Re: [積分] 三角函數代換

[bighead0720 (ocean)]

※ 引述《gogoabc (小多愛喝養樂多)》之銘言：
:      x^3
: ∫----------dx
:   √(x^2+9)

:           1
: 答案是   --- (x^2-18)√(x^2+9)+c
:           3

set x = 3tanu , dx = 3sec^udu

     x^3
∫----------dx = 27∫(tanu)^3 * secudu
  √(x^2+9)

               = 27∫tanu[(secu)^2 -1]secudu

               = 27∫[(secu)^3 - secu]tanudu

               = 27∫[(secu)^2 - 1]secu*tanudu

(set t = secu , dt = secu*tanudu)

               = 27∫(t^2 - 1)dt

               = 27(t^3/3 -t) + C

               = 9t^3 - 27t + C

               = 9(secu)^3 - 27secu + C

               = 9secu[(secu)^2 - 3] + C

(since x=3tanu ,  secu = √(x^2 + 3) / 3  )

               = 3√(x^+3) * [(x^2+3)/9 -3] + C

               = (1/3)* √(x^+3) * (x^2 -18) + C

--
※ 發信站: 批踢踢實業坊(ptt.cc) 
◆ From: 219.91.87.248
推 gogoabc:謝謝