Re: [積分] 三角函數代換
※ 引述《gogoabc (小多愛喝養樂多)》之銘言：
: ∫x √(1-x^4) dx
: 1 1
: 答案是 --- arc(sin x^2) + --- x^2 √( 1-x^4)
: 4 4
set x^2 = sinu , 2xdx = cosudu
∫x √(1-x^4) dx = (1/2)∫(cosu)^2du
= (1/4)∫(1+cos2u)du
= u/4 + sin2u/8 + C
= u/4 + sinu*cosu/4 + C
(since x^2 = sinu , cosu = √(1-x^4) )
= (1/4)*sin^(-1)(x^2) + (1/4)*x^2*√(1-x^4) + C
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02/25 18:15, , 1F
02/25 18:15, 1F
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