int_(pi/4)^( pi/2)cos2x log(3sin x)dx

Evaluate: int_(pi/4)^( pi/2)cos2x log sin xdx

int_(pi//4)^(pi//2) cos 2 x log sin x " " dx=

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

[int_(-pi/4)^( pi/4)log(cos x+sin x)dx=(1)/(2)],[[" 1) "pi log2," 2) "-pi log2," 3) "-(pi)/(4)log2," 4) "-(pi)/(2)log2]]

Evaluate: int_(-pi/4)^( pi/4)log(sin x+cos x)dx

int_(0)^((pi)/(2))cos2x log sin xdx

Prove that int_(-pi/4)^( pi/4)log(sin x+cos x)backslash dx=-(pi)/(4)backslash log2

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

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

int _(-pi //2) ^(pi//2) log ((2-sin x)/(2+sin x))dx=