JEE Main 2026 — 5 April, Evening Shift — Question 48

$\frac{d y}{d x}+y \tan x=\frac{\sec ^{3} x}{\sqrt{\tan x}}$ $\mathrm{IF}=\mathrm{e}^{\int \tan \mathrm{x} \mathrm{dx}}=\sec \mathrm{x}$ $y \sec x=\int \frac{\sec ^{4} x}{\sqrt{\tan x}} d x$ $\mathrm{y} \sec \mathrm{x}=\int \frac{\left(1+\tan ^{2} \mathrm{x}\right)}{\sqrt{\tan \mathrm{x}}} \sec ^{2} \mathrm{xdx}$ $\tan \mathrm{x}=\mathrm{t}$ $\mathrm{y} \sec \mathrm{x}=2 \sqrt{\tan \mathrm{x}}+\frac{2}{5}(\tan \mathrm{x})^{5 / 2}+\mathrm{c}$ y $\left(\frac{\pi}{4}\right)=\frac{6 \sqrt{2}}{5}$ $\frac{6 \sqrt{2}}{5} \times \sqrt{2}=2+\frac{2}{5}+\mathrm{c} \Rightarrow \frac{12}{5}=\frac{12}{5}+\mathrm{c}$ $\Rightarrow \mathrm{c}=0$ $\mathrm{y}(\pi / 3)=\frac{8}{5} \cdot 3^{1 / 4} \Rightarrow \alpha=2 \times 3^{\frac{1}{4}}$ $\therefore \alpha^{4}=48$