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int0 ^ (pi/2) log(cosx) dx=...

`int_0 ^ (pi/2) log(cosx) dx=`

A

`pi/2 log(1/2)`

B

`1-pi/2 log(1/2)`

C

`1+pi/2 log (1/2)`

D

`pi/2 log 2`

Text Solution

Verified by Experts

The correct Answer is:
D

Given, `int_(0)^(pi/2) log cos x dx = (pi)/(2) log '1/2 "…"(i)`
Let `I = int_(0)^(pi/2) log sec x dx = int_0^(pi/2) log(1/(cosx)) dx`
`= int_(0)^(pi/2) log(cosx)^(-1) dx = -int_(0)^(pi/2) log (cos x) dx`
`= I int_(0)^(pi/2) log (cos x) dx [ :' log m^(-n) = - nlogm]`
`= - pi/2 log(1/2) ( :' log(1/a) = - log a)`
` = pi/2 log 2`
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