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Question：Use ANY method to evaluate the integral.∫ 8 dw / [w²√(4 - w²)] = ?🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘Solution：∫ 8 dw / [w²√(4 - w²)]Trigonometric substitutionLet 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