三角函數有理式的定積分求值

(1)∫[0,π] 1/(a+ b cosx+ c sinx) dx
Sol:
令z=tan(x/2)
則dx=2dz/(1+z^2)
_sin(x)=2z/(1+z^2)
_cos(x)=(1-z^2)/(1+z^2)
上下限:
[0,π]---->[0,∞ ]

_∫[0,π] 1/(a+ b cosx+ c sinx) dx
&
=∫[0,∞] 2/{[a+b(1-z^2)/(1+z^2)+c(2z)/(1+z^2)]*(1+z^2)}dz
&
=∫[0,∞] 2/[(a-b)(z^2)+2cz+(a+b)]dz
&
=2/(a-b)∫[0,∞] 1/[(z^2)+(2c/(a-b))z+(a+b)/(a-b)]dz
&
=2/(a-b)∫[0,∞] 1/{[(z+(c/(a-b))^2]+(a^2-b^2-c^2)/(a-b)^2}dz
再令
A^2=(a^2-b^2-c^2)/(a-b)^2
&
=2/(a-b)∫[0,∞] 1/{[(z+(c/(a-b))^2]+A^2}dz
&
=2/[(A^2)(a-b)]∫[0,∞] 1/{[((z+(c/(a-b))/A)^2]+1}dz
&
=2/[(A^2)(a-b)]∫[0,∞] 1/{[((z+(c/(a-b))/A)^2]+1}d[(z+(c/(a+b)))/A]
&
=2Arctan[z+(c/(a+b)))/A]/[(A^2)(a-b)]｜[0,∞]
&
=2*[(π/2)-Arctan[c/(a-b)/A]]/[(A^2)(a-b)]
&
=2*{(π/2)-Arctan[c/√ (a^2-b^2-c^2)]}/√ (a^2-b^2-c^2)
for a^2>=b^2+c^2
======================================================
(2)∫[0,π] 1/[a+ b (cosx)^2+ c (sinx)^2] dx------(A)
Sol:
令f(x)=1/[a+ b (cosx)^2+ c (sinx)^2]
因f(x)=f(-x)
(A)=(1/2)∫[-π,π]1/[a+ b (cosx)^2+ c (sinx)^2] dx------(B)
令
z=x+iy ; i=√(-1)
c={z｜|z|=1,θ:-π-->π}
cosx=[z+1/z]/2
sinx=[z-1/z]/2i
dθ=dx=dz/(iz)
則
(B)=(1/2)∮c 1/[a+ b [[z+1/z]^2]/4+ c [[z-1/z]^2/(-4)] dz/(iz)
化簡可得:
=2/[i(b-c)]∮c z/{z^4+[(4a+2b+2c)/(b-c)]z^2+1} dz
其中
G(z)=z/{z^4+[(4a+2b+2c)/(b-c)]z^2+1}
=4π/(b-c)Σ(c內){ResG(z)}

2009-04-05 02:45:52 補充：
&conint=封閉路徑積分符號

2009-04-05 04:49:16 補充：
大師:
我獻醜了

第一題不知道我這樣解是否是最好的方法,感覺好繁瑣!
第二題光是要判斷四個pole有沒有在c內就夠複雜了,
殘值的部份明天我再試試看.

現在太累了實在想睡覺了~

2009-04-09 17:20:00 補充：
第二題,用複變解,但是計算殘值的部份太複雜,後來想到改用變數變換的方式:

2009-04-09 17:21:45 補充：
_∫[0,α] 1/[a+ b (cosx)^2+ c (sinx)^2] dx

(cosx)^2=(1+cos2x)/2 ; (sinx)^2=(1-cos2x)/2 ;帶入可降低分母次方

=∫[0,α] 1/[a+ b(1+cos2x)/2+ c(1-cos2x)/2] dx

=∫[0,α] 1/[(a+b/2+c/2)+((b-c)/2)*cos2x] dx

=[2/(b-c)]∫[0,α] 1/[(2a+b+c)/(b-c) + cos2x] dx

2009-04-09 17:22:28 補充：
令A=(2a+b+c)/(b-c) 且u=2x , du=2dx

=[1/(b-c)]∫[0,2α] 1/[A + cosu] du

令y=tan(u/2)
du=2dy/(1+y^2)
cosu=(1-y^2)/(1+y^2)

2009-04-09 17:23:54 補充：
=[1/(b-c)]∫[0,tan(α)] 1/[A + (1-y^2)/(1+y^2)] 2dy/(1+y^2)

=[2/(b-c)]∫[0,tan(α)] 1/[A(1+y^2) + (1-y^2)] dy

=[2/(b-c)]∫[0,tan(α)] 1/[(A-1)*y^2 + (A+1)] dy

2009-04-09 17:24:25 補充：
=[2/(b-c)/(A+1)]∫[0,tan(α)] 1/[ [(A-1)/(A+1)]*y^2 + 1 ] dy

=[2/(b-c)/√(A^2 - 1)]∫[0,tan(α)] 1/[ [√[(A-1)/(A+1)]*y]^2 + 1 ] d{√[(A-1)/(A+1)]y}

=[2/(b-c)/√(A^2 - 1)]*arctan[(A-1)/(A+1)]y | [0,tan(α)]

2009-04-09 17:25:08 補充：
=[2/(b-c)/√(A^2 - 1)]*arctan[(A-1)/(A+1)]*tan(α) ,因為α=π

=[2/(b-c)/√(A^2 - 1)]*arctan[(A-1)/(A+1)]*tan(π)

=0

2009-04-09 17:56:02 補充：
有道理,我再想想!

2009-04-09 18:43:04 補充：
改正如下:

2009-04-09 18:44:42 補充：
貼在意見中

2009-04-09 18:44:55 補充：
改正如下:
http://tw.myblog.yahoo.com/jw!T9PfnVCLEUQeT40Ir.s-/article?mid=3&prev=-1&next=2

應該可以改為 ∫1/(a+b sin x) dx 與 ∫1/(a+b sin^2 x) dx 的形式吧

建議用微積分方法求之!

2009-04-09 17:43:47 補充：
Q2:函數恆正,為何定積分會為0, 不合理!

太兇險啦