JEE Main 2026 — 22 January, Morning Shift — Question 22

$\quad \int(\tan \mathrm{x})^{-1 / 2} \cdot \sec ^{8} \mathrm{x} \mathrm{dx}$

$=\int(\tan \mathrm{x})^{-11 / 2}\left(1+\tan ^{2} \mathrm{x}\right) 3 \sec ^{2} \mathrm{xdx}$

Put $\tan \mathrm{x}=\mathrm{t}$

$\Rightarrow \int \mathrm{t}^{-11 / 2}\left(1+\mathrm{t}^{2}\right)^{3} \mathrm{dx}=\int \mathrm{t}^{-11 / 2}\left(1+\mathrm{t}^{6}+3 \mathrm{t}^{2}+3\right) \mathrm{dt}$

$=\int\left(\mathrm{t}^{-11 / 2}+\mathrm{t}^{1 / 2}+3 . \mathrm{t}^{-7 / 2}+3 . \mathrm{t}^{-3 / 2}\right) \mathrm{dt}$ $=-\frac{2}{9}(\cot \mathrm{x})^{9 / 2}-\frac{6}{5}(\cot )^{5 / 2}-6(\cot \mathrm{x})^{1 / 2}+\frac{2}{3}(\cot \mathrm{x})^{-3 / 2}+\mathrm{C}$

$\Rightarrow \mathrm{p}_{1}=2, \mathrm{p}_{2}=6, \mathrm{p}_{3}=6, \mathrm{p}_{4}=2$

$\mathrm{q}_{1}=9, \mathrm{q}_{2}=5, \mathrm{q}_{3}=1, \mathrm{q}_{4}=3$

$\frac{15 p_{1} p_{2} p_{3} p_{4}}{q_{1} q_{2} q_{3} q_{4}}=\frac{15 \cdot 2 \cdot 6 \cdot 6 \cdot 2}{9 \cdot 5 \cdot 1 \cdot 3}=16$