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JEE MAIN 2026 · Integral Calculus

JEE Main 2026 Session 2 Question Paper: 8 Apr, Shift 2 (B.E./B.Tech), question 22

Updated
Numerical answer+4 / -1Answer matches NTA’s final key

If ∫π/6π/4(cot⁡(x−π3)cot⁡(x+π3)+1)dx=αlog⁡e(3−1)\displaystyle\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\left(\sqrt3 - 1\right), then 9α29\alpha^2 is equal to ______.

Your answer

Solution

Solution

Rewrite the integrand. With A=x+π3A = x + \frac{\pi}{3}, B=x−π3B = x - \frac{\pi}{3}, A−B=2π3A - B = \frac{2\pi}{3} and

cot⁡(A−B)=cot⁡Acot⁡B+1cot⁡B−cot⁡A  ⇒  cot⁡Acot⁡B+1=cot⁡2π3 (cot⁡B−cot⁡A)=−13(cot⁡B−cot⁡A).\cot(A - B) = \frac{\cot A\cot B + 1}{\cot B - \cot A} \;\Rightarrow\; \cot A\cot B + 1 = \cot\tfrac{2\pi}{3}\,(\cot B - \cot A) = -\tfrac{1}{\sqrt3}(\cot B - \cot A).

Integrate.

I=−13[ln⁡∣sin⁡(x−π3)∣−ln⁡∣sin⁡(x+π3)∣]π/6π/4.I = -\frac{1}{\sqrt3}\Big[\ln\left|\sin\left(x - \tfrac{\pi}{3}\right)\right| - \ln\left|\sin\left(x + \tfrac{\pi}{3}\right)\right|\Big]_{\pi/6}^{\pi/4}.

  • At x=π4x = \frac{\pi}{4}: sin⁡(−π12)\sin\left(-\frac{\pi}{12}\right) and sin⁡7π12\sin\frac{7\pi}{12}, ratio of sizes tan⁡π12=2−3\tan\frac{\pi}{12} = 2 - \sqrt3.
  • At x=π6x = \frac{\pi}{6}: sin⁡(−π6)\sin\left(-\frac{\pi}{6}\right) and sin⁡π2\sin\frac{\pi}{2}, ratio 12\frac12.

I=−13ln⁡2−31/2=−13ln⁡(2(2−3))=−13ln⁡((3−1)2)=−23ln⁡(3−1).I = -\frac{1}{\sqrt3}\ln\frac{2 - \sqrt3}{1/2} = -\frac{1}{\sqrt3}\ln\big(2(2 - \sqrt3)\big) = -\frac{1}{\sqrt3}\ln\big((\sqrt3 - 1)^2\big) = -\frac{2}{\sqrt3}\ln(\sqrt3 - 1).

So α=−23\alpha = -\frac{2}{\sqrt3} and 9α2=9⋅43=129\alpha^2 = 9 \cdot \frac43 = 12.

Answer: 12.

Answer checked against the official answer key. The solution is our own working.

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