∫x^2/(1-x^2)^1/2 dx 用第二类换 - 知识问答

∫x^2/(1-x^2)^1/2 dx 用第二类换元法求不定积分过程,

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这个不用二类换元法反而简单些吧
设x=sint,dx=costdt,√(1-x^2)=cost
原式=∫sin²t/cost*costdt
=∫sin²tdt
=∫(1-cos2t)/2dt
=1/2∫dt-1/2∫cos2tdt
=t/2+1/4∫cos2td2t
=t/2+1/4*sin2t+C
=t/2+1/2*sintcost+C
=1/2*arcsinx+1/2*x/√(1-x^2)+C

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