急 求此題數學題目?

Question：

Use ANY method to evaluate the integral.

∫ 8 dw / [w²√(4 - w²)] = ?

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Solution：

∫ 8 dw / [w²√(4 - w²)]

Trigonometric substitution

Let w = 2 sinθ, dw = 2 cosθ dθ.

= ∫ 8( 2 cosθ dθ )/[4 sin²θ √(4 - 4sin²θ)]

= ∫ 16 cosθ dθ / (8 sin²θ cosθ)

= 2 ∫ csc²θ dθ

= -2 cotθ + C, where C is a constant.

= -2 cot[sin⁻¹(w/2)] + C

[ = -2√(4 - w²)/w  + C ]

Other method

= ∫ [2w² + 2(4 - w²)] dw / [w²√(4 - w²)]

= ∫ [2w²/√(4 - w²) + 2√(4 - w²)]/w² dw 

= ∫ { d[-2√(4 - w²)/w]/dw } dw

= -2√(4 - w²)/w + C

使用三角函數代換的方法: w = 2sin(x). 則

  ∫8dw/[w^2√(4-w^2)]

    = ∫16cos(x)dx/[4sin^2(x)2cos(x)]

    = ∫2dx/sin^2(x)

    = -2cot(x) + C

    = -2√(4-w^2)/w + C

另法;

  1/[w^2√(4-w^2)] = A√(4-w^2)/w^2 + B/√(4-w^2)

    = [A(4-w^2)+Bw^2]/[w^2√(4-w^2)]

故, 可假設

    A(4-w^2)+Bw^2 ≡ 1

即 B = A = 1/4.

所以,

  ∫8/[w^2√(4-w^2)] dw

    = ∫2√(4-w^2)/w^2 dw + ∫2/√(4-w^2) dw ... (1)

其中, 因

  d/dw √(4-w^2) = -w/√(4-w^2)

故

  ∫2/√(4-w^2) dw

    = ∫2(1/w)[w/√(4-w^2)] dw

    = -2(1/w)√(4-w^2) + ∫2(-1/w^2)√(4-w^2) dw

    = -2√(4-w^2)/w - ∫2√(4-w^2)/w^2 dw .......(2)

把 (2) 代入 (1), 兩不定積分一正一負抵消,

留下積分常數, 得:

  ∫8/[w^2√(4-w^2)] dw = -2√(4-w^2)/w + C

這題簡單跳過!!!!!!!!!!!!!!!!

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