求解一个不定积分x^2/根号下(a^2-x^2)

Question

求解一个不定积分
x^2 / 根号下(a^2-x^2)

Answer 1

令x=asint,(t在正负0.5pi之间） ,dx=acostdt,cost=√1-x^2/a^2
则∫x^2、√（a^2-x^2)dx
=∫a^2sin^2tacost/|a|costdt
=∫|a|sin^2t
=|a|∫(1-cos2t)/2
=|a|/2*∫(1-cos2t)
=|a|/2*[t-(sin2t)/2]
=|a|/2*[arcsinx/a-x/a*√(1-x^2/a^2)]
=|a|/2arcsinx/a-x/2√(1-x^2/a^2)