indefinite intergral
回答 (1)
2008-02-16 11:30 pm
✔ 最佳答案
∫sin4(x/2)cos2(x/2) dx= ∫sin2(x/2) (sin2(x/2)cos2(x/2)) dx
= ∫((1-cosx)/2 ) (sinx /2)2 dx
= (1/8)∫(1-cosx)sin2x dx
= (1/8) (∫sin2x dx - ∫sin2x cosx dx)
= (1/8) (∫ (1-cos2x)/2 dx - ∫sin2x d(sinx))
=(1/8) (x/2 - (sin2x /4) - (sin3x /3)) + C
= x/16 - (sin2x /32) - (sin3x /24) + C
收錄日期: 2021-04-13 15:09:21
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https://hk.answers.yahoo.com/question/index?qid=20080216000051KK02078