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JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Indefinite Integrals

If

\int (sin x)^{\frac{- 11}{2}} (cos x)^{\frac{- 5}{2}} dx = - \frac{p _{1}}{q _{1}} (cot x)^{\frac{9}{2}} - \frac{p _{2}}{q _{2}} (cot x)^{\frac{5}{2}} - \frac{p _{3}}{q _{3}} (cot x)^{\frac{1}{2}} + \frac{p _{4}}{q _{4}} (cot x)^{\frac{- 3}{2}} + C

, where

p_{i}

and

q_{i}

are positive integers with

gcd (p_{i}, q_{i}) = 1

for

i = 1, 2, 3, 4

and C is the constant of integration, then

\frac{15 p _{1} p _{2} p _{3} p _{4}}{q _{1} q _{2} q _{3} q _{4}}

is equal to

Answer: 16

Solution

Write

(sin x)^{- 11/2} (cos x)^{- 5/2} = (tan x)^{- 11/2} sec^{8} x

and substitute

t = tan x

,

d t = sec^{2} x d x

, with

sec^{6} x = (1 + t^{2})^{3}

:

\int t^{- 11/2} (1 + t^{2})^{3} d t = \int (t^{- 11/2} + 3 t^{- 7/2} + 3 t^{- 3/2} + t^{1/2}) d t = - \frac{2}{9} t^{- 9/2} - \frac{6}{5} t^{- 5/2} - 6 t^{- 1/2} + \frac{2}{3} t^{3/2} + C .

Since

t = tan x

gives

t^{- 1} = cot x

, this equals

- \frac{2}{9} (cot x)^{9/2} - \frac{6}{5} (cot x)^{5/2} - 6 (cot x)^{1/2} + \frac{2}{3} (cot x)^{- 3/2} + C

. Thus

(p_{i}, q_{i}) = (2, 9), (6, 5), (6, 1), (2, 3)

and

\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.

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Integer answer

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