作者Frobenius (i^(-i)= e^(π/2))
看板trans_math
标题Re: [积分]
时间Wed Feb 7 00:48:56 2007
1
∫---------- dx
1 + cosx
(1 - cosx) (1 - cosx)
= ∫-------------------- dx = ∫----------- dx
(1 + cosx)(1 - cosx) (sinx)^2
= ∫(cscx)^2 - (cscx)(cotx) dx = - cotx + cscx + C
cosx 1 1 - cosx 1 - cosx
= - ---- + ---- + C = ---------- = ------------------- + C
sinx sinx sinx 2sin(x/2)cos(x/2)
sin(x/2)sin(x/2)
= ------------------ + C = tan(x/2) + C
sin(x/2)cos(x/2)
1 2du 1-u^2
∫---------- dx ( u = tan(x/2)，dx = -------，cosx = ------- )
1 + cosx 1+u^2 1+u^2
1 2du 2du 2
= ∫----------- ------- = ∫------------------ = ∫--- du = ∫du
1-u^2 (1+u^2) + (1-u^2) 2
(1 + -----) (1+u^2)
1+u^2
= tan(x/2) + C
∫ 〔(√3 ) sinx + cosx 〕^2 dx
= ∫ 4〔sinx (√3/2 ) + cosx (1/2) 〕^2 dx
= ∫ 4〔sin(x + π/6)〕^2 dx
= ∫ 2 2〔sin(x + π/6)〕^2 dx = 2∫ 2〔sin(x + π/6)〕^2 dx
= 2 ∫〔1 - cos(2x + π/3)〕dx = ∫ 2 dx - ∫ cos(2x + π/3) d(2x + π/3)
= 2x - sin(2x + π/3) + C
= 2x - sin(2x) (1/2) - cos(2x) (√3/2 ) + C
= 2x - (√3/2) cos(2x) - (1/2) sin(2x) + C
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1F：推 GayerDior:如此高度智慧的文章 应该让所有人都读到 61.229.147.158 02/07 01:01
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