int_(0)^((pi)/(2))log cos xdx=(pi)/(2)log(1)/(2)

If int_(0)^((pi)/(2))logcosxdx=(pi)/(2)log((1)/(2)) , then int_(0)^((pi)/(2))logsecdx=

int_(0)^((pi)/(2))log(sinx)dx=int_(0)^((pi)/(2))log(cosx)dx=(pi)/(2)log.(1)/(2)

int_(0)^((pi)/(2))log(sinx)dx=int_(0)^((pi)/(2))log(cosx)dx=(pi)/(2)log.(1)/(2)

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

Show that int_(0)^((pi)/(2))logsinxdx=(pi)/(2)log((1)/(2))=(-pi)/(2)log2

If int_(0)^(pi//2) log cos x dx =(pi)/(2)log ((1)/(2)), then int_(0)^(pi//2) log sec x dx =

If int_(0)^(pi//2) log cos x dx =(pi)/(2)log ((1)/(2)), then int_(0)^(pi//2) log sec x dx =

int_(0)^((pi)/(2))log sin xdx=int_(0)^((pi)/(2))log cos xdx=(1)/(2)(pi)log((1)/(2))

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) .

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