(cosx)^8 =[( cosx)^2]^4 = (1/16) (1 + cos2x)^4 = (1/16) [ (1 + cos2x)^2 ]^2= (1/16) [ 1 + 2 cos2x +( cos2x)^2 ]^2 = (1/4) [ 3/2 + 2 cos2x + (1/2)cos4x ]^2= (1/16) [ 9/4 + 4 (cos2x)^2 + (1/4) ( cos4x)^2 + 6 cos2x +(3/2) cos4x + 2 cos2x cos4x ]= (1/16)[9/4 + 2 + 2 cos4x + 1/8 + (1/8) cos8x + 6cos2x +(3/2) cos4x + cos6x + cos2x ]= 35/128 + (7/16)cos2x + (7/32)cos4x + (1/16)cos6x + (1/128)cos8x原式= 35 x /128 + (7/32)sin2x +(7/128)sin4x + (1/96)sin6x + (1/1024)sin8x + C
方法就是用降次公式化简,过程有点复杂,耐心一点化就可以了
另外课本上我记得应该有公式∫(cosx)^ndx谢谢大神