If Iₙ = ∫_(π/4)^(π/2) cot ⁿx dx, then :

Solution

Maths sir · 3 steps

Pair Iₙ with Iₙ₊₂: then 1 + cot² x = csc² x turns the sum into one easy integral.

1. Pair the terms. Add Iₙ and Iₙ₊₂ under one integral sign. The integrand becomes cotⁿ x (1 + cot² x) = cotⁿ x csc² x.
2. Substitute t. Put t = cot x, so dt = -csc² x dx. When x = π/4, t = 1, and when x = π/2, t = 0. The minus sign flips the limits back to 0 to 1.

   Iₙ + Iₙ₊₂ = ∫_(π/4)^(π/2) cotⁿ x csc² x dx = ∫₀¹ tⁿ dt = 1/(n+1)

   Option	The three terms	Holds?
   (a) reciprocals in A.P.	3, 4, 5	yes, common difference 1
   (b) sums in A.P.	1/3, 1/4, 1/5	no, these are in H.P.
   (c) reciprocals in G.P.	3, 4, 5	no, 4² ≠ 3 × 5
   (d) G.P. with middle squared	1/3, 1/16, 1/5	no, 1/256 ≠ 1/15

3. So. With n = 2, 3, 4 the sums are 1/3, 1/4, 1/5. Their reciprocals 3, 4, 5 are in A.P., so the answer is (a).

The idea to keep. For integrals of tanⁿ x or cotⁿ x, add Iₙ and Iₙ₊₂. The identity gives sec² or csc², which is exactly the derivative you need for the substitution.

Answer from the NTA final answer key. Solution by PrepWiser.