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Q.0π/2dxa2cos2x+b2sin2x\displaystyle \int_{0}^{\pi/2} \dfrac{dx}{a^2 \cos^2 x + b^2 \sin^2 x} equals

a
π2ab\dfrac{\pi}{2ab}
b
πa+b\dfrac{\pi}{a+b}
c
π2(a+b)\dfrac{\pi}{2(a+b)}
d
π2ab\dfrac{\pi}{2\sqrt{ab}}

Correct Answer: Option A

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

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