JEE Main 2026 — Mathematics Calculus

Given integral is $\displaystyle I = \displaystyle\int_{\pi/6}^{\pi/4}\left(\cot\left(x-\dfrac{\pi}{3}\right)\cot\left(x+\dfrac{\pi}{3}\right)+1\right)dx$I=∫π/6π/4​(cot(x−3π​)cot(x+3π​)+1)dx

Using the identity $\cot A \cot B + 1 = \dfrac{\cos(A-B)}{\sin A \sin B}$cotAcotB+1=sinAsinBcos(A−B)​, the integrand simplifies to:

$$f(x) = \dfrac{\cos\left(\left(x-\dfrac{\pi}{3}\right) - \left(x+\dfrac{\pi}{3}\right)\right)}{\sin\left(x-\dfrac{\pi}{3}\right)\sin\left(x+\dfrac{\pi}{3}\right)}$$f(x)=sin(x−3π​)sin(x+3π​)cos((x−3π​)−(x+3π​))​

$$f(x) = \dfrac{\cos\left(-\dfrac{2\pi}{3}\right)}{\sin^2 x - \sin^2\left(\dfrac{\pi}{3}\right)}$$f(x)=sin2x−sin2(3π​)cos(−32π​)​

$$f(x) = \dfrac{-1/2}{\sin^2 x - 3/4} = \dfrac{2}{3 - 4\sin^2 x}$$f(x)=sin2x−3/4−1/2​=3−4sin2x2​

Dividing the numerator and the denominator by $\cos^2 x$cos2x:

$$f(x) = \dfrac{2\sec^2 x}{3\sec^2 x - 4\tan^2 x} = \dfrac{2\sec^2 x}{3(1+\tan^2 x) - 4\tan^2 x} = \dfrac{2\sec^2 x}{3 - \tan^2 x}$$f(x)=3sec2x−4tan2x2sec2x​=3(1+tan2x)−4tan2x2sec2x​=3−tan2x2sec2x​

The integral becomes:

$$I = \displaystyle\int_{\pi/6}^{\pi/4} \dfrac{2\sec^2 x}{3 - \tan^2 x} dx$$I=∫π/6π/4​3−tan2x2sec2x​dx

Substituting $\tan x = t \Rightarrow \sec^2 x dx = dt$tanx=t⇒sec2xdx=dt.

When $x = \dfrac{\pi}{6}$x=6π​, $t = \dfrac{1}{\sqrt{3}}$t=3​1​. When $x = \dfrac{\pi}{4}$x=4π​, $t = 1$t=1.

$$I = \displaystyle\int_{1/\sqrt{3}}^{1} \dfrac{2}{3 - t^2} dt$$I=∫1/3​1​3−t22​dt

Using the standard integral $\displaystyle \displaystyle\int \dfrac{1}{a^2 - x^2} dx = \dfrac{1}{2a}\ln\left|\dfrac{a+x}{a-x}\right|$∫a2−x21​dx=2a1​ln​a−xa+x​​:

$$I = 2 \left[ \dfrac{1}{2\sqrt{3}} \ln\left|\dfrac{\sqrt{3}+t}{\sqrt{3}-t}\right| \right]_{1/\sqrt{3}}^{1}$$I=2[23​1​ln​3​−t3​+t​​]1/3​1​

$$I = \dfrac{1}{\sqrt{3}} \left( \ln\left(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\right) - \ln\left(\dfrac{\sqrt{3}+\dfrac{1}{\sqrt{3}}}{\sqrt{3}-\dfrac{1}{\sqrt{3}}}\right) \right)$$I=3​1​​ln(3​−13​+1​)−ln​3​−3​1​3​+3​1​​​​

$$I = \dfrac{1}{\sqrt{3}} \left( \ln\left(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\right) - \ln\left(\dfrac{3+1}{3-1}\right) \right)$$I=3​1​(ln(3​−13​+1​)−ln(3−13+1​))

$$I = \dfrac{1}{\sqrt{3}} \left( \ln\left(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\right) - \ln 2 \right) = \dfrac{1}{\sqrt{3}} \ln\left(\dfrac{\sqrt{3}+1}{2(\sqrt{3}-1)}\right)$$I=3​1​(ln(3​−13​+1​)−ln2)=3​1​ln(2(3​−1)3​+1​)

Rationalizing the term inside the logarithm by multiplying the numerator and the denominator by $(\sqrt{3}-1)$(3​−1):

$$\dfrac{\sqrt{3}+1}{2(\sqrt{3}-1)} = \dfrac{(\sqrt{3}+1)(\sqrt{3}-1)}{2(\sqrt{3}-1)^2} = \dfrac{3-1}{2(\sqrt{3}-1)^2} = \dfrac{1}{(\sqrt{3}-1)^2} = (\sqrt{3}-1)^{-2}$$2(3​−1)3​+1​=2(3​−1)2(3​+1)(3​−1)​=2(3​−1)23−1​=(3​−1)21​=(3​−1)−2

Thus,

$$I = \dfrac{1}{\sqrt{3}} \ln\left((\sqrt{3}-1)^{-2}\right) = -\dfrac{2}{\sqrt{3}} \ln(\sqrt{3}-1)$$I=3​1​ln((3​−1)−2)=−3​2​ln(3​−1)

Comparing this with $I = \alpha\log_e(\sqrt{3}-1)$I=αloge​(3​−1), we get:

$$\alpha = -\dfrac{2}{\sqrt{3}}$$α=−3​2​

Therefore, $9\alpha^2 = 9 \left(-\dfrac{2}{\sqrt{3}}\right)^2 = 9 \times \dfrac{4}{3} = 12$9α2=9(−3​2​)2=9×34​=12.

Answer: $12$12