`int_(0)^(pi//2)(2logcosx-logsin2x)\ dx=-(pi)/(2)log2`

Prove that int_(0)^(pi//2)(2logsinx-logsin2x)dx=(pi)/(2)(log2) .

int_(0)^((pi)/(2))(2log cos x-log sin2x)dx

int_(0)^(pi//2) (2logsin x - log sin 2x) dx=

int_(0)^(pi//2)log(cosx)dx=

Show that int_(0)^((pi)/(2))log(sin2x)dx=-(pi)/(2)(log2)

Using property of define integrals, prove that : int_(0)^(pi//2)logcos x dx=(-pi)/2log2

Statement-1: int_(0)^(pi//2) x cot x dx=(pi)/(2)log2 Statement-2: int_(0)^(pi//2) log sin x dx=-(pi)/(2)log2