Evaluate : `int_0^(pi/2)(2logsinx-logsin2x)\ dx`

Explanation

`"Let "I= int_(0)^(pi//2) (log sin x-log sin 2x )dx`
`=int_(0)^(pi//2)log ((sin^(2)x)/(sin2x))dx`
`=int_(0)^(pi//2)log ((sin^(2)x)/(2sin x cos x))dx`
`=int_(0)^(pi//2)log ((tanx)/(2))dx`
`=int_(0)^(pi//2)log (tan x)-log 2dx`
`=int_(0)^(pi//2)log (tan x)dx -int_(0)^(pi//2)log 2 dx`
`rArr I=I_(1)-log 2[x]_(0)^(pi/2)=I_(1)-((pi)/(2)-0)log 2.......(1)`
`" where " I_(1)=int_(0)^(pi//2)log(tan x)dx" ".......(2)`
`rArr I_(1)=int_(0)^(pi//2) log [tan.((pi)/(2)-x)]dx`
`[ :' int_(0)^(a) f(x) dx= int_(0)^(a) f(a-x)dx]`
`rArr I_(1)=int_(0)^(pi//2)log (cot x) dx" ".....(3)`
Adding equations (2) and (3)
`2I_(1) =int_(0)^(pi//2) {(log (tan x)+log (cot x)}dx`
` =int_(0)^(pi//2) log (tan x cot x) dx`
`=int_(0)^(pi//2) log 1 dx=0`
`rArr I_(1)=0`
Put the value of `I_(1)` in equation (1)
`I=0-(pi)/(2) log 2=-(pi)/(2) log 2`
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