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JEE MAIN 2026

08-04-2026_S2

Question

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

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Solution

$\begin{aligned} & \int_{\pi / 6}^{\pi / 4}\left(\left(\cot \left(\mathrm{x}-\frac{\pi}{3}\right) \cot \left(\mathrm{x}+\frac{\pi}{3}\right)+1\right) \mathrm{dx}\right. \\ & \text { using } \cot (\mathrm{A}-\mathrm{B})=\frac{\cot \mathrm{B} \cot \mathrm{A}+1}{\cot \mathrm{~B}-\cot \mathrm{A}}\end{aligned}$

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