सिद्ध कीजिये -
`int_(0)^(pi//2)log(sinx)dx=int_(0)^(pi//2)log(cosx)dx=(pi)/(2)log((1)/(2))`

Explanation

माना कि `int_(0)^(pi//2) log sin xdx " "` …(1)
तो , ` I = int_(0)^(pi//2) log sin (pi/2 - x) dx `
` = int_(0)^(pi//2) log cos xdx " "` …(2)
(1) और (2) को जोड़ने पर हमे मिलता है ,
` = int_(0)^(pi//2) log ((sin2x)/2) dx = int_(0)^(pi//2) (log sin 2x - log 2) dx `
` = int_(0)^(pi//2) log sin 2x dx - log 2 int_(0)^(pi//2) dx `
` = int _(0)^(pi//2) log sin 2x dx - pi/2 log 2 `
अब ` int _(0)^(pi//2) log sin 2x dx = 1/2 int_(0)^(pi) log sin z dz" "` [ z = 2x रखने पर]
` =1/2 2.int_(0)^(pi//2) log sin z dz`
` [ :' " यदि " f(2a-x) = f(x), " तो " int_(0)^(2a)f(x) dx = 2 int_(0)^(a) f(x) dx]`
` = int _(0)^(pi//2) log sin zdz = int _(0)^(pi//2) log sin xdx = I`
` :. ( 3 ) " से, " 2I = I - pi/2 log 2 " ":. I = pi/2 log 2 `
Note :
(1) ` int_(0)^(pi//2) log cos x dx = int _(0)^(pi//2) log sin x dx = -pi/2 log 2 `
(2) `int_(0)^(pi//2) log "cosec " dx = - int _(0)^(pi//2) log sin x dx = pi/2 log 2 `
(3) `int_(0)^(pi//2) log "cosec " dx = - int _(0)^(pi//2) log cos x dx = pi/2 log 2 `