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1/(1+cosx)的不定积分是怎么算啊
∫dx/1+cosx=?
人气:384 ℃ 时间:2020-03-29 18:14:09
解答
1+cosx=2[cos(x/2)]^2
1/(1+cosx)=0.5[sec(x/2)]^2
∫dx/(1+cosx)
=∫0.5[sec(x/2)]^2dx
=∫[sec(x/2)]^2d0.5x
=∫dtan(x/2)
=tan(x/2)+c
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