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Q.0π/2x2sin2x+cos2xdx\displaystyle \int_{0}^{\pi/2} \dfrac{x^2}{\sin^2 x + \cos^2 x} \, dx equals

a
π24\dfrac{\pi^2}{4}
b
π22\dfrac{\pi^2}{2}
c
π2\pi^2
d
π/2\pi/2

Correct Answer: Option A

The correct solution involves applying the fundamental concept to derive the final value step by step...

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