積sin^4x dx
積sin 四次X dx
若果不用什麼reduction formula
可以怎樣做
無論有沒有out syll 也請教一下我
中四AMATHS 不定積分
2008-06-13 4:57 am
回答 (2)
2008-06-13 5:23 am
✔ 最佳答案
sin^4x = (sin^2x)(sin^2x) = (sin^2x)(1 - cos^2x) = sin^2x - sin^2xcos^2x .(1)since cos2x = 1 - 2sin^2x, therefore, sin^2x = (1 - cos2x)/2.
(2)sin^2xcos^2x = sin^2(2x)/4 = (1- cos4x)/8.
Adding (1) and (2) together, we get
sin^4x = 1/2 - cos2x/2 + 1/8 - cos4x/8.
Therefore, Ssin^4xdx = S[5/8 - cos2x/2 - cos4x/8]dx = 5x/8 -sin2x/4 - sin4x/32 + C.
2008-06-13 5:41 am
∫sin^4xdx
=∫(sin²x)²dx
=∫[(1-cos2x)/2]²dx
=∫[(1-2cos2x+cos²2x)/4]dx
=(1/4)x-sin2x+∫(cos²2x)/4dx
=(1/4)x-sin2x+∫(1+cos4x)/8dx
=(3/8)x-sin2x+(1/32)sin4x+C, where C is a constant.
=∫(sin²x)²dx
=∫[(1-cos2x)/2]²dx
=∫[(1-2cos2x+cos²2x)/4]dx
=(1/4)x-sin2x+∫(cos²2x)/4dx
=(1/4)x-sin2x+∫(1+cos4x)/8dx
=(3/8)x-sin2x+(1/32)sin4x+C, where C is a constant.
收錄日期: 2021-04-16 12:46:11
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080612000051KK03756