int_(0)^((pi)/(2))sin xf(sin2x)dx=int_(0)^((pi)/(2))cos xf(sin2x)dx

Show : int_(0)^((pi)/(2))sinxf(sin2x)dx=int_(0)^((pi)/(2))cosxf(sin2x)dx

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

If A=int_((rho)/(2))^((pi)/(2))cos(sin x)dxB=int_(0)^((pi)/(2))sin(cos x)dx and C=int_(0)^((pi)/(2))cos(x)dx

int_(0)^((pi)/(2))(x sin x*cos x)dx

int_(0)^(pi//2)sin x.sin 2x dx=

int_(0)^((pi)/(2))(2log cos x-log sin2x)dx

int_(0)^( pi/2)|cos x-sin x|dx

int_(0)^((pi)/(2))(x)/(sin x+cos x)dx

int_(0)^( pi/2)(sin x)*dx

underset is If int_(0)^( pi)xf(sin x)dx=A int_(0)^((pi)/(2))f(sin x)dx, then A