JEE Main 2026 — 24 January, Morning Shift — Question 11

Let $\ln t = x$, then $t = e^x$, $dt = e^x dx$. $f(t) = \int \frac{1 - \sin x}{1 - \cos x} e^x dx$. Simplify: $\frac{1 - \sin x}{1 - \cos x} = \frac{1 - 2\sin(x/2)\cos(x/2)}{2\sin^2(x/2)} = \frac{1}{2}\csc^2(x/2) - \cot(x/2)$. Thus $f(t) = \int \left( \frac{1}{2}\csc^2(x/2) - \cot(x/2) \right) e^x dx$. Let $g(x) = -\cot(x/2)$, then $g'(x) = \frac{1}{2}\csc^2(x/2)$. So $f(t) = \int (g'(x) + g(x)) e^x dx = g(x) e^x + C = -e^x \cot(x/2) + C$. Hence $f(t) = -t \cot\left( \frac{\ln t}{2} \right) + C$. Given $f(e^{\pi/2}) = -e^{\pi/2} \cot(\pi/4) + C = -e^{\pi/2}$, so $C = 0$. Then $f(e^{\pi/4}) = -e^{\pi/4} \cot(\pi/8) = -e^{\pi/4} (\sqrt{2}+1)$. Thus $\alpha = -(\sqrt{2}+1) = -1 - \sqrt{2}$.