3條intergals 數 20marks

[匿名]

2011-05-02 7:19 pm

1. ∫sin^2x cos^2x dx

2. ∫dx / √(x^2-4x+3)

3. By using the substitution u=sin^x , find ∫ [(sinxcosxdx) / (4sin^x - 9cos^x)].

更新1:

第2條點解 ∫ dx/√[(x - 2)2 - 1] = ∫ d(x - 2)/√[(x - 2)2 - 1] why無拿拿會dx變左d(x-2) 既?

回答 (1)

用戶25975

2011-05-02 8:21 pm

✔ 最佳答案

1) ∫ sin2 x cos2 x dx

= (1/4) ∫ sin2 2x dx

= (1/8) ∫ (1 - cos 4x) dx

= x/8 - (sin 4x)/32 + C

2) ∫ dx/√(x2 - 4x + 3)

= ∫ dx/√[(x - 2)2 - 1]

= ∫ d(x - 2)/√[(x - 2)2 - 1]

= ln |(x - 2) + √(x2 - 4x + 3)| + C

3) Sub u = sin2 x, then du = 2 sin x cos x dx

∫ (sin x cos x dx)/∫(4 sin2 x - 9 cos2 x)

= (1/2) ∫ du/[4u - 9(1 - u)]

= (1/2) ∫ du/(13u - 9)

= (1/26) ln |13u - 9| + C

= (1/26) ln |13sin2 x - 9| + C

2011-05-02 21:25:19 補充：
因為 d(k) = 0 for all constant k

So:

d(x - 2) = dx - d(-2) = dx

參考: 原創答案

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