3} )}{sin ( x + \frac{π}{3} )}]_{π /6}^{π /4} .

Evaluating the bracket at the two limits reduces it to

2 ln (3 - 1)

, so the integral equals

- \frac{2}{3} ln (3 - 1)

. Thus

α = - \frac{2}{3}

and

9 α^{2} = 9 \cdot \frac{4}{3} = 12.

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Integer answer