Verified Solution • 2025–26 Session moderate

QJEEM19Mae936: JEE Main 2019 Mathematics - Class 12 JEEM19Mae936 Solution: For x 2 n + 1, n N (the set of natural numbers), the integral…

For x2 ≠\ne n π\pi + 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} is equal to :

(where c is a constant of integration)

Multiple Choice Options

  • A log⁡e∣12sec⁡2(x2−1)∣+c{\log _e}\left| {{1 \over 2}{{\sec }^2}\left( {{x^2} - 1} \right)} \right| + c
  • B 12log⁡e∣sec⁡(x2−1)∣+c{1 \over 2}{\log _e}\left| {\sec \left( {{x^2} - 1} \right)} \right| + c
  • C 12log⁡e∣sec⁡2(x2−12)∣+c{1 \over 2}{\log _e}\left| {{{\sec }^2}\left( {{{{x^2} - 1} \over 2}} \right)} \right| + c
  • D log⁡e∣sec⁡(x2−12)∣+c{\log _e}\left| {\sec \left( {{{{x^2} - 1} \over 2}} \right)} \right| + c

Step-by-Step Verified Solution

∫x2sin⁡(x2−)−sin⁡2(x2−1)2sin⁡(x2−1)+sin⁡2(x2−1)  dx\int {x\sqrt {{{2\sin \left( {{x^2} - } \right) - \sin 2\left( {{x^2} - 1} \right)} \over {2\sin \left( {{x^2} - 1} \right) + \sin 2\left( {{x^2} - 1} \right)}}} } \,\,dx

=∫x2sin⁡(x2−1)−2sin(x2−1)cos⁡(x2−1)2sin⁡(x2−1)+2sin⁡(x2−1)cos⁡(x2−1)  dx= \int {x\sqrt {{{2\sin \left( {{x^2} - 1} \right) - 2sin\left( {{x^2} - 1} \right)\cos \left( {{x^2} - 1} \right)} \over {2\sin \left( {{x^2} - 1} \right) + 2\sin \left( {{x^2} - 1} \right)\cos \left( {{x^2} - 1} \right)}}} } \,\,dx

=∫x1−cos⁡(x2−1)1+cos⁡(x2−1) dx= \int {x\sqrt {{{1 - \cos \left( {{x^2} - 1} \right)} \over {1 + \cos \left( {{x^2} - 1} \right)}}} } \,dx

=∫x2sin⁡2(x2−12)2cos⁡2(x2−12) dx= \int {x\sqrt {{{2{{\sin }^2}\left( {{{{x^2} - 1} \over 2}} \right)} \over {2{{\cos }^2}\left( {{{{x^2} - 1} \over 2}} \right)}}} } \,dx

=∫xtan⁡(x2−12)dx= \int {x\tan } \left( {{{{x^2} - 1} \over 2}} \right)dx

put    x2−12=t{{{x^2} - 1} \over 2} = t

⇒x2−1=2t\Rightarrow {x^2} - 1 = 2t

⇒2xdx=2dt\Rightarrow 2xdx = 2dt

⇒xdx=dt\Rightarrow xdx = dt

∴\therefore    ∫tan⁡t dt\int {\tan t\,dt}

=ln⁡∣sec⁡t∣+C= \ln \left| {\sec t} \right| + C

=ln⁡∣sec⁡(x2−12)∣+C= \ln \left| {\sec \left( {{{{x^2} - 1} \over 2}} \right)} \right| + C

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