Integration
回答 (2)
2011-11-19 8:35 am
✔ 最佳答案
∫ tan(x)tan(2x)tan(3x) dx=∫ [2sin(3x)sin(x)sin(2x)]/[2cos(3x)cos(x)cos(2x)] dx
=∫ {[cos(2x)-cos(4x)]sin(2x)}/{[cos(4x)+cos(2x)]cos(2x)} dx
=(-1/2) ∫ (u-2u²+1)/[(2u²-1+u)u] du
=(1/2) ∫ [ 1/u + (2/3)/(u+1) -(4/3)/(2u-1) ] du
=(1/2)ln|cos(2x)| +(1/3)ln|cos(2x) +1 | - (1/3) ln| 2cos(2x)-1| + C
2011-11-19 8:56 pm
(1/3)ln|cos(2x) +1 | = (1/3)ln|2cos^2 x| = (2/3)ln|cos x| +1/3 ln(2) and (1/3)ln(2) can be absorbed into the constant term
收錄日期: 2021-04-16 13:45:31
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https://hk.answers.yahoo.com/question/index?qid=20111119000051KK00016