$int sin^4 dx$

A cura di: Stefano Sannella

Calcolare

$int sin^4 dx$

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$sin^4x=sin^2x*sin^2x=sin^2x*(1-cos^2x)=$

$=sin^2x-sin^2x*cos^2x=sin^2x-(sinx*cosx)^2=sin^2x-(1/2*2sinxcosx)^2$

= $sin^2x-1/4*sin^2(2x)=1/2*(1-cos2x)-1/4*1/2*(1-cos4x)=3/8-1/2*cos2x+1/8cos4x$

per cui

$intsin^4xdx=int(3/8-1/2*cos2x+1/8cos4x)dx=3/8x-1/4*sin2x+1/32*sin4x+K$