[積分] 兩題外表類似...

看板trans_math作者 (莫忘初衷)時間20年前 (2005/07/19 17:33), 編輯推噓9(904)
留言13則, 4人參與, 最新討論串1/1
1. ∞ lnx ∫ ─────dx 0 1 + x^2 2. 1 ln (x+1) ∫ ────── dx 0 1 + x^2 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 218.187.240.201

218.166.215.192 07/19, , 1F
請問一下第一題目有打錯嗎?
218.166.215.192 07/19, 1F

60.198.69.14 07/19, , 2F
算出第一題是0...第二題是(pi/8)*ln2...
60.198.69.14 07/19, 2F

59.115.74.213 07/19, , 3F
有算出來的給點提示吧 ORz
59.115.74.213 07/19, 3F

60.198.69.14 07/19, , 4F
兩題做法都蠻類似的...令x=tanA,dx=sec^2AdA
60.198.69.14 07/19, 4F

60.198.69.14 07/19, , 5F
ln(tanA)=ln(sinA)-ln(cosA),再另其中一積分角度
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60.198.69.14 07/19, , 6F
A=(pi/2)-B,dA=-dB,使得兩積分恰好消掉為0...
60.198.69.14 07/19, 6F

60.198.69.14 07/19, , 7F
2前面做法同1...ln(tanA+1)=ln(sinA+cosA)-ln(cosA)
60.198.69.14 07/19, 7F

60.198.69.14 07/19, , 8F
ln(sinA+cosA)=ln{√2cos[(pi/4)-A]}...
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60.198.69.14 07/19, , 9F
ln{√2cos[(pi/4)-A]}=ln(√2)+ln{cos[(pi/4)-A]}
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60.198.69.14 07/19, , 10F
再另B=(pi/4)-A,dA=-dB,之後可將同樣兩項消去...
60.198.69.14 07/19, 10F

59.115.74.213 07/19, , 11F
好神奇的做法喔 謝啦
59.115.74.213 07/19, 11F

59.113.139.64 07/19, , 12F
是很神~不過我蠻奇你從那學來的?
59.113.139.64 07/19, 12F

60.198.69.14 07/20, , 13F
圖書館找書看到的阿...
60.198.69.14 07/20, 13F
文章代碼(AID): #12tCZmG3 (trans_math)