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数学
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计算不定积分:(1)∫tan^2 xdx ; (2).∫dx/(1-2x)^(1/2)
人气:393 ℃ 时间:2020-04-06 06:11:00
解答
(1)∫tan^2 xdx ;
=∫(sec²x-1)dx
=tanx-x+c
(2).∫dx/(1-2x)^(1/2)
=-1/2 ∫1/(1-2x)^(1/2) d(1-2x)
=-(1-2x)^(1/2)+c∫(sec²x-1)dx里面的“²”是怎样输上的搜狗拼音里,输pingfang
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