cosx的四次方的原函数是什么

∫(cosx)^4dx
=∫[(1+cos2x)/2]^2dx
=1/4∫[1+2cos2x+(cos2x)^2]dx
=1/4∫dx+1/4∫2cos2xdx+1/4∫(cos2x)^2dx
=x/4+C+1/4∫cos2xd(2x)+1/4∫[(1+cos4x)/2]dx
=x/4+(sin2x)/4+C+1/4∫1/2dx+1/4∫(cos4x)/2dx
=3x/8+(sin2x)/4+C+1/32∫4cos4xdx
=3x/8+(sin2x)/4+C+1/32∫cos4xd(4x)
=3x/8+(sin2x)/4+(sin4x)/32+C