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怎么求x√(4x-x方)的积分?
题目详情
怎么求x√(4x-x方)的积分?
▼优质解答
答案和解析
令x=2+2sinu,则:sinu=(x-2)/2,u=arcsin[(x-2)/2],dx=2cosudu.
∴∫x√(4x-x^2)dx
=∫(2+2sinu)√[4-(x^2-4x+4)](2cosu)du
=4∫(1+sinu)cosu√[4-(x-2)^2]du
=4∫(1+sinu)cosu√[4-4(sinu)^2]du
=8∫(1+sinu)(cosu)^2du
=8∫(cosu)^2du+8∫sinu(cosu)^2du
=4∫(1+cos2u)du-8∫(cosu)^2d(cosu)
=4∫du+2∫cos2ud(2u)-(8/3)(cosu)^3
=4u+2sin2u-(8/3)[1-(sinu)^2]^(3/2)+C
=4arcsin[(x-2)/2]+4sinucosu-(8/3)[1-(x-2)^2/4]^(3/2)+C
=4arcsin[(x-2)/2]+4[(x-2)/2]√[1-(sinu)^2]
-(2/3)(4-x^2+4x-4)^(3/2)+C
=4arcsin[(x-2)/2]+2(x-2)√[1-(x-2)^2/4]-(2/3)(4x-x^2)^(3/2)+C
=4arcsin[(x-2)/2]+(x-2)√(4-x^2+4x-4)-(2/3)(4x-x^2)√(4x-x^2)+C
=4arcsin[(x-2)/2]+(1/3)(3x-6-8x+2x^2)√(4x-x^2)+C
=4arcsin[(x-2)/2]+(1/3)(2x^2-5x-6)√(4x-x^2)+C
∴∫x√(4x-x^2)dx
=∫(2+2sinu)√[4-(x^2-4x+4)](2cosu)du
=4∫(1+sinu)cosu√[4-(x-2)^2]du
=4∫(1+sinu)cosu√[4-4(sinu)^2]du
=8∫(1+sinu)(cosu)^2du
=8∫(cosu)^2du+8∫sinu(cosu)^2du
=4∫(1+cos2u)du-8∫(cosu)^2d(cosu)
=4∫du+2∫cos2ud(2u)-(8/3)(cosu)^3
=4u+2sin2u-(8/3)[1-(sinu)^2]^(3/2)+C
=4arcsin[(x-2)/2]+4sinucosu-(8/3)[1-(x-2)^2/4]^(3/2)+C
=4arcsin[(x-2)/2]+4[(x-2)/2]√[1-(sinu)^2]
-(2/3)(4-x^2+4x-4)^(3/2)+C
=4arcsin[(x-2)/2]+2(x-2)√[1-(x-2)^2/4]-(2/3)(4x-x^2)^(3/2)+C
=4arcsin[(x-2)/2]+(x-2)√(4-x^2+4x-4)-(2/3)(4x-x^2)√(4x-x^2)+C
=4arcsin[(x-2)/2]+(1/3)(3x-6-8x+2x^2)√(4x-x^2)+C
=4arcsin[(x-2)/2]+(1/3)(2x^2-5x-6)√(4x-x^2)+C
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