Question

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 ________.

Answer 1

Correct Answer: 16

Let \[ I=\int_{\pi/6}^{\pi/4}\left(\cot\left(x-\frac{\pi}{3}\right)\cot\left(x+\frac{\pi}{3}\right)+1\right)\,dx \] Step 1: Simplify integrand Using identity \[ \cot A\cot B+1=\frac{\cos(A-B)}{\sin A\sin B} \] Take \[ A=x-\frac{\pi}{3},\quad B=x+\frac{\pi}{3} \] Then \[ A-B=-\frac{2\pi}{3} \] \[ \cos(A-B)=\cos\left(\frac{2\pi}{3}\right)=-\frac12 \] Also, \[ \sin A\sin B=\sin^2x-\sin^2\frac{\pi}{3} \] \[ =\sin^2x-\frac34 \] Hence \[ I=\int_{\pi/6}^{\pi/4}\frac{-1/2}{\sin^2x-3/4}\,dx \] \[ =\int_{\pi/6}^{\pi/4}\frac{-2}{4\sin^2x-3}\,dx \] Step 2: Put \(t=\tan x\) Multiply numerator and denominator by \(\sec^2x\): \[ I=\int_{\pi/6}^{\pi/4}\frac{-2\sec^2x}{\tan^2x-3}\,dx \] Let \[ t=\tan x,\quad dt=\sec^2x\,dx \] Limits: \[ x=\frac{\pi}{6}\Rightarrow t=\frac{1}{\sqrt3} \] \[ x=\frac{\pi}{4}\Rightarrow t=1 \] Thus \[ I=2\int_{1/\sqrt3}^{1}\frac{dt}{3-t^2} \] \[ =\frac{1}{\sqrt3}\left[\ln\left|\frac{\sqrt3+t}{\sqrt3-t}\right|\right]_{1/\sqrt3}^{1} \] \[ I=-\frac{2}{\sqrt3}\ln(\sqrt3-1) \] Comparing with \[ I=\alpha\ln(\sqrt3-1) \] \[ \alpha=-\frac{2}{\sqrt3} \] \[ 9\alpha^2=9\cdot\frac{4}{3}=12 \] \[ \boxed{12} \]