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Indefinite Integrals question

2019 · 9 Jan · Shift 1 · Q39
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  5. /2019 · 9 Jan · Shift 1 · Q39

Indefinite Integrals question

2019 · 9 Jan · Shift 1 · Q39

JEE MainMathematicsIndefinite IntegralsMCQ+4 / −1
For x2 eee n π\piπ+ 1, n ∈\in∈ N (the set of natural numbers), the integral ∫x2sin⁡(x2−1)−sin⁡2(x2−1)2sin⁡(x2−1)+sin⁡2(x2−1)dx\int {x\sqrt {{{2\sin ({x^2} - 1) - \sin 2({x^2} - 1)} \over {2\sin ({x^2} - 1) + \sin 2({x^2} - 1)}}} dx}∫x2sin(x2−1)+sin2(x2−1)2sin(x2−1)−sin2(x2−1)​​dx is equal to : (where c is a constant of integration)
  1. A
    log⁡e∣12sec⁡2(x2−1)∣+c{\log _e}\left| {{1 \over 2}{{\sec }^2}\left( {{x^2} - 1} \right)} \right| + cloge​​21​sec2(x2−1)​+c
  2. B
    12log⁡e∣sec⁡(x2−1)∣+c{1 \over 2}{\log _e}\left| {\sec \left( {{x^2} - 1} \right)} \right| + c21​loge​​sec(x2−1)​+c
  3. C
    12log⁡e∣sec⁡2(x2−12)∣+c{1 \over 2}{\log _e}\left| {{{\sec }^2}\left( {{{{x^2} - 1} \over 2}} \right)} \right| + c21​loge​​sec2(2x2−1​)​+c
  4. D
    log⁡e∣sec⁡(x2−12)∣+c{\log _e}\left| {\sec \left( {{{{x^2} - 1} \over 2}} \right)} \right| + cloge​​sec(2x2−1​)​+c
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