JEE Advanced 2025 Paper 2 · Q16 · Definite Integrals

Question

If $\alpha = \displaystyle \int_{1/2}^{2} \frac{	an^{-1}(x)}{2x^2 - 3x + 2}\, dx$, then the value of $\sqrt{7} \cdot 	an\left(\dfrac{2\alpha\sqrt{7}}{\pi}ight)$ is __________.

(Here, the inverse trigonometric function $	an^{-1} x$ assumes values in $(-\pi/2, \pi/2)$.)

SolveKar AI · Accepted Answer

Substitute $x 	o 1/x$ in $\alpha$. The substitution maps $[1/2, 2]$ to $[2, 1/2]$; use $	an^{-1}(1/x) = \pi/2 - 	an^{-1}(x)$ for $x > 0$, and $\frac{2}{x^2} - \frac{3}{x} + 2 = \frac{2x^2 - 3x + 2}{x^2}$. After simplification: $\alpha = \int_{1/2}^{2} \frac{\pi/2 - 	an^{-1} x}{2x^2 - 3x + 2}\, dx$. Adding to the original: $2\alpha = \frac{\pi}{2} \cdot \int_{1/2}^{2} \frac{dx}{2x^2 - 3x + 2}$. Complete the square: $2x^2 - 3x + 2 = 2(x - 3/4)^2 + 7/8$. Integral $J = \int \frac{dx}{2(x-3/4)^2 + 7/8} = \frac{2}{\sqrt{7}} \arctan\left(\frac{4x - 3}{\sqrt{7}}ight) + C$. Evaluate from $1/2$ to $2$: $\arctan(5/\sqrt{7}) + \arctan(1/\sqrt{7}) = \arctan(3\sqrt{7})$ (by $	an(A+B)$ formula since $(5/\sqrt{7} + 1/\sqrt{7}) / (1 - 5/7) = (6/\sqrt{7}) / (2/7) = 21/\sqrt{7} = 3\sqrt{7}$). So $2\alpha = \frac{\pi}{2} \cdot \frac{2}{\sqrt{7}} \arctan(3\sqrt{7}) = \frac{\pi}{\sqrt{7}} \arctan(3\sqrt{7})$. Therefore $\frac{2\alpha\sqrt{7}}{\pi} = \arctan(3\sqrt{7})$, and $\sqrt{7} \cdot 	an(\arctan(3\sqrt{7})) = \sqrt{7} \cdot 3\sqrt{7} = 21$.

JEE Advanced 2025 Paper 2 · Q16 · Definite Integrals

Solution