高数不定积分∫(cosx)^8dx求详解,是(cosx)^8而不是cos(x^8)

豆叮99 1年前 已收到1个回答 举报

ghost1122 春芽

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(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

1年前

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