Re: [積分] 三角函數代換
※ 引述《gogoabc (小多愛喝養樂多)》之銘言：
: √(x^2-9)
: ∫-----------dx
: x^3
: 答案是 1/6 sec^-1(x/3)-√(x^2-9)/(2x^2)+c
set x = 3secu , dx = 3secu*tanudu
√(x^2-9) (tanu)^2
∫-------------dx = (1/3)*∫-----------du
x^3 (secu)^2
= (1/3)*∫(sinu)^2du
= (1/6)*∫(1-cos2u)du
= u/6 - (sin2u)/12 + C
= u/6 - (sinu*cosu)/6 + C
(since x=3secu , sinu*cosu = 3√(x^2 -9) / (x^2) )
= (1/6)*sec^(-1)(x/3) - √(x^2-9)/(2x^2) + C
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