int_(0)^( pi y)log(1+sin x)dx=pi log2

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) .

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

int_(0)^(pi) x log sinx dx

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

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

int_ (0) ^ (pi) x log sin xdx

int_(0)^(pi) x log sinx\ dx

int_(0)^(pi) x log sinx\ dx

int_(0)^(pi)log(1+cosx)dx=-pi(log2)