integration

kjying

这个怎样做@_@

intergrate  sin^-1  ( x )

大家有什么网站有提供INTEGRATION /DIFFERENTIATION RULES吗?_?

找来找去找不到~>_<


顺便分享下刚找到的INTEGRATION SOFTWARE:

MULTIPLE INTEGRATION
LIMITED INTEGRATION

到www.download.com就有的下了~~

[ 本帖最后由 kjying 于 12-7-2008 01:13 PM 编辑 ]

hamilan911

用integration by parts
int arc sin x
= x (arc sin x) - int [x/sqrt{1-x^2}]
= x (arc sin x) + sqrt{1-x^2} + c

kjying

原帖由 hamilan911 于 12-7-2008 02:53 PM 发表
用integration by parts
int arc sin x
= x (arc sin x) - int [x/sqrt{1-x^2}]
= x (arc sin x) + sqrt{1-x^2} + c

谢谢...明白了....

顺便~~
http://en.wikipedia.org/wiki/Integration_by_parts

在这里找到了ARC TAN的例子~~

~Lucifer~

借帖问下。
double integrate
xcos（xy）cos^2(pai*x) dx dy=?
0<x<1/2
0<y<pai

~HeBe~_@

我的Int是integrate

Int( 0 to pi) Int( 0 to 1/2)     x cos(xy)  cos^2(pi x) dxdy
=Int( 0 to pi) Int( 0 to 1/2)   x cos(xy) (1+ cos (pi x))/2  dxdy
=(1/2) Int( 0 to pi) Int( 0 to 1/2)   x cos(xy) (1+ cos( 2 pi x))  dxdy
=(1/2) Int( 0 to pi) Int( 0 to 1/2)   x cos(xy) + x cos(xy) cos (2 pi x)  dxdy
=(1/2) Int( 0 to pi) Int( 0 to 1/2)   x cos(xy) + x cos(xy) cos (2 pi x)  dxdy
=(1/4) Int( 0 to pi) Int( 0 to 1/2)  2 x cos(xy) + x.[2 cos(xy) cos (2 pi x)] dxdy
=(1/4) Int( 0 to pi) Int( 0 to 1/2)  2 x cos(xy) + x.[ cos(2pi x + yx) + cos(2 pi x - yx) ] dxdy
=(1/4) Int( 0 to pi) Int( 0 to 1/2)  2 x cos(xy) + xcos(2pi x + yx) + xcos(2 pi x - yx)  dxdy
Integration by parts
=(1/4) Int( 0 to pi)   2x sin(yx) - 2(1/y^2) cos(yx) + x[ sin(2 pi x+ yx)/(2pi +y) + sin(2 pi x- yx)/(2pi -y) ]  - [ -cos (2 pi x+ yx)/(2pi +y)^2  - cos(2 pi x- yx)/(2pi -y)^2  ] (0 to pi)     dy

然后，就这样子做下去应该是没有问题。。。