我練習積分時做到一題
∫arcsin(x)dx
我看詳解上寫
設 sin(u)=x dx=cos(u)du
然後
原式=∫ucos(u)du
=usin(u)+cos(u)+C
我太笨這一步我完全不知道怎麼算
我想請問一下∫ucos(u)du 這個要怎麼算呢??
三角函數積分
2009-07-04 4:04 am
回答 (2)
2009-07-04 5:07 am
✔ 最佳答案
∫ucos(u)du令 p=u，dq=cos(u)du
dp=du，q=sin(u)
∫ucos(u)du
=∫pdq
=pq－∫qdp
=usin(u)－ ∫sin(u)du
=usin(u)+cos(u)+c
2009-07-04 4:17 am
是用份部積分的(integration by parts)
∫ ucos(u)du
= ∫ ud(sin(u)) [dsin(u) = cos(u)du]
= usin(u) - ∫ sin(u)du (份部積分)
= usin(u) + cos(u) + C
= xsin-1x + sqrt(1 - x2) + C (由於sin(u) = x)
∫ ucos(u)du
= ∫ ud(sin(u)) [dsin(u) = cos(u)du]
= usin(u) - ∫ sin(u)du (份部積分)
= usin(u) + cos(u) + C
= xsin-1x + sqrt(1 - x2) + C (由於sin(u) = x)
收錄日期: 2021-04-19 14:54:55
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090703000016KK09290