如题所述
第1个回答 2022-08-12
令x=2sinu,则:sinu=x/2,u=arcsin(x/2),dx=2cosudu.
∴∫x^2√(4-x^2)dx
=2∫(sinu)^2√[4-4(sinu)^2]cosudu
=4∫(sinu)^2(cosu)^2du
=∫(sin2u)^2du
=(1/2)∫(1-cos4u)du
=(1/2)∫du-(1/8)∫cos4ud(4u)
=(1/2)u-(1/8)sin4x+C
=(1/2)u-(1/4)sin2ucos2u+C
=(1/2)arcsin(x/2)-(1/2)sinucosu[1-2(sinu)^2]+C
=(1/2)arcsin(x/2)-(1/2)(x/2)√[1-(sinu)^2][1-2(x/2)^2]+C
=(1/2)arcsin(x/2)-(1/4)x(1-x^2/4)√(1-x^2/4)+C
=(1/2)arcsin(x/2)-(1/32)x(4-x^2)√(4-x^2)+C
∴∫x^2√(4-x^2)dx
=2∫(sinu)^2√[4-4(sinu)^2]cosudu
=4∫(sinu)^2(cosu)^2du
=∫(sin2u)^2du
=(1/2)∫(1-cos4u)du
=(1/2)∫du-(1/8)∫cos4ud(4u)
=(1/2)u-(1/8)sin4x+C
=(1/2)u-(1/4)sin2ucos2u+C
=(1/2)arcsin(x/2)-(1/2)sinucosu[1-2(sinu)^2]+C
=(1/2)arcsin(x/2)-(1/2)(x/2)√[1-(sinu)^2][1-2(x/2)^2]+C
=(1/2)arcsin(x/2)-(1/4)x(1-x^2/4)√(1-x^2/4)+C
=(1/2)arcsin(x/2)-(1/32)x(4-x^2)√(4-x^2)+C