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根号下1+sinx dx 求积分
人气:317 ℃ 时间:2020-05-21 07:59:57
解答
1+sinx=(sin(x/2)+cos(x/2))^2
即原式=∫(sin(x/2)+cos(x/2))dx
=2∫sin(x/2)d(x/2)+2∫cos(x/2)d(x/2)
=2sin(x/2)-2cos(x/2)
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