cos2x=1-2sin^2x, so sin^2x=(1-cos2x)/2.
sin^4x=(sin^2x)^2=(1-cos2x)^2/4=(1-2cos2x+cos^2(2x))/4.
cos4x=2cos^2(2x)-1, so cos^2(2x)=(1+cos4x)/2.
sin^4x=(1-2cos2x+(1+cos4x)/2)/4=(2-4cos2x+1+cos4x)/8=(3-4cos2x+cos4x)/8.
(integral wrt x of)sin^4x.dx=3x/8-sin2x/4+sin4x/32+k, where k is a constant.