Re: [積分] 積分求救
※ 引述《metastable (清士)》之銘言：
: (1 - X^2 )^(1/2)
: ∫---------------------dx
: 2 X^2 - 1
: a b
: 我試過用分式去看 ---------------- + --------------
: √2 X^2 + 1 √2 X^2 -1
: 找不到答案
: -1 x -1
: ANS 是 - (1/2)tanh --------------- - (1/2) sin x
: (1-x^2)^(1/2)
: 有誰能告訴我怎麼積分?
上下同乘 √( 1 - x^2 )
x^2 - 1
原式 = ∫----------------------------- dx
(√( 1 - x^2 )) ( 1 - 2x^2 )
= (1/2)[ arctanh(x/sqrt(1-x^2)) - arcsinx ]
- arctanh(x/sqrt(1-x^2))
= (1/2)[ -arctanh(x/sqrt(1-x^2)) - arcsinx ] + c
-----
不負責任的話可以丟多項式讓你去從答案湊,
但是還是留一下過程好了,
至於分子是 x^2 那項, 就乘 2/2, 再加一減一拆項就可以了
Let x = sint dx = cost dt
√( 1 - x^2 ) = cost
1
Then ∫-------------------------- dx
( 1 - 2x^2)√( 1 - x^2 )
cost
= ∫------------------------ dt
[ 1 - 2(sint)^2 ] cost
1 (sect)^2
= ∫--------------- dt = ∫---------------------- dt
1 - 2(sint)^2 (sect)^2 - 2(tant)^2
(sect)^2 1
= ∫-------------- dt = ∫-------------- d(tant)
1 - (tant)^2 1 - (tant)^2
= arctanh(tant) + c
x
tant = ---------------
√( 1 - x^2 )
--
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◆ From: 114.41.128.6
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哪裡來的(-cos)^2咧@@?
※ 編輯: BaBi 來自: 114.41.128.6 (04/01 00:00)
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x^2
∫----------------------------- dx
(√( 1 - x^2 )) ( 1 - 2x^2 )
( 2x^2 - 1 ) + 1
= (1/2) ∫----------------------------- dx
(√( 1 - x^2 )) ( 1 - 2x^2 )
1
= (-1/2) ∫-------------- dx
√( 1 - x^2 )
1
+ (1/2) ∫----------------------------- dx
(√( 1 - x^2 )) ( 1 - 2x^2 )
= (-1/2)(arcsinx) + (1/2)[arctanh(x/sqrt(1-x^2)]
※ 編輯: BaBi 來自: 114.41.128.6 (04/01 00:18)
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