原式=∫[(1+cos2t)/2]²dt
=1/4∫(1+2cos2t+cos²2t)dt
=1/4∫1dt+1/2∫cos2tdt+1/4∫cos²2tdt
=1/4t+1/4*sin2t+1/4∫(1+cos4t)/2dt
=1/4t+1/4*sin2t+1/32∫(1+cos4t)d4t
=1/4t+(sin2t)/4+t/8+(sin4t)/32+C
原式=∫[(1+cos2t)/2]²dt
=1/4∫(1+2cos2t+cos²2t)dt
=1/4∫1dt+1/2∫cos2tdt+1/4∫cos²2tdt
=1/4t+1/4*sin2t+1/4∫(1+cos4t)/2dt
=1/4t+1/4*sin2t+1/32∫(1+cos4t)d4t
=1/4t+(sin2t)/4+t/8+(sin4t)/32+C