JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Definite Integrals

$f(x)={\int }_0^{x}g(t){\log }_e\left(\frac{1-t}{1+t}\right)dt$, with $g$ odd.

The factor ${\log }_e\left(\frac{1-t}{1+t}\right)$ is itself odd (replacing $t\to -t$ gives ${\log }_e\left(\frac{1+t}{1-t}\right)=-{\log }_e\left(\frac{1-t}{1+t}\right)$), so $g(t){\log }_e\left(\frac{1-t}{1+t}\right)$ is a product of two odd functions, hence even. The integral of an even function from $0$ to $x$ is odd in $x$, so $f(-x)=-f(x)$.

Since $f$ is odd, ${\int }_{-\pi /2}^{\pi /2}f(x)dx=0$, so


Replacing $x$ by $-x$ in this integral (limits unchanged, $x^2\cos x$ even):


Adding the two expressions for $I$:


So $I={\int }_0^{\pi /2}{x}^2\cos xdx$. Integrating by parts twice:


Comparing with ${\left(\frac{\pi }{\alpha }\right)}^2-\alpha$, we get $\alpha =2$.