`int_0^(pi/6) cosx. cos2x dx=`

int_(0)^(pi/6) sin2x . cosx dx

int_(0)^(pi/6) sin2x . cosx dx

int_(0)^(pi/6) sin2x . cosx dx

int_(0)^(pi/6) sin2x . cosx dx

int_(0)^(pi//6) cos x cos 3x dx

int_(0)^(pi//6) cos x cos 3x dx

The value of the integral sqrt2 int_0^(pi/2) f(sin2x) sinx dx=A (sqrt2/9) int_0^(pi/4) f(cos2x)cosx dx then the value of A is

int_0^(pi/4) (cos x- sin x) dx + int_(pi/4)^((5pi)/4) (sinx-cosx) dx + int_(2pi)^(pi/4) (cosx- sinx) dx =

Evaluate int_0^(pi//2) (cosx)/ (cos^2x+ sin^2 x) dx

int_(0)^(2pi) x cos^(6) x dx=