Question

$int (cos2x*cos4x - sin2x*cos4x)dx$

интерграл от 0 до $frac{pi}{2}$ 

 

с решением плиз

Answer 1

cos2x*cos4x=1/2[cos6x+cos2x]

sin2xcos4x=1/2[sin6x-sin2x]

1/2[1/6sin6x+1/2sin2x]-1/2[-1/6cos6x+1/2cos2x]=

=1/12sin6x+1/4sin2x+1/12cos6x-1/4cos2x

sinПn=0

cos0=1

1/12cos3П-1/4cosП-1/12+1/4=-1/12+1/4-1/12+1/4=1/2-1/6=1/3