定積分求解

這是我的標記, 這是不正式, 請不要在其他地方使用
1. (2∫1)(x)dx=積分x由1到2
2. 2[x]1=(2-1)
3. k的開方根=√k

設x = (√k)tanZ---①
tanZ = x / (√k)
sinZ = x / √(k+x^2)
cosZ = (√k) / √(k+x^2)

a -> arctan [a/(√k)] (代x=a入①)
b -> arctan [b/(√k)] (代x=b入①)
tanA = a/(√k) -> A = arctan [a/(√k)]
tanB = b/(√k) -> B = arctan [b/(√k)]
sinA = a / √(k+a^2)
sinB = b / √(k+b^2)
cosA = (√k) / √(k+a^2)
cosB = (√k) / √(k+b^2)
dx = d [(√k)tanA]

積分從a積到b x^2/(k+x^2)^(3/2)
=(b∫a) [x^2/(k+x^2)^(3/2)] dx
=(arctan [b/(√k)] ∫ arctan [a/(√k)])
{ [(√k) tanZ]^2 / {k + [(√k) tanZ]^2}^(3/2) } d [(√k)tanZ]
=(B ∫ A) {k(tanZ)^2 / [k+k(tanZ)^2]^(3/2)}d[(√k)tanZ]
=(B ∫ A) { (tanZ)^2 /(√k)[1+(tanZ)^2]^(3/2)}d[(√k)tanZ]
=(B ∫ A) { (tanZ)^2 / [1+(tanZ)^2]^(3/2)}d(tanZ)
=(B ∫ A) { (tanZ)^2 / [(secZ)^2]^(3/2) }d(tanZ)
=(B ∫ A) [ (tanZ)^2 / (secZ)^3 ]d(tanZ)
=(B ∫ A) [ (tanZ)^2 / (1/cosZ)^3 ]d(tanZ)
=(B ∫ A) { [(cosZ)^3](tanZ)^2 }d(tanZ)
=(B ∫ A) [ (cosZ) (sinZ)^2 ]d(tanZ)
=(B ∫ A) [ (cosZ) (sinZ)^2 ] (secZ)^2 dZ
=(B ∫ A) [ (cosZ) (sinZ)^2 ] (1/cosZ)^2 dZ
=(B ∫ A) [ (sinZ) (tanZ) ] dZ
=(B ∫ A) [ - (tanZ) ] d(cosA)
=B[ - (tanZ)(cosZ) ]A - (B ∫ A) [ - (cosZ) ] d(tanA)
=B[ - sinZ ]A + (B ∫ A) (cosZ) d(tanZ)
=B[ - sinZ ]A + (B ∫ A) [ (cosZ)(secZ)^2 ] dZ
=B[ - sinZ ]A + (B ∫ A) (secZ) dZ
=B[ - sinZ ]A + B[ ln |sec x + tan x| ]A
= - sinB + sinA + ln |sec B + tan B| - ln |sec A + tan A|
= - sinB + sinA + ln |1/cos B + tan B| - ln |1/cos A + tan A|
=a / √(k+a^2) - b / √(k+b^2)
+ ln |√(k+b^2) / (√k) + b/(√k)| - ln |√(k+a^2) / (√k) + a/(√k)|
=a / √(k+a^2) - b / √(k+b^2)
+ ln | [b+√(k+b^2)] / (√k) | - ln | [a+√(k+a^2)] / (√k) |
=a / √(k+a^2) - b / √(k+b^2)
+ ln | { [b+√(k+b^2)] / (√k) } / { [a+√(k+a^2)] / (√k) } |
=a / √(k+a^2) - b / √(k+b^2) + ln | [b+√(k+b^2)] / [a+√(k+a^2)] |
=a√(k+a^2) / (k+a^2) - b√(k+b^2) / (k+b^2)
+ ln | [b+√(k+b^2)][a - √(k+a^2)] / [a^2 - (k+a^2)] | (有理化)
=a√(k+a^2) / (k+a^2) - b√(k+b^2) / (k+b^2)
+ ln | [b+√(k+b^2)][a - √(k+a^2)] / k | (因為 |-k| = |k|)

2013-06-19 06:57:59 補充：
可不可解釋一下為什麼要設x=n*sec(y)?
x=n*tan(y)可以嗎?
請指教,謝謝!

w=∫(a~b)x^2*dx/(k+x^2)^(3/2)(1) 如果k>0 => k=m^2令 x=m*tan(y) => dx=m*sec^2(y)dyw1=∫m^2*tan^2(y)*m*sec^2(y)dy/[m^3*sec^3(y)]=∫tan^2(y)dy/sec(y)=∫(sec^2(y)-1)dy/sec(y)=∫dy/sec(y)-∫sec(y)dy=∫cos(y)dy-∫cos(y)dy/cos^2(y)=sin(y)-∫d(sin(y))/(1-sin^2(y))=sin(y)-∫du/(1-u^2)........u=sin(y)=sin(y)-∫du/2(1+u)-∫du/2(1-u)=sin(y)-0.5*ln(1+u)-0.5ln(1-u)=sin(y)-0.5*ln(1-u^2)=sin(y)-0.5*ln(cos^2(y))=sin(y)-ln(cos(y))=x/√(m^2+x^2)-ln(m/√(m^2+x^2))=x/√(k+x^2)-ln[√(k/(k+x^2))]=b/√(k+b^2)-a/√(k+a^2)-ln[√(k/(k+b^2))]+ln[√(k/(k+a^2))]=b/√(k+b^2)-a/√(k+a^2)+ln{√[(k+b^2)/(k+a^2)]}
(2) 如果k<0 => k=-n^2令 x=n*sec(y) => dx=n*sec(y)tan(y)dyw2=∫n^2*sec^2(y)*nsec(y)*tan(y)dy/[n^3*tan^3(y)]=∫sec^3(y)dy/tan(y)=∫(1+tan^2(y))sec(y)dy/tan(y)=∫dy/tan(y)cos(y)+∫tan(y)sec(y)dy=∫dy/sin(y)+∫d(sec(y))=∫sin(y)dy/sin^2(y)+sec(y)=∫d(cos(y))/(1-cos^2(y))+sec(y)=∫du/(1-u^2)+sec(y).......u=cos(y)=∫du/2(1+u)+∫du/2(1-u)+sec(y)=0.5*[ln(1+u)(1-u)]+sec(y)=0.5*ln(1-u^2)+sec(y)=0.5*ln(sin^2(y))+sec(y)=ln(sin(y))+sec(y)=ln(√(x^2-n^2)/x)+x/n=ln(√(x^2+k)/x)+x/√k=ln(√(b^2+k)/b)-ln(√(a^2+k)/a)+b/√k-a/√k=ln{a√(b^2+k)/[b√(a^2+k)]}+(b-a)/√k


2013-06-18 19:24:05 補充：
修正(2): k<0部份

w2=ln(√(x^2+k)/x)+x/√-k

=ln(√(b^2+k)/b)-ln(√(a^2+k)/a)+b/√-k-a/√-k

=ln{a√(b^2+k)/[b√(a^2+k)]}+(b-a)/√-k.........ans