Question

不定积分的一题~

∫x^2/√(9-x^2)后面dx.麻烦写下过程,谢谢.

Answer 1

设x=3sint
∫x^2/根号下(9-X^2) dx
=∫9(sint)^2*3costdt/3cost
=(9/2)∫(1-cos2t)dt
=(9/4)∫(1-cos2t)d(2t)
=(9/4)(2t-sin2t)+C
=(9/2)(t-sintcost)+C
=(9/2)[arcsin(x/3)-x/3*根号(9-x^2)]+C

Answer 2

里面x^2=-[(9-x^2)-9]
所以原式=-∫dx + 3∫1/√[1-(x/3)^2] dx
=-∫dx + ∫1/√[1-(x/3)^2] d(x/3)
=-x + arcsin(x/3)