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Q.0π/2log(sinx)+log(cosx)dx\displaystyle \int_{0}^{\pi/2} \log(\sin x) + \log(\cos x) \, dx equals

a
πlog2\pi \log 2
b
πlog2-\pi \log 2
c
00
d
π/2\pi/2

Correct Answer: Option B

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

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