A cura di: Stefano Sannella
Calcolare
$int sin^4 dx$
$sin^4x=sin^2x*sin^2x=sin^2x*(1-cos^2x)=$
$=sin^2x-sin^2x*cos^2x=sin^2x-(sinx*cosx)^2=sin^2x-(1/2*2sinxcosx)^2$
= $sin^2x-1/4*sin^2(2x)=1/2*(1-cos2x)-1/4*1/2*(1-cos4x)=3/8-1/2*cos2x+1/8cos4x$
per cui
$intsin^4xdx=int(3/8-1/2*cos2x+1/8cos4x)dx=3/8x-1/4*sin2x+1/32*sin4x+K$
- Integrali