Re: [積分] 三角函數代換
※ 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
--
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02/25 18:15, , 1F
02/25 18:15, 1F
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