作者j0958322080 (Tidus)
看板Math
標題Re: [微積] 幾題微積分
時間Thu Nov 29 21:07:43 2018
※ 引述《ttPttPtt (ttPtt)》之銘言:
: 想請問以下幾個形式的微積分如何解?
C is constant of the integration
: (1)∫√(a+bx^2)dx
x = tanu√(a/b), du = (secu)*(secu)*√(a/b)
cosu = √(a/b)/√[x*x + a/b] = 1/secu
∫√(a+bx^2)dx = [a/√b]*∫(secu)^3 du
∫(secu)^3 du = ∫secu*d(tanu)
= secu*tanu - ∫secu*tanu*tanu*du
∫secu*tanu*tanu*du = ∫secu*(secu*secu-1)*du
= ∫(secu)^3 du - ∫(secu) du
so ∫(secu)^3 du = secu*tanu - ∫(secu)^3 du + ∫(secu) du
-->∫(secu)^3 du = 0.5*(secu*tanu + ∫(secu) du)
= 0.5*(secu*tanu + ln|secu + tanu|)
[a/√b]*∫(secu)^3 du
= 0.5*(a/√b)*[x√(x*x + a/b)/(a/b) + ln|ax/b + √(x*x + a/b)/(a/b)|]
x√(a+bx^2) a*(√b)*ln|√(a+bx^2) + bx|
= ----------- + --------------------------- + C
2 2
: (2)∫dx/(√(x*(a+b*x^2)))
: (3)∫dx/(√(a*x^2+b*x+c))
a*x^2+b*x+c = a(x-b/2a)^2 + [c-(b/2a)^2]
^^^^^^^^^ + ^^^^^^^^^^^^
y^2 k
y = tanu√(k/a), dy = du*(secu)*(secu)*√(k/a)
cosu = [√(k/a)]/(y*y+k/a)
∫dx/(√(a*x^2+b*x+c)) = ∫dy/√(ay^2 + k)
= ∫secudu = ln|secu + tanu|
so turn into x, then get the answer
: (4)∫√(a*x^2+b*x+c)dx
simila as (1)
: 感謝各位駐版版友
: -----
: Sent from JPTT on my HTC_X10u.
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推 ttPttPtt : 謝謝您 11/29 23:02