JEE Main 2026 — 8 April, Evening Shift — Question 39

$\int_{\pi / 6}^{\pi / 4}\left(\left(\cot \left(\mathrm{x}-\frac{\pi}{3}\right) \cot \left(\mathrm{x}+\frac{\pi}{3}\right)+1\right) \mathrm{dx}\right.$ using $\cot (\mathrm{A}-\mathrm{B})=\frac{\cot \mathrm{B} \cot \mathrm{A}+1}{\cot \mathrm{~B}-\cot \mathrm{A}}$ $\Rightarrow \int_{\pi / 6}^{\pi / 4}-\frac{1}{\sqrt{3}}\left[\cot \left(\mathrm{x}-\frac{\pi}{3}\right)-\cot \left(\mathrm{x}+\frac{\pi}{3}\right)\right] \mathrm{dx}$ $\Rightarrow \frac{-1}{\sqrt{3}}\left[\left.\ln \left|\frac{\sin \left(x-\frac{\pi}{3}\right)}{\sin \left(x+\frac{\pi}{3}\right)}\right|\right|_{\frac{\pi}{6}} ^{\frac{\pi}{4}}\right.$

$$\begin{aligned} & =\frac{-1}{\sqrt{3}}\left[\ln \left|\left(\frac{\sin 15^{\circ}}{\left.\sin 105^{\circ}\right)}\right)\right|-\ln \left|\frac{\sin 30^{\circ}}{\sin 90^{\circ}}\right|\right] \\& =\frac{-1}{\sqrt{3}}\left[\ln \left(\tan 15^{\circ}\right)-\ln \left(\frac{1}{2}\right)\right. \\& =\frac{-1}{\sqrt{3}}[\ln (2-\sqrt{3})+\ln 2] \\& \Rightarrow \frac{-1}{\sqrt{3}} \ln (4-2 \sqrt{3})=\frac{-2}{\sqrt{3}} \ln (\sqrt{3}-1) \\& \therefore \alpha=\frac{-2}{\sqrt{3}} \therefore 9 \alpha^{2}=12 \end{aligned}$$