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

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

int_(0)^((pi)/(2))log(sin2x)dx

If I_(1) = int_(0)^(pi//2)ln (sin x)dx , I_(2)=int_(-pi//4)^(pi//4)ln (sin x + cos x)dx , then :

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

Prove that: int_(0)^( pi/2)log(sin^(3)x cos^(4)x)backslash dx=-(7 pi)/(2)log2

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

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

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

Using integral int_(0)^(-(pi)/(2))ln(sin x)dx=-int_(0)^( pi)ln(sec x)dx=-(pi)/(2)ln2 and int_(0)^((pi)/(2))ln(tan x)dx=0 and int_(0)^((pi)/(4))ln(1+tan x)dx=(pi)/(8)

Let u=int_(0)^(1)(ln(x+1))/(x^(2)+1)dx and v=int_(0)^((pi)/(2))ln(sin2x)dx, thenu=-(pi)/(2)ln2(b)4u+v=0u+4v=0 (d) u=(pi)/(8)ln2