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

int_(0)^(pi//2)log(cosx)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) .

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

int _(0)^(pi/2) log(cotx)dx

If int_(0)^((pi)/2)log(cosx)dx=-(pi)/2log2 , then int_(0)^((pi)/2)log(cosecx)dx=

If int_(0)^((pi)/2)log(cosx)dx=-(pi)/2log2 , then int_(0)^((pi)/2)log(cosecx)dx=

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

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

Prove that int_(0)^(pi//2)log (sinx)dx=int_(0)^(pi//2) log (cosx)dx=-(pi)/(2) log 2 .

Prove that int_(0)^(pi//2)log (sinx)dx=int_(0)^(pi//2) log (cosx)dx=-(pi)/(2) log 2 .