求微积分 ∫sin^2(x)cos^4(x) dx

sin^2(x)cos^4(x)
=1/4*sin²2xcos²x
=1/4*(1-cos4x)/2*(1+cos2x)/2
=1/16*(1+cos2x-cos4x-cos2xcos4x)
=1/16*(1+cos2x-cos4x-cos2xcos4x)
=1/16*[1+cos2x-cos4x-1/2*(cos3x+cos6x)]
=1/16+1/16*cos2x-1/16*cos4x-1/32*cos3x-1/32*cos6x
所以原式=∫dx/16+1/16*∫cos2xdx-1/16*∫cos4xdx-1/32*∫cos3xdx-1/32*∫cos6xdx
=x/16+1/32*∫cos2xd2x-1/64*∫cos4xd4x-1/96*∫cos3xd3x-1/192*∫cos6xd6x
=x/16+1/32*sin2x-1/64*sin4x-1/96*sin3x-1/192*sin6x+C