`int_(0)^(pi/4)log sin2xdx`

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

Evaluate int_(0)^(pi//2)lnsin2xdx

Evaluate int_0^(pi/2) log sin xdx

Statement-1: int_(0)^(pi//2) x cot x dx=(pi)/(2)log2 Statement-2: int_(0)^(pi//2) log sin x dx=-(pi)/(2)log2

Evaluate int_(-pi/4)^(pi/4) sin^2xdx

The value int_(0)^(pi//2) sin^(2)xdx will be :

Prove that: int_(0)^(pi//2) log (sin x) dx =int_(0)^(pi//2) log (cos x) dx =(-pi)/(2) log 2

The value of int_(0)^(pi//2) (2log sin x-log sin 2x)dx , is

The value of int_0^(pi/4)sin2xdx is

int_(0)^( pi/4)sec^(2)xdx