∫x^2⼀(1-x^2)^1⼀2 dx 用第二类换元法求不定积分过程,

2024-11-17 00:37:41
推荐回答(1个)
回答1:

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