सिद्ध कीजिए कि
`int_(0)^(pi//2)log(sinx)dx=int_(0)^(pi//2)log(cosx)dx=(-pi)/(2)log2`

Explanation

माना `I=int_(0)^(pi//2)log(sinx)dx" …(1)"`
`rArr" "I=int_(0)^(pi//2)log{sin((pi)/(2)-x)}dx" "` प्रगुण (4 ) से
`rArr" "I=int_(0)^(pi//2)log cos x dx" ..(2)"`
समीकरण (1 ) और (2 ) को जोड़ने पर
`2I=int_(0)^(pi//2)[log(sinx)+log(cosx)]dx`
`=int_(0)^(pi//2)log(sinx.cosx)dx`
`=int_(0)^(pi//2)log((sinx)/(2))dx`
`=int_(0)^(pi//2)log(sin2x)dx-int_(0)^(pi//2)log2 dx`
`=int_(0)^(pi)log(sint)(dt)/(2)-log2int_(0)^(pi//2)1.dx`
माना `2x=t`
`rArr 2dx=dt`
`rArr dx=(dt)/(2)`
`=(1)/(2)int_(0)^(pi)log(sint)dt-log2[x]_(0)^(pi//2)x=0" पर "t=0`
`=(1)/(2)xxint_(0)^(pi//2)log(sint)dt-(pi)/(2)log2" "x=(pi)/(2)" पर " t=pi`
`=int_(0)^(pi//2)log(sinx)dx-(pi)/(2)log2`
`rArr" "2I=I-(pi)/(2)log2`
`rArr" "I=(-pi)/(2)log2`
`therefore int_(0)^(pi//2)log(sinx)dx`
`=int_(0)^(pi//2)log(cosx)dx=-(pi)/(2)log2" "`