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求不定积分sin根号下x
人气:434 ℃ 时间:2020-01-29 04:02:48
解答
设√x=t,x=t^2,dx=2tdt,
原式=∫sint*2tdt=2∫t*sintdt
=2∫td(-cost)
=-2tcost+2∫costdt
=-2tcost+2sint+C
=-2√xcos√x+2sin√x+C.
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