Question

求不定积分∫x^2√(4-x^2)dx，求详细过程，急

Answer 1

∫x^2√(4-x^2)dx

let
x= 2sina
dx=2cosada
∫x^2√(4-x^2)dx

=∫(2sina)^2. (2cosa)(2cosa)da
=∫ (4sinacosa)^2 da
=∫ 4(sin2a)^2 da
= 2 ∫ ( 1-cos4a) da
= 2 ( a- (sin4a) /4 ) + C

consider
(cosa + isina)^4 = cos4a+ isin4a
sin4a= 3(cosa)^3sina+ (sina)^4
sina = x/2
cosa =√(4-x^2)/2

∫x^2√(4-x^2)dx
= 2 ( a- (sin4a) /4 ) + C
= 2[ arcsin(x/2) - (1/4)( (3/16)x(4-x^3)^(3/2)+ x^4/16 ) ]+C