JEE Main 2022 · 29 July · evening · Single correct · +4 / −1

For I(x)=∫ (sec² x-2022)/(sin²⁰²² x)dx, if I(π/4)=2¹⁰¹¹, then

  1. (a) 3¹⁰¹⁰I(π/3)-I(π/6)=0
  2. (b) 3¹⁰¹⁰I(π/6)-I(π/3)=0
  3. (c) 3¹⁰¹¹I(π/3)-I(π/6)=0
  4. (d) 3¹⁰¹¹I(π/6)-I(π/3)=0
Answer and solution

Answer (a) 3¹⁰¹⁰I(π/3)-I(π/6)=0

Solution

Maths sir · 4 steps

Split the integral and do parts on the sec² x piece; the -2022 piece then cancels completely.

  1. Parts on one piece. Write I = ∫ sec² x sin⁻²⁰²² x dx - 2022∫ sin⁻²⁰²² x dx. In the first integral take sin⁻²⁰²² x as the first function and sec² x as the second, since ∫ sec² x dx = tan x.

    ∫ (sec² x)/(sin²⁰²² x) dx = (tan x)/(sin²⁰²² x) + 2022∫ tan x · (cos x)/(sin²⁰²³ x) dx = (tan x)/(sin²⁰²² x) + 2022∫ dx/(sin²⁰²² x)

  2. Cancel and fix C. The leftover integral cancels the -2022 piece, so I(x) = (tan x)/(sin²⁰²² x) + C. At x = π/4, sin²⁰²² x = (1/√2)²⁰²² = 2⁻¹⁰¹¹, so I(π/4) = 2¹⁰¹¹ + C = 2¹⁰¹¹, which gives C = 0.
  3. Evaluate both values. I(π/3) = √3 × 2²⁰²²/3¹⁰¹¹ and I(π/6) = 1/√3 × 2²⁰²² = 2²⁰²²/√3. Then 3¹⁰¹⁰ I(π/3) = (√3 × 2²⁰²²)/3 = 2²⁰²²/√3.
  4. So. 3¹⁰¹⁰ I(π/3) = I(π/6), so 3¹⁰¹⁰ I(π/3) - I(π/6) = 0. The answer is (a).

The idea to keep. When an integrand has the form f'(x)g(x) + f(x)g'(x) hidden inside it, do parts on just one piece. The other piece cancels and you are left with f(x)g(x) + C.

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

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Asked about 60 times in JEE Main since 2020. This question is from 2022.