int_(0)^( pi)x log(sin x)dx=-(pi^(2))/(2)(log2)

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

int_(0)^( pi)cos2x*log(sin x)dx

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

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

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

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

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

Prove that, int_(0)^(pi)log(1+cos x)dx=-pi log2 , given int_(0)^((pi)/(2))log((sin x))dx=(pi)/(2)"log"(1)/(2) .

Prove that: int_(0)^(pi//2) log (sin x) dx =int_(0)^(pi//2) log (cos x) dx =(-pi)/(2) log 2