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int(0)^((pi)/(2))log(sinx)dx=int(0)^((pi...

`int_(0)^((pi)/(2))log(sinx)dx=int_(0)^((pi)/(2))log(cosx)dx=(pi)/(2)log.(1)/(2)`

A

`-((pi)/2)log2`

B

`pi"log"1/2`

C

`-pi "log"1/2`

D

`(pi)/2log2`

Text Solution

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The correct Answer is:
A
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