推 llww : 謝謝L大 07/01 14:37
※ 編輯: LPH66 (140.112.30.32), 07/01/2015 14:45:03
推 Philethan : 第一行應該是 sec^2 t / tan t = 1 / (sint*cost) 07/01 14:45
→ LPH66 : 補在下面了 XD 07/01 14:47
推 Philethan : 嗯嗯XD！ 07/01 14:49
※ 引述《llww (開心渡過每一天)》之銘言：
: http://imgur.com/CJJ2IXa
: 如上面網址圖片, 填充第6,8,9題以及計算第3題,
: 拜託各位高手了,想好久都算不出來.
計算3
令 x = tan t, 則 1+x^2 = sec^2 t, dx = sec^2 t dt
即 dx/(1+x^2) = dt
x + 1/x = (x^2+1)/x = sec^2 t / tan t = sin t / cos^3 t
原式成為
π/2
∫ ln (sin t / cos^3 t) dt
0
π/2
= ∫ (ln sin t - 3 ln cos t) dt
0
π/2 π/2
由變數變換可知 I = ∫ ln sin t dt = ∫ ln cos t dt
0 0
而此積分可由此兩形式相加算得 I = -πln2 / 2 (過程略)
故原式 = -2I = πln2 #
---
剛才想了一下, x + 1/x 那裡可以這樣接:
x + 1/x = sin t / cos t + cos t / sin t = 1 / (sin t cos t)
π/2 π/2
原式變成 ∫ ln (1 / sin t cos t) dt = - ∫ ln (sin t cos t) dt
0 0
這即是上面 I 的兩形式相加的積分, 所以可以從那裡直接算得原式 = -2I = πln2
--
'You've sort of made up for it tonight,' said Harry. 'Getting the
sword. Finishing the Horcrux. Saving my life.'
'That makes me sound a lot cooler then I was,' Ron mumbled.
'Stuff like that always sounds cooler then it really was,' said
Harry. 'I've been trying to tell you that for years.'
-- Harry Potter and the Deathly Hollows, P.308
--
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