Amaths integration --urgent

1
∫(sin^3xcos^5x)dx
=∫(sin^3xcos^4x)dsinx
=∫(sin^3x)(1-sin^2x)^2 dsinx
=∫(sin^3x)(1-2sin^2x+sin^4x) dsinx
=∫(sin^3x-2sin^5x+sin^7x) dsinx
=sin^4x/4-sin^6x/3+sin^8x/8+C
2
∫(sin^5xcos^2x)dx
=-∫(sin^4xcos^4x)dcosx
=-∫(1-cos^2x)^2 cos^4x dcosx
=-∫(1-2cos^2x+cos^4x) cos^4x dcosx
=-∫(cos^4x-2cos^6x+cos^8x) dcosx
=-cos^5x/5+2cos^7x/7-cos^9x/9+C

2009-01-24 16:12:16 補充：
不會。有NEW WAY ADDITIONAL MATHEMATICS 為証

2009-01-24 16:13:19 補充：
而且ADDITIONAL MATHEMATICS 都有Reduction Formula EMK兄你忘記了?

因為A.Maths冇教Substitution，請問題目有冇Reduction Formula？

2009-01-23 22:08:05 補充：
如果咩都冇，正常呢D數應該係P.Maths先學架喎……

2009-01-24 06:20:02 補充：
myisland8132兄，你一寫出 cos x dx = d(sin x)，
呢個已經超出左A.Maths既範圍架啦喎。