Prove that: `int_(0)^(pi//2) log (sin x) dx =int_(0)^(pi//2) log (cos x) dx =(-pi)/(2) log 2`

Explanation

`"Let I"=int_(0)^(pi//2) log (sin x) dx " ".......(1)`
`I=int_(0)^(pi//2) log {sin((pi)/(2)-x)} dx`
`rArr I=int_(0)^(pi//2) log cos x dx" ".......(2)`
Adding eqs. (1) and(2)
`2I=int_(0)^(pi//2) [log (sin x)+ log (cos x)]dx`
`=int_(0)^(pi//2) log (sin x. cos x) dx`
`=int_(0)^(Pi//2) log ((sin 2x)/(2))dx`
`=int_(0)^(pi//2) log (sin 2x) dx-int_(0)^(pi//2) log 2dx " "Let 2x =t`
`=int_(0)^(pi//2) log (sin t) (dt)/(2) -log 2int_(0)^(pi//2) 1.dx rArr dx =(dt)/(2)`
` =1/2int_(0)^(pi//2) log (sin t) dt -log 2[x]_(0)^(pi//2) At x=0,t=0`
` =1/2 xx 2 int_(0)^(pi//2)log (sint) dt -(pi)/(2) log 2 " "At x=(pi)/(2) ,t=pi`
`=int_(0)^(pi//2) log (sin x) dx -(pi)/(2) log 2`
`rArr 2I= -(pi)/(2) log 2`
`rArr I= -(pi)/(2) log 2`
`:. int_(0)^(pi//2) log (sin x) dx =int_(0)^(pi//2) log (cos x) dx`
`=-(pi)/(2) log 2.`
Hence Proved.