Question

∫﹛1÷[x+(1-x²)½]﹜dx 如题，求指导。

Answer 1

方法一:
令x=siny, (1-x²)½=cosy
原式=∫﹛1/(siny+cosy)dsiny
==∫﹛cosy/(siny+cosy)dy
=∫cosy[(cosy)^2-(siny)^2]/(siny+cosy)cos2ydy [上下同乘以cos2y]
=∫cosy(cosy-siny)/cos2ydy
这里分成两部分,先看后一部分

∫-cosysiny/cos2ydy
∫-0.5sin2y/cos2ydy
=-0.5tan2ydy
=-1/40.5∫tan2yd(2y)
到这步套公式就可以了,然后把y换回x的函数,你应该会了,不会可以追问

方法二:
令x=sinu，dx=cosudu，(1-x²)½=cosu
原式=∫ cosu/[sinu+cosu]du
=1/2∫ [(cosu-sinu)+(cosu+sinu)]/[sinu+cosu]du
=1/2∫ (cosu-sinu)/[sinu+cosu]du+1/2∫ (cosu+sinu)/[sinu+cosu]du
=1/2∫ 1/[sinu+cosu]d(sinu+cosu)+1/2∫ 1du
=1/2ln|sinu+cosu|+1/2u+C
=1/2ln|x+(1-x²)½|+1/2arcsinu+C