∫(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 解题的思路是将高次幂转换为低次幂!
追问是x的四次方,看清楚
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