tF}[ΜΕIθΜΨΎ (997Ϊ½)
γΊO1-V
oπ KΑͺ―Ά°({Ζ) (Χ) ©ID Ϊ½x Ϊ[ρ
945: 09/29()22:06 ID:tsfKlyQm(1/8) AAS
AAΘ
946: 09/29()22:07 ID:tsfKlyQm(2/8) AAS
AAΘ
947: 09/29()22:09 ID:tsfKlyQm(3/8) AAS
AAΘ
948: 09/29()22:10 ID:tsfKlyQm(4/8) AAS
AAΘ
949: 09/29()22:12 ID:tsfKlyQm(5/8) AAS
AAΘ
950: 09/29()22:13 ID:tsfKlyQm(6/8) AAS
η[0¨] (sin(x)/x)^2 dx
@@η[0¨] (sin(x)/x)^2 dx
@= η[0¨] (1/x^2) sin(x)^2 dx
@= η[0¨] (- 1/x)'sin(x)^2 dx
@= [(- 1/x) sin(x)^2][0¨] - η[0¨] (- 1/x) (sin(x)^2)'dx
@@[(- 1/x) sin(x)^2][0¨]
@= [- sin(x)^2/x][0¨]
Θ18
951: 09/29()22:15 ID:tsfKlyQm(7/8) AAS
I=η[0¨] sin2x cosx (1/x2)dx = (1/2)η[0¨] (sin2x sinx)(1/x2)dx
=(1/2){ [sin2x sinx(-1/x)][,0]
- η[0¨] (2cos2x sinx + sin2x cosx) (-1/x)dx }
=(1/2)[ 0+η[0¨] (1/2){(sin3x-sinx) + (sin3x+sinx)}/x dx ]
=(1/4)η[0¨] (2sin3x)/x dx = (1/2)η[0¨] (sin3x)/(3x) d(3x)
=Ξ/4EEEEE?
η[0¨] sin3x (1/x3)dx
Θ5
952: 09/29()22:16 ID:tsfKlyQm(8/8) AAS
@?[0¨]x^2/(x^2+1)^3dx
@f(z) = z^2/(z^2+1)^3
@@@ Res[f(z),i]
@@ = (1/2)lim_{z¨i}{z^2/(z+i)^3}"
@@ = (1/2)lim_{z¨i}[2{z/(z+i)^3}'-3{z^2/(z+i)^4}']
@@ = (1/2)lim_{z¨i}[2{1/(z+i)^3-3z/(z+i)^4}-3{2z/(z+i)^4-4z^2/(z+i)^5}]
@@ = (1/2)lim_{z¨i}[2/(z+i)^3-12z/(z+i)^4+12z^2/(z+i)^5]
Θ11
γΊO1-VΦΚΒυέxπ
½Ϊξρ ΤΪ½o ζΪ½o πΜ’Η½Ϊ AA»ΡΘ²Ω
Κ±Μθ Κ±TOP 2.162s*