關於不定積分的一條難問題.

如下:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Jan08/Crazyint4.jpg

∫sin^4(x)dx
=∫((1-cos2x)/2)^2 dx
=∫((1-2cos2x+cos^2 (2x))/4) dx
=∫(1/4 - (cos2x)/2 +1/4(((cos2(2x))+1)/2))
=∫(1/4)dx-∫((1/2)cos2x)dx + ∫(1/8(cos4x+1))dx
=(x/4) -(1/4(sin2x)) + (x/8) + (1/32(sin4x)) + C
=((3x)/8)-(1/4)(sin2x + (1/8)(sin4x)) + C

唔知仲使唔使simplfy個answer!!
Hope that I can help you!!

2008-01-17 22:55:41 補充：
THIS IS THE CORRECT ONE!!∫sin^4(x)dx=∫((1-cos2x)/2)^2 dx=∫((1-2cos2x+cos^2 (2x))/4) dx=∫(1/4 - (cos2x)/2 +1/4(((2cos2(2x))+1)/2))=∫(1/4)dx-∫((1/2)cos2x)dx + ∫(1/8(2cos4x+1))dx=(x/4) -(1/4(sin2x)) + (x/8) + (1/16(sin4x)) + C=((3x)/8)-(1/4)(sin2x + (1/4)(sin4x)) + C

∫sin^4(x)dx
= ∫((sinX)^2)^2dx 張佢拆開做2個2次,咁就計到喇
= ∫(1/2(1-cos4X))^2dx 計左仕面嗰舊先
= 1/2∫ (1-2(cos4X)+(cos4X)^2)dx
= 1/2 (X-2(1/4(sin4X))+∫1/2(1+cos8x)dx)
= 1/2X-1/4(sin4x) + (1/2)(1/2)(x+1/8(sin8x))
= 1/2X-1/4(sin4X) + 1/4x+ 1/32(sin8x)
= 3/4X -1/4(sin4x) + 1/32(sin8X)
可能會有d careless mistake 令到個答案唔係好岩(希望你唔好介意啦!)
但係個方法係冇錯既!

2008-01-17 22:42:45 補充：
sorry 溜左個c3/4X -1/4(sin4x) 1/32(sin8X) c c is a constant