ज्ञात करे [Evaluate ] `int_(0)^(pi) log ( 1+cosx)dx `

Explanation

माना कि `I = int _(0)^(pi) log (1+ cos x ) dx " "` …(1)
तो , ` I = int _(0)^(pi) log [ 1 + cos (pi - x ) ] dx `
या , ` I = int _(0)^(pi) log (1- cos x) dx " "` …(2)
अब `(1)+(2) rArr 2I = int_(0)^(pi) log [( 1+ cos x) ( 1- cos x) ] dx`
` = int_(0)^(pi) log (1- cos^(2)x ) dx = int _(0)^(pi) log sin ^(2) x dx `
` = 2 int _(0)^(pi) log sin xdx = 2.2 int_(0)^(pi//2) log sin xdx`
` = 4 int_(0)^(pi//2) log sin xdx `
` [ :' " यदि " f(2a-x) = f(x) ," तो " int_(0)^(2a) f(x) dx = 2 int_(0)^(a) f(x) dx ] `
` :. I = 2 int_(0)^(pi//2) log sin xdx = 2 ( -pi/2 log 2)`
` = - pi log 2`