Correct Answer – `-(pi)/(2)`

माना `” “I=int_(0)^(pi//2)(2log sin x-log sin 2x)dx`
`=int_(0)^(pi//2)(log sin^(2)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(tanx)-log 2dx`
`=int_(0)^(pi//2)log(tanx)dx-int_(0)^(pi//2)log2dx`
`rArr” “I=I_(1)-log2[x]_(0)^((pi)/(2))=I_(1)-((pi)/(2)-0)log 2” …(1)”`
जहाँ,`” “I_(1)=int_(0)^(pi//2)log(tan x)dx” …(2)”`
`rArr” “I_(1)=int_(0)^(pi//2)log[tan((pi)/(2)-x)]dx`
`[because int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx]`
`rArr” “I_(1)=int_(0)^(pi//2)log(cotx)dx” …(3)”`
समीकरण (2 ) और (3 ) को जोड़ने पर ,
`2I_(1)=int_(0)^(pi//2){(log(tanx)+log(cotx)}dx`
`int_(0)^(pi//2)log(tanxcotx)dx`
`=int_(0)^(pi//2)log 1dx=0`
`rArr” “I_(1)=0`
`I_(1)` का मान समीकरण (1 ) में रखने पर,
`I=0-(pi)/(2)log 2=-(pi)/(2)log 2`