搜搜考试网求不定积分∫ dx/(a+cosx)
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求不定积分∫ dx/(a+cosx)
设：t = tan(x/2) ,x = 2arctant ,dx = 2dt/(1+t^2) ,cosx = (1-t^2)/(1+t^2)
I=∫ dx/(a+cosx)
= ∫ 2dt/(a(1+t^2)+(1-t^2)
=2∫dt/[(a-1)t^2+(a+1)]
① a1 时：
I = 2/(a+1) ∫dt/{[(a-1)/(a+1)]t^2+1}
= 2/√(a^2-1) ∫d[√[(a-1)/(a+1)t] /{(√[(a-1)/(a+1)t)^2+1}
= 2/√(a^2-1) arctan[√[(a-1)/(a+1)t] + C
I = 2/√(a^2-1) arctan[√[(a-1)/(a+1)tan(x/2)] + C
I=∫ dx/(a+cosx)
= ∫ 2dt/(a(1+t^2)+(1-t^2)
=2∫dt/[(a-1)t^2+(a+1)]
① a1 时：
I = 2/(a+1) ∫dt/{[(a-1)/(a+1)]t^2+1}
= 2/√(a^2-1) ∫d[√[(a-1)/(a+1)t] /{(√[(a-1)/(a+1)t)^2+1}
= 2/√(a^2-1) arctan[√[(a-1)/(a+1)t] + C
I = 2/√(a^2-1) arctan[√[(a-1)/(a+1)tan(x/2)] + C