JEE Main 2025 — 22 January, Evening Shift — Question 11

Identify the standard form We look for a function $f(x)$ such that the integrand is in the form $e^x [f(x) + f'(x)]$. Let us consider:

$$f(x) = \frac{x \sin^{-1} x}{\sqrt{1-x^{2}}}$$

Differentiate $f(x)$ Using the quotient rule $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2}$: Let $u = x \sin^{-1} x$ and $v = \sqrt{1-x^2}$. $u' = \sin^{-1} x + \frac{x}{\sqrt{1-x^2}}$ $v' = \frac{-2x}{2\sqrt{1-x^2}} = \frac{-x}{\sqrt{1-x^2}}$

Substituting these into the derivative:

$$f'(x) = \frac{\sqrt{1-x^2} \left( \sin^{-1} x + \frac{x}{\sqrt{1-x^2}} \right) - (x \sin^{-1} x) \left( \frac{-x}{\sqrt{1-x^2}} \right)}{1-x^2}$$ $$f'(x) = \frac{(1-x^2)\sin^{-1} x + x\sqrt{1-x^2} + x^2 \sin^{-1} x}{(1-x^2)\sqrt{1-x^2}}$$ $$f'(x) = \frac{\sin^{-1} x - x^2 \sin^{-1} x + x\sqrt{1-x^2} + x^2 \sin^{-1} x}{(1-x^2)^{3/2}}$$ $$f'(x) = \frac{\sin^{-1} x + x\sqrt{1-x^2}}{(1-x^2)^{3/2}} = \frac{\sin^{-1} x}{(1-x^2)^{3/2}} + \frac{x}{1-x^2}$$

Determine $g(x)$ Since the integrand is exactly $e^x [f(x) + f'(x)]$, the integral evaluates to:

$$g(x) = e^x f(x) = \frac{e^x \cdot x \sin^{-1} x}{\sqrt{1-x^{2}}}$$

Evaluate $g\left(\frac{1}{2}\right)$ Substitute $x = \frac{1}{2}$: $\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$ $\sqrt{1 - \left(\frac{1}{2}\right)^2} = \frac{\sqrt{3}}{2}$

$$g\left(\frac{1}{2}\right) = e^{1/2} \left( \frac{\frac{1}{2} \cdot \frac{\pi}{6}}{\frac{\sqrt{3}}{2}} \right) = \sqrt{e} \left( \frac{\pi}{6\sqrt{3}} \right)$$

Rationalizing the denominator:

$$g\left(\frac{1}{2}\right) = \frac{\pi \sqrt{3e}}{18}$$

$\frac{\pi}{6} \sqrt{\frac{e}{3}}$