JEE Main 2026 Apr 8 Shift 2, Mathematics Q22: Integration by Parts

Using the identity $\cot (A-B)=\frac{\cot A\cot B+1}{\cot B-\cot A}$ with $A=x+\frac{\pi }{3}$ and $B=x-\frac{\pi }{3}$, then $A-B=\frac{2\pi }{3}$:

$\cot \frac{2\pi }{3}=-\frac{1}{\sqrt{3}}=\frac{\cot (x-\pi /3)\cot (x+\pi /3)+1}{\cot (x-\pi /3)-\cot (x+\pi /3)}$

Hence $\cot (x-\pi /3)\cot (x+\pi /3)+1=-\frac{1}{\sqrt{3}}[\cot (x-\pi /3)-\cot (x+\pi /3)]$.

$I=-\frac{1}{\sqrt{3}}{\int }_{\pi /6}^{\pi /4}[\cot (x-\pi /3)-\cot (x+\pi /3)]dx=-\frac{1}{\sqrt{3}}{\left[\ln \left|\frac{\sin (x-\pi /3)}{\sin (x+\pi /3)}\right|\right]}_{\pi /6}^{\pi /4}$

At $x=\pi /4$: ratio $=\frac{\sin (-\pi /12)}{\sin (5\pi /12)}=-\tan 15^{\circ }=-(2-\sqrt{3})$.

At $x=\pi /6$: ratio $=\frac{\sin (-\pi /6)}{\sin (\pi /2)}=-\frac{1}{2}$.

$I=-\frac{1}{\sqrt{3}}[\ln (2-\sqrt{3})-\ln \frac{1}{2}]=-\frac{1}{\sqrt{3}}\ln [2(2-\sqrt{3})]=-\frac{1}{\sqrt{3}}\ln (4-2\sqrt{3})$

Now $4-2\sqrt{3}=(\sqrt{3}-1{)}^2$, so $\ln (4-2\sqrt{3})=2\ln (\sqrt{3}-1)$.

$I=-\frac{2}{\sqrt{3}}\ln (\sqrt{3}-1)⟹\alpha =-\frac{2}{\sqrt{3}}$.

$9{\alpha }^2=9\cdot \frac{4}{3}=12$.