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Q.0π/2sin2xsinx+cosxdx\displaystyle \int_{0}^{\pi/2} \dfrac{\sin^2 x}{\sin x + \cos x} \, dx equals

a
π412\dfrac{\pi}{4} - \dfrac{1}{2}
b
π4\dfrac{\pi}{4}
c
π21\dfrac{\pi}{2} - 1
d
π2\dfrac{\pi}{2}

Correct Answer: Option C

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

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