求不定积分~如题，见附件~
如题，见附件~
解：原式=∫[(1+cos2x)/2]²dx =(1/4)∫(1+cos2x)²dx =(1/4)∫(1+2cos2x+cos²2x)dx =(1/4)∫[1+2cos2x+(1+cos4x)/2]dx =(1/4)∫[(3/2)+2cos2x+(1/2)cos4x]dx =(1/4)[(3/2)x+sin2x+(1/8)sin4x]+C =(3/8)x+(1/4)sin2x+(1/32)sin4x+C