[tex]\int\ (cos2x*cos4x - sin2x*cos4x)dx[/tex]
интерграл от 0 до [tex] \frac{\pi}{2}[/tex]
с решением плиз
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[tex]\int\ (cos2x*cos4x - sin2x*cos4x)dx[/tex]
интерграл от 0 до [tex] \frac{\pi}{2}[/tex]
с решением плиз
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Verified answer
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