Question

∫ x^2 (4-x^2)^1/2 dx 用换元法怎麼解

Answer 1

let
x= 2sinu
dx =2cosu du

∫ x^2. (4-x^2)^(1/2) dx
=∫ (2sinu)^2. (2cosu) (2cosudu)
=16∫ (sinu)^2. (cosu)^2 du
=4∫ (sin2u)^2 du
=(1/8) ∫ (1-cos4u) du
=(1/8)[u -(1/4)sin4u] + C
=(1/8)[ arcsin(x/2) -(1/8)x(4-x^2)^(1/2) . [4- x^2(4-x^2) ]^(1/2) ] + C

where
sinu = x/2
cosu = (4-x^2)^(1/2) /2
sin2u = (1/2)x(4-x^2)^(1/2)
cos2u = [4- x^2(4-x^2) ]^(1/2) / 2
sin4u = x(4-x^2)^(1/2) . [4- x^2(4-x^2) ]^(1/2) /2

Answer 2

Let x = 2siny and dx = 2cosydy
then √(4-x²) = 2cosy
cosy = (1/2)√(4-x²)
∫ dx/[x²√(4-x²)]
= ∫ 2cosydy/(4sin²y*2cosy)
= (1/4)∫ csc²y dy
= (-1/4)coty + C
= (-1/4)*(1/2)√(4-x²) / (x/2) + C
= -√(4-x²) / (4x) + C