Re: [微積] 大學積分

[ctchang34 ()]

w = ∫dy/(c-siny)  請問這怎麼積??

Set y = 2*atan(x) => dy = 2dx/(1+x^2)

=> x = tan(y/2)

=> siny = sin[2*atan(x)]

= 2*sin[atan(x)]*cos[atan(x)]

= 2x/(1+x^2)


分母 = c - siny

= c - 2x/(1+x^2)

= (c - 2x + cx^2)/(1+x^2)

= [(cx)^2 - 2cx + 1) + (c - 1)]/c(1 + x^2)

= [(cx - 1)^2 + (c - 1)]/c(1 + x^2)


w = 2c∫dx/[(cx - 1)^2 + (c - 1)]

= 2∫d(cx - 1)/[(cx - 1)^2 + (c - 1)];;; u = cx - 1

= 2∫du/[u^2 + (c - 1)]

= [2/√(c - 1)]*atan[u/√(c - 1)] + C

= [2/√(c - 1)]*atan[(cx - 1)/√(c - 1)] + C

= [2/√(c - 1)]*atan{[c*tan(y/2) - 1]/√(c - 1)} + C

= ans


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推 CyCls       : 感謝～ 但為何要設y=2atan(x)  怎麼知道要乘2？
推 LPH66       : 這其實就只是半角代換 x = tan(y/2)
→ LPH66       : 也就是前篇推文講的萬能代換
→ LPH66       : 之所以如此稱呼是因為 sinθ cosθ tanθ
→ LPH66       : 皆可表為 tan(θ/2) 的有理式, 積分會方便很多
推 G41271      : 考試時這題要特別分開討論c=1時的情況，不然會被扣
→ G41271      : 分