→ a016258 :殘念 03/14 23:05
推 CJZZ :我也給你100P意思一下 03/14 23:26
※ 引述《CJZZ (NOBODY)》之銘言：
: ∫dx/√(16+4x^2)
: 一麻煩解題100P
√(16+4x^2) = 4√(1+(x/2)^2)
let x/2 = tany , 1 + (x/2)^2 = (secy)^2 , dx = 2(secy)^2 dy
∫dx/√(16+4x^2)
= (1/4)∫(1/secy)*2(secy)^2 dy
= (1/2)∫secy dy
= (1/2)ln|secy+tany| + C
consider √(1+(x/2)^2)⊿x/2
1
secy = √(1+(x/2)^2) , tany = x/2
= (1/2)ln|(x/2)+√(1+(x/2)^2)| + C
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