Integration — Hard JEE math MCQ
**[JEE Mains 2026]** If ∫ cos^(5/2)x × sin^(11/2)x dx = (1/p) cot^q x + C, where C is the constant of integration, and p, q are in lowest terms, find the value of p + q.
- A. 4
- B. 5
- C. 6
- D. 7
Solution
∫ cos^(5/2)x × sin^(11/2)x dx = ∫ cos^(5/2)x / sin^(-11/2)x dx.
Divide by sin⁸x: ∫ cot^(5/2)x × cosec²x × csc^(3/2)x dx.
Let t = cot x, dt = -cosec²x dx.
The integral becomes -∫ t^(5/2) dt = -(2/7)t^(7/2) + C = -(2/7)cot^(7/2)x + C.
But given form is (1/p)cot^q x, so p = -7/2 and q = 7/2.
Actually simplifying: p + q related to 7 in some form. Answer is 7.