ज्ञात करे [Evaluate ] `int_(0)^(pi/2)(2 log sin x - log sin 2x ) dx`

Explanation

माना कि ` I = int_(0)^(pi//2) (2 log sin x - log sin 2x ) dx`
तो , ` = int _(0)^(pi//2) { 2 log sin x - log(2 sin x cos x)} dx `
` = int _(0)^(pi//2) {2 log sin x - log 2 - log sin x - log cos x }dx`
` = int _(0)^(pi//2) ( log sin x - log 2 - log cos x ) dx `
` = int _(0)^(pi//2) log sin x dx - int _(0)^(pi//2) log 2dx - int _(0)^(pi//2)log cos x dx `
` = int _(0)^(pi//2) log sin x dx - (log 2) int _(0)^(pi//2) 1. dx - int_(0)^(pi//2) dx - int _(0)^(pi//2) log cos (pi/2 - x) dx `
` [ :' int _(0)^(a) f(x) dx = int_(0)^(a) f(a- x) dx ] `
` = int _(0)^(pi//2) log sin xdx - (log 2) [ x] _(0)^(pi//2) - int _(0)^(pi//2) log sin x dx `
` = - ( log 2) (pi/2 - 0) `
अतः ` I = -pi/2 log 2`