求教一道積分題
回答 (2)
2008-01-29 1:22 am
參考: ME
2008-01-29 1:00 am
Integrate sinx dx/[(cosx)^2+cosx-2]
=Integrate -dcos x/[(cosx)^2+cosx-2] since sinx dx = -dcos x
Let y=cosx
Integrate -dcos x/[(cosx)^2+cosx-2]
=Integrate -dy/(y^2+y-2)
=Integrate -dy/(y+2)(y-1)
=Integrate -1/(y+2)(y-1) dy
=Integrate [1/(y+2)-1/(y-1)] dy
=Integrate dy/(y+2) - Integrate dy/(y-1)
=ln(y+2)-ln(y-1)+c where c is a constant
=ln[(y+2)/(y-1)]+c
=ln[(cosx+2)/(cosx-1)]
=Integrate -dcos x/[(cosx)^2+cosx-2] since sinx dx = -dcos x
Let y=cosx
Integrate -dcos x/[(cosx)^2+cosx-2]
=Integrate -dy/(y^2+y-2)
=Integrate -dy/(y+2)(y-1)
=Integrate -1/(y+2)(y-1) dy
=Integrate [1/(y+2)-1/(y-1)] dy
=Integrate dy/(y+2) - Integrate dy/(y-1)
=ln(y+2)-ln(y-1)+c where c is a constant
=ln[(y+2)/(y-1)]+c
=ln[(cosx+2)/(cosx-1)]
收錄日期: 2021-04-23 12:42:31
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https://hk.answers.yahoo.com/question/index?qid=20080128000051KK01470