Question

求不定积分∫x^2√（4-x^2）dx，急，求详细解答过程

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

Answer 2

用三角换元法 -x=2-x,令x=2sinu则dx=2cosudu,sinu=x/2 √(4-x)=√(4-4sinu)=2cosu,cosu=(1/2)√(4-x) ∴∫x^2√(4-x) dx =∫(4sinu)2cosu*2cosu du =16∫sinucosu du =16∫sinucosu du =16∫2cos^3u dcosU =16cos^6U