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根号下(1+sinx) 不定积分
人气:426 ℃ 时间:2020-03-26 20:13:39
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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