令x=2sint,dx=2costdt
代入原式得:=∫16sin^4t.2cost.2costdt
=64∫sin^4t.cos^2tdt
=64∫sin²t*(sin2t/2)²dt
=16∫(1-cos2t)/2*(1-cos4t)/2dt
=4∫(1-cos2t)*(1-cos4t)dt
=4∫(cos4tcos2t-cos2t-cos4t+1)dt
=4∫((cos6t+cos2t)/2-cos2t-cos4t+1)dt
=4[sin6t/12-sin2t/4-sin4t/4+t]+C
=sin6t/3-sin2t-sin4t+4t+C
t=arcsinx/2代入,得
∫x^4*√(4-x^2)dx=sin6√(4-x^2)/3-sin2√(4-x^2)-sin4√(4-x^2)+4√(4-x^2)+C