作者dreamroyc ()
看板trans_math
標題[積分]
時間Fri Mar 23 23:26:36 2007
Prove that,for even powers of sine,
pi/2 1*3*5*…(2n-1) pi
∫ sin^2n(x)=------------------ * ----
0 2*4*6*…2n 2
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◆ From: 61.230.2.97
※ 編輯: dreamroyc 來自: 61.230.2.97 (03/23 23:28)
※ 編輯: dreamroyc 來自: 61.230.2.97 (03/23 23:29)
※ 編輯: dreamroyc 來自: 61.230.2.97 (03/23 23:29)
※ 編輯: dreamroyc 來自: 61.230.2.97 (03/23 23:30)
1F:推 GayerDior:令t=2n如何?且I_t=∫(sinx)^t ,x=0~pi/2 61.229.145.39 03/23 23:33
2F:→ GayerDior:加油你會的!! 61.229.145.39 03/23 23:34
3F:推 dreamroyc:我試試看 謝謝妳 61.230.2.97 03/23 23:39
4F:→ GayerDior:更正:且I_t=∫(sinx)^t dx ,x=0~pi/2 61.229.145.39 03/23 23:45
5F:推 dreamroyc:會不會用黎曼合比較好解? 61.230.2.97 03/23 23:49
6F:→ dreamroyc:老實說令2n=t 有點算不出來 61.230.2.97 03/23 23:50
7F:推 GayerDior:Integration By Part 加油你會的!!!! 61.229.145.39 03/23 23:56
8F:→ GayerDior:其實我懶ㄉ打= =而且我要睡ㄌ881 61.229.145.39 03/23 23:57
9F:推 dreamroyc:好的,我努力,感謝回覆。 61.230.2.97 03/23 23:57
10F:推 vu3cj0su3:reduction? 218.160.159.43 03/24 01:48