1/(1+sinx)的不定积分怎么求?arctan根号下(根号x-1)的不定积分怎么求?
人气:391 ℃ 时间:2020-02-03 19:17:29
解答
我来帮你!楼主 1.三角换元 + 万能公式 令tan(x/2)=t ,则sinx=2t/(1+t^2),dx=2dt/(1+t^2),带入整理,∫1/(1+sinx)dx =∫2dt/(1+2t+t^2)= 2∫dt/(1+t)^2 = -2/(1+t)+ C = -2/[1+tan(x/2)]+ C 2.直接整体换元 令arctan√...
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