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三角换元脱根号,换元x=cosu,
=∫ucos³u/sinudcosu
=-∫ucos³udu
=∫u(sin²u-1)dsinu
=1/3∫udsin³u-(usinu-∫sinudu)
=usin³u/3-1/3∫sin³udu-usinu-cosu
=usin³u/3-usinu-cosu-1/3∫(cos²u-1)dcosu
=usin³u/3-usinu-cosu-cos³u/9+cosu/3+C
=∫ucos³u/sinudcosu
=-∫ucos³udu
=∫u(sin²u-1)dsinu
=1/3∫udsin³u-(usinu-∫sinudu)
=usin³u/3-1/3∫sin³udu-usinu-cosu
=usin³u/3-usinu-cosu-1/3∫(cos²u-1)dcosu
=usin³u/3-usinu-cosu-cos³u/9+cosu/3+C
更多追问追答
追答
=∫1/sin³u(1-sin²u)dsinu
=∫1/sin³u+1/sinu(1-sin²u)dsinu
=∫1/sin³u+1/sinu+1/2(1-sinu)-1/2(1+sinu)dsinu
=-1/2sin²u+ln|sinu|-(ln|1-sinu|+ln|1+sinu|)/2+C
=-csc²u/2+ln|sinu|-ln|cosu|+C
=-csc²u/2+ln|tanu|+C
或者=∫csc³usecudu
=-∫tanucsc²udcotu
=-∫(cotu+1/cotu)dcotu
=-cot²u/2-ln|cotu|+C
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