`int_0^(pi/2)log(sin2x)dx`

int_0^(pi//2)log(tanx)dx

Let u=int_0^1("ln"(x+1))/(x^2+1)dx a n d v=int_0^(pi/2)ln(sin2x)dx ,t h e n (a) u=-pi/2ln2 (b) 4u+v=0 (c) u+4v=0 (d) u=pi/8ln2

Let u=int_0^1("ln"(x+1))/(x^2+1)dxa n dv=int_0^(pi/2)ln(sin2x)dx ,t h e n u=-pi/2ln2 (b) 4u+v=0 u+4v=0 (d) u=pi/8ln2

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 .

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

If I_(1)=int_(0)^(pi//2)log (sin x)dx and I_(2)=int_(0)^(pi//2)log (sin 2x)dx , then

Evaluate : int_0^(pi/2) log sin x dx .

int_0^(pi//2) log(tan x)dx =